Creating a mean variable, from variable names and weights supplied by vectors












1















Suppose I want to create a mean variable in a given dataframe based on two vectors, one specifying the names of the variables to use, and one specifying weights by which these variables should go into the mean variable:



vars <- c("a", "b", "c","d"))
weights <- c(0.5, 0.7, 0.8, 0.2))
df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
colnames(df) <- c("a","b","c","d","e","f")


How could I use dplyr::mutate() to create a mean variable that uses vars and weights to calculate a rowwise score? mutate() should specifically use the variables supplied by vars
The result should basically do the following:



df <- df %>% 
rowwise() %>%
mutate(comp = mean(c(vars[1]*weights[1], vars[2]*weights[2], ...)))


Written out:



df2 <- df %>% 
rowwise() %>%
mutate(comp = mean(c(0.5*a, 0.7*b, 0.8*c, 0.2*d)))


I can't figure out how to do this because, although vars contains the exact variable names that I want to use for mutate in my df, inside vars they are strings. How could I make mutate() understand that the strings vars contains relate to columns in my df? If you know another procedure not using mutate() that's fine also. Thanks!










share|improve this question





























    1















    Suppose I want to create a mean variable in a given dataframe based on two vectors, one specifying the names of the variables to use, and one specifying weights by which these variables should go into the mean variable:



    vars <- c("a", "b", "c","d"))
    weights <- c(0.5, 0.7, 0.8, 0.2))
    df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
    c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
    colnames(df) <- c("a","b","c","d","e","f")


    How could I use dplyr::mutate() to create a mean variable that uses vars and weights to calculate a rowwise score? mutate() should specifically use the variables supplied by vars
    The result should basically do the following:



    df <- df %>% 
    rowwise() %>%
    mutate(comp = mean(c(vars[1]*weights[1], vars[2]*weights[2], ...)))


    Written out:



    df2 <- df %>% 
    rowwise() %>%
    mutate(comp = mean(c(0.5*a, 0.7*b, 0.8*c, 0.2*d)))


    I can't figure out how to do this because, although vars contains the exact variable names that I want to use for mutate in my df, inside vars they are strings. How could I make mutate() understand that the strings vars contains relate to columns in my df? If you know another procedure not using mutate() that's fine also. Thanks!










    share|improve this question



























      1












      1








      1








      Suppose I want to create a mean variable in a given dataframe based on two vectors, one specifying the names of the variables to use, and one specifying weights by which these variables should go into the mean variable:



      vars <- c("a", "b", "c","d"))
      weights <- c(0.5, 0.7, 0.8, 0.2))
      df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
      c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
      colnames(df) <- c("a","b","c","d","e","f")


      How could I use dplyr::mutate() to create a mean variable that uses vars and weights to calculate a rowwise score? mutate() should specifically use the variables supplied by vars
      The result should basically do the following:



      df <- df %>% 
      rowwise() %>%
      mutate(comp = mean(c(vars[1]*weights[1], vars[2]*weights[2], ...)))


      Written out:



      df2 <- df %>% 
      rowwise() %>%
      mutate(comp = mean(c(0.5*a, 0.7*b, 0.8*c, 0.2*d)))


      I can't figure out how to do this because, although vars contains the exact variable names that I want to use for mutate in my df, inside vars they are strings. How could I make mutate() understand that the strings vars contains relate to columns in my df? If you know another procedure not using mutate() that's fine also. Thanks!










      share|improve this question
















      Suppose I want to create a mean variable in a given dataframe based on two vectors, one specifying the names of the variables to use, and one specifying weights by which these variables should go into the mean variable:



      vars <- c("a", "b", "c","d"))
      weights <- c(0.5, 0.7, 0.8, 0.2))
      df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
      c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
      colnames(df) <- c("a","b","c","d","e","f")


      How could I use dplyr::mutate() to create a mean variable that uses vars and weights to calculate a rowwise score? mutate() should specifically use the variables supplied by vars
      The result should basically do the following:



      df <- df %>% 
      rowwise() %>%
      mutate(comp = mean(c(vars[1]*weights[1], vars[2]*weights[2], ...)))


      Written out:



      df2 <- df %>% 
      rowwise() %>%
      mutate(comp = mean(c(0.5*a, 0.7*b, 0.8*c, 0.2*d)))


      I can't figure out how to do this because, although vars contains the exact variable names that I want to use for mutate in my df, inside vars they are strings. How could I make mutate() understand that the strings vars contains relate to columns in my df? If you know another procedure not using mutate() that's fine also. Thanks!







      r vector mean mutate weighted






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      edited Nov 26 '18 at 10:46







      nklsstll

















      asked Nov 24 '18 at 16:28









      nklsstllnklsstll

      105




      105
























          3 Answers
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          1














          You may use



          df %>% mutate(wmean = apply(.[vars], 1, weighted.mean, weights))
          # a b c d e f mean
          # 1 1 2 1 4 3 5 1.590909
          # 2 4 3 1 5 2 5 2.681818
          # 3 5 7 2 3 2 7 4.363636
          # 4 7 5 3 3 1 1 4.545455


          but there is not much to gain with tidyverse as base R approaches can be almost the same and end up being shorter:



          df$wmean <- apply(df[vars], 1, weighted.mean, weights)


          or one of the following:



          df$wmean <- colSums(t(df[vars]) * weights) / sum(weights)
          df$wmean <- as.matrix(df[vars]) %*% weights / sum(weights)
          df$wmean <- rowSums(sweep(df[vars], 2, weights, `*`)) / sum(weights)





          share|improve this answer

































            0














            Row-wise operations can be a bit tricky in the tidyverse. This is a case where some base R knowledge can be really handy. For example, you can do it in one line with apply (note that I corrected a typo in the line that creates weights and drop columns e and f, which do not have weights):



            vars <- c("a", "b", "c","d")
            weights <- c(0.5, 0.7, 0.8, 0.2)
            df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
            c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
            colnames(df) <- c("a","b","c","d","e","f")

            df$weighted.mean <- apply(df %>% select(-e, -f), 1, weighted.mean, weights)

            a b c d e f weighted.mean
            1 1 2 1 4 3 5 1.590909
            2 4 3 1 5 2 5 2.681818
            3 5 7 2 3 2 7 4.363636
            4 7 5 3 3 1 1 4.545455


            If you really wanted to do it in the tidyverse, this should get you started:



            library(tidyverse)

            df.weights <- data.frame(vars, weights)

            df.new <- df %>%
            mutate(row.num = 1:n()) %>%
            gather(variable, value, -row.num) %>%
            left_join(df.weights, by = c(variable = 'vars')) %>%
            filter(variable %in% vars) %>%
            group_by(row.num) %>%
            mutate(weighted.mean = weighted.mean(value, weights))





            share|improve this answer


























            • Repeating the OP's as.vector call to surround a c encourages wasteful operations. That code looks pretty convoluted.

              – 42-
              Nov 24 '18 at 16:48













            • Didn't catch that, thank you.

              – jdobres
              Nov 24 '18 at 16:51











            • His code has a spelling error in the second (unnecessary) as.vectors.

              – 42-
              Nov 24 '18 at 16:53











            • Thank you, this also works but I ended up choosing Julius Vainoras answer because it uses the vars vector and is even shorter. Better catch up on some of that basic R knowledge!

              – nklsstll
              Nov 24 '18 at 17:23



















            0














            There should be a tidyverse solution using pmap, but it eludes me. Here's another approach using tidyverse packages purrr and tibble





            library(tidyverse)

            vars <- c("a", "b", "c", "d")
            weights <- c(0.5, 0.7, 0.8, 0.2)
            df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
            c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
            colnames(df) <- c("a","b","c","d","e","f")

            df %>%
            transpose() %>%
            simplify_all() %>%
            map_dbl(~weighted.mean(.x[vars], weights)) %>%
            add_column(df, wmean = .)
            #> a b c d e f wmean
            #> 1 1 2 1 4 3 5 1.590909
            #> 2 4 3 1 5 2 5 2.681818
            #> 3 5 7 2 3 2 7 4.363636
            #> 4 7 5 3 3 1 1 4.545455


            Created on 2018-11-24 by the reprex package (v0.2.1)






            share|improve this answer























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              3 Answers
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              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

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              active

              oldest

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              active

              oldest

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              1














              You may use



              df %>% mutate(wmean = apply(.[vars], 1, weighted.mean, weights))
              # a b c d e f mean
              # 1 1 2 1 4 3 5 1.590909
              # 2 4 3 1 5 2 5 2.681818
              # 3 5 7 2 3 2 7 4.363636
              # 4 7 5 3 3 1 1 4.545455


              but there is not much to gain with tidyverse as base R approaches can be almost the same and end up being shorter:



              df$wmean <- apply(df[vars], 1, weighted.mean, weights)


              or one of the following:



              df$wmean <- colSums(t(df[vars]) * weights) / sum(weights)
              df$wmean <- as.matrix(df[vars]) %*% weights / sum(weights)
              df$wmean <- rowSums(sweep(df[vars], 2, weights, `*`)) / sum(weights)





              share|improve this answer






























                1














                You may use



                df %>% mutate(wmean = apply(.[vars], 1, weighted.mean, weights))
                # a b c d e f mean
                # 1 1 2 1 4 3 5 1.590909
                # 2 4 3 1 5 2 5 2.681818
                # 3 5 7 2 3 2 7 4.363636
                # 4 7 5 3 3 1 1 4.545455


                but there is not much to gain with tidyverse as base R approaches can be almost the same and end up being shorter:



                df$wmean <- apply(df[vars], 1, weighted.mean, weights)


                or one of the following:



                df$wmean <- colSums(t(df[vars]) * weights) / sum(weights)
                df$wmean <- as.matrix(df[vars]) %*% weights / sum(weights)
                df$wmean <- rowSums(sweep(df[vars], 2, weights, `*`)) / sum(weights)





                share|improve this answer




























                  1












                  1








                  1







                  You may use



                  df %>% mutate(wmean = apply(.[vars], 1, weighted.mean, weights))
                  # a b c d e f mean
                  # 1 1 2 1 4 3 5 1.590909
                  # 2 4 3 1 5 2 5 2.681818
                  # 3 5 7 2 3 2 7 4.363636
                  # 4 7 5 3 3 1 1 4.545455


                  but there is not much to gain with tidyverse as base R approaches can be almost the same and end up being shorter:



                  df$wmean <- apply(df[vars], 1, weighted.mean, weights)


                  or one of the following:



                  df$wmean <- colSums(t(df[vars]) * weights) / sum(weights)
                  df$wmean <- as.matrix(df[vars]) %*% weights / sum(weights)
                  df$wmean <- rowSums(sweep(df[vars], 2, weights, `*`)) / sum(weights)





                  share|improve this answer















                  You may use



                  df %>% mutate(wmean = apply(.[vars], 1, weighted.mean, weights))
                  # a b c d e f mean
                  # 1 1 2 1 4 3 5 1.590909
                  # 2 4 3 1 5 2 5 2.681818
                  # 3 5 7 2 3 2 7 4.363636
                  # 4 7 5 3 3 1 1 4.545455


                  but there is not much to gain with tidyverse as base R approaches can be almost the same and end up being shorter:



                  df$wmean <- apply(df[vars], 1, weighted.mean, weights)


                  or one of the following:



                  df$wmean <- colSums(t(df[vars]) * weights) / sum(weights)
                  df$wmean <- as.matrix(df[vars]) %*% weights / sum(weights)
                  df$wmean <- rowSums(sweep(df[vars], 2, weights, `*`)) / sum(weights)






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 24 '18 at 16:54

























                  answered Nov 24 '18 at 16:49









                  Julius VainoraJulius Vainora

                  37.8k76584




                  37.8k76584

























                      0














                      Row-wise operations can be a bit tricky in the tidyverse. This is a case where some base R knowledge can be really handy. For example, you can do it in one line with apply (note that I corrected a typo in the line that creates weights and drop columns e and f, which do not have weights):



                      vars <- c("a", "b", "c","d")
                      weights <- c(0.5, 0.7, 0.8, 0.2)
                      df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
                      c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
                      colnames(df) <- c("a","b","c","d","e","f")

                      df$weighted.mean <- apply(df %>% select(-e, -f), 1, weighted.mean, weights)

                      a b c d e f weighted.mean
                      1 1 2 1 4 3 5 1.590909
                      2 4 3 1 5 2 5 2.681818
                      3 5 7 2 3 2 7 4.363636
                      4 7 5 3 3 1 1 4.545455


                      If you really wanted to do it in the tidyverse, this should get you started:



                      library(tidyverse)

                      df.weights <- data.frame(vars, weights)

                      df.new <- df %>%
                      mutate(row.num = 1:n()) %>%
                      gather(variable, value, -row.num) %>%
                      left_join(df.weights, by = c(variable = 'vars')) %>%
                      filter(variable %in% vars) %>%
                      group_by(row.num) %>%
                      mutate(weighted.mean = weighted.mean(value, weights))





                      share|improve this answer


























                      • Repeating the OP's as.vector call to surround a c encourages wasteful operations. That code looks pretty convoluted.

                        – 42-
                        Nov 24 '18 at 16:48













                      • Didn't catch that, thank you.

                        – jdobres
                        Nov 24 '18 at 16:51











                      • His code has a spelling error in the second (unnecessary) as.vectors.

                        – 42-
                        Nov 24 '18 at 16:53











                      • Thank you, this also works but I ended up choosing Julius Vainoras answer because it uses the vars vector and is even shorter. Better catch up on some of that basic R knowledge!

                        – nklsstll
                        Nov 24 '18 at 17:23
















                      0














                      Row-wise operations can be a bit tricky in the tidyverse. This is a case where some base R knowledge can be really handy. For example, you can do it in one line with apply (note that I corrected a typo in the line that creates weights and drop columns e and f, which do not have weights):



                      vars <- c("a", "b", "c","d")
                      weights <- c(0.5, 0.7, 0.8, 0.2)
                      df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
                      c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
                      colnames(df) <- c("a","b","c","d","e","f")

                      df$weighted.mean <- apply(df %>% select(-e, -f), 1, weighted.mean, weights)

                      a b c d e f weighted.mean
                      1 1 2 1 4 3 5 1.590909
                      2 4 3 1 5 2 5 2.681818
                      3 5 7 2 3 2 7 4.363636
                      4 7 5 3 3 1 1 4.545455


                      If you really wanted to do it in the tidyverse, this should get you started:



                      library(tidyverse)

                      df.weights <- data.frame(vars, weights)

                      df.new <- df %>%
                      mutate(row.num = 1:n()) %>%
                      gather(variable, value, -row.num) %>%
                      left_join(df.weights, by = c(variable = 'vars')) %>%
                      filter(variable %in% vars) %>%
                      group_by(row.num) %>%
                      mutate(weighted.mean = weighted.mean(value, weights))





                      share|improve this answer


























                      • Repeating the OP's as.vector call to surround a c encourages wasteful operations. That code looks pretty convoluted.

                        – 42-
                        Nov 24 '18 at 16:48













                      • Didn't catch that, thank you.

                        – jdobres
                        Nov 24 '18 at 16:51











                      • His code has a spelling error in the second (unnecessary) as.vectors.

                        – 42-
                        Nov 24 '18 at 16:53











                      • Thank you, this also works but I ended up choosing Julius Vainoras answer because it uses the vars vector and is even shorter. Better catch up on some of that basic R knowledge!

                        – nklsstll
                        Nov 24 '18 at 17:23














                      0












                      0








                      0







                      Row-wise operations can be a bit tricky in the tidyverse. This is a case where some base R knowledge can be really handy. For example, you can do it in one line with apply (note that I corrected a typo in the line that creates weights and drop columns e and f, which do not have weights):



                      vars <- c("a", "b", "c","d")
                      weights <- c(0.5, 0.7, 0.8, 0.2)
                      df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
                      c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
                      colnames(df) <- c("a","b","c","d","e","f")

                      df$weighted.mean <- apply(df %>% select(-e, -f), 1, weighted.mean, weights)

                      a b c d e f weighted.mean
                      1 1 2 1 4 3 5 1.590909
                      2 4 3 1 5 2 5 2.681818
                      3 5 7 2 3 2 7 4.363636
                      4 7 5 3 3 1 1 4.545455


                      If you really wanted to do it in the tidyverse, this should get you started:



                      library(tidyverse)

                      df.weights <- data.frame(vars, weights)

                      df.new <- df %>%
                      mutate(row.num = 1:n()) %>%
                      gather(variable, value, -row.num) %>%
                      left_join(df.weights, by = c(variable = 'vars')) %>%
                      filter(variable %in% vars) %>%
                      group_by(row.num) %>%
                      mutate(weighted.mean = weighted.mean(value, weights))





                      share|improve this answer















                      Row-wise operations can be a bit tricky in the tidyverse. This is a case where some base R knowledge can be really handy. For example, you can do it in one line with apply (note that I corrected a typo in the line that creates weights and drop columns e and f, which do not have weights):



                      vars <- c("a", "b", "c","d")
                      weights <- c(0.5, 0.7, 0.8, 0.2)
                      df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
                      c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
                      colnames(df) <- c("a","b","c","d","e","f")

                      df$weighted.mean <- apply(df %>% select(-e, -f), 1, weighted.mean, weights)

                      a b c d e f weighted.mean
                      1 1 2 1 4 3 5 1.590909
                      2 4 3 1 5 2 5 2.681818
                      3 5 7 2 3 2 7 4.363636
                      4 7 5 3 3 1 1 4.545455


                      If you really wanted to do it in the tidyverse, this should get you started:



                      library(tidyverse)

                      df.weights <- data.frame(vars, weights)

                      df.new <- df %>%
                      mutate(row.num = 1:n()) %>%
                      gather(variable, value, -row.num) %>%
                      left_join(df.weights, by = c(variable = 'vars')) %>%
                      filter(variable %in% vars) %>%
                      group_by(row.num) %>%
                      mutate(weighted.mean = weighted.mean(value, weights))






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Nov 24 '18 at 16:55

























                      answered Nov 24 '18 at 16:44









                      jdobresjdobres

                      4,8861522




                      4,8861522













                      • Repeating the OP's as.vector call to surround a c encourages wasteful operations. That code looks pretty convoluted.

                        – 42-
                        Nov 24 '18 at 16:48













                      • Didn't catch that, thank you.

                        – jdobres
                        Nov 24 '18 at 16:51











                      • His code has a spelling error in the second (unnecessary) as.vectors.

                        – 42-
                        Nov 24 '18 at 16:53











                      • Thank you, this also works but I ended up choosing Julius Vainoras answer because it uses the vars vector and is even shorter. Better catch up on some of that basic R knowledge!

                        – nklsstll
                        Nov 24 '18 at 17:23



















                      • Repeating the OP's as.vector call to surround a c encourages wasteful operations. That code looks pretty convoluted.

                        – 42-
                        Nov 24 '18 at 16:48













                      • Didn't catch that, thank you.

                        – jdobres
                        Nov 24 '18 at 16:51











                      • His code has a spelling error in the second (unnecessary) as.vectors.

                        – 42-
                        Nov 24 '18 at 16:53











                      • Thank you, this also works but I ended up choosing Julius Vainoras answer because it uses the vars vector and is even shorter. Better catch up on some of that basic R knowledge!

                        – nklsstll
                        Nov 24 '18 at 17:23

















                      Repeating the OP's as.vector call to surround a c encourages wasteful operations. That code looks pretty convoluted.

                      – 42-
                      Nov 24 '18 at 16:48







                      Repeating the OP's as.vector call to surround a c encourages wasteful operations. That code looks pretty convoluted.

                      – 42-
                      Nov 24 '18 at 16:48















                      Didn't catch that, thank you.

                      – jdobres
                      Nov 24 '18 at 16:51





                      Didn't catch that, thank you.

                      – jdobres
                      Nov 24 '18 at 16:51













                      His code has a spelling error in the second (unnecessary) as.vectors.

                      – 42-
                      Nov 24 '18 at 16:53





                      His code has a spelling error in the second (unnecessary) as.vectors.

                      – 42-
                      Nov 24 '18 at 16:53













                      Thank you, this also works but I ended up choosing Julius Vainoras answer because it uses the vars vector and is even shorter. Better catch up on some of that basic R knowledge!

                      – nklsstll
                      Nov 24 '18 at 17:23





                      Thank you, this also works but I ended up choosing Julius Vainoras answer because it uses the vars vector and is even shorter. Better catch up on some of that basic R knowledge!

                      – nklsstll
                      Nov 24 '18 at 17:23











                      0














                      There should be a tidyverse solution using pmap, but it eludes me. Here's another approach using tidyverse packages purrr and tibble





                      library(tidyverse)

                      vars <- c("a", "b", "c", "d")
                      weights <- c(0.5, 0.7, 0.8, 0.2)
                      df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
                      c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
                      colnames(df) <- c("a","b","c","d","e","f")

                      df %>%
                      transpose() %>%
                      simplify_all() %>%
                      map_dbl(~weighted.mean(.x[vars], weights)) %>%
                      add_column(df, wmean = .)
                      #> a b c d e f wmean
                      #> 1 1 2 1 4 3 5 1.590909
                      #> 2 4 3 1 5 2 5 2.681818
                      #> 3 5 7 2 3 2 7 4.363636
                      #> 4 7 5 3 3 1 1 4.545455


                      Created on 2018-11-24 by the reprex package (v0.2.1)






                      share|improve this answer




























                        0














                        There should be a tidyverse solution using pmap, but it eludes me. Here's another approach using tidyverse packages purrr and tibble





                        library(tidyverse)

                        vars <- c("a", "b", "c", "d")
                        weights <- c(0.5, 0.7, 0.8, 0.2)
                        df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
                        c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
                        colnames(df) <- c("a","b","c","d","e","f")

                        df %>%
                        transpose() %>%
                        simplify_all() %>%
                        map_dbl(~weighted.mean(.x[vars], weights)) %>%
                        add_column(df, wmean = .)
                        #> a b c d e f wmean
                        #> 1 1 2 1 4 3 5 1.590909
                        #> 2 4 3 1 5 2 5 2.681818
                        #> 3 5 7 2 3 2 7 4.363636
                        #> 4 7 5 3 3 1 1 4.545455


                        Created on 2018-11-24 by the reprex package (v0.2.1)






                        share|improve this answer


























                          0












                          0








                          0







                          There should be a tidyverse solution using pmap, but it eludes me. Here's another approach using tidyverse packages purrr and tibble





                          library(tidyverse)

                          vars <- c("a", "b", "c", "d")
                          weights <- c(0.5, 0.7, 0.8, 0.2)
                          df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
                          c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
                          colnames(df) <- c("a","b","c","d","e","f")

                          df %>%
                          transpose() %>%
                          simplify_all() %>%
                          map_dbl(~weighted.mean(.x[vars], weights)) %>%
                          add_column(df, wmean = .)
                          #> a b c d e f wmean
                          #> 1 1 2 1 4 3 5 1.590909
                          #> 2 4 3 1 5 2 5 2.681818
                          #> 3 5 7 2 3 2 7 4.363636
                          #> 4 7 5 3 3 1 1 4.545455


                          Created on 2018-11-24 by the reprex package (v0.2.1)






                          share|improve this answer













                          There should be a tidyverse solution using pmap, but it eludes me. Here's another approach using tidyverse packages purrr and tibble





                          library(tidyverse)

                          vars <- c("a", "b", "c", "d")
                          weights <- c(0.5, 0.7, 0.8, 0.2)
                          df <- data.frame(cbind(c(1,4,5,7), c(2,3,7,5), c(1,1,2,3),
                          c(4,5,3,3), c(3,2,2,1), c(5,5,7,1)))
                          colnames(df) <- c("a","b","c","d","e","f")

                          df %>%
                          transpose() %>%
                          simplify_all() %>%
                          map_dbl(~weighted.mean(.x[vars], weights)) %>%
                          add_column(df, wmean = .)
                          #> a b c d e f wmean
                          #> 1 1 2 1 4 3 5 1.590909
                          #> 2 4 3 1 5 2 5 2.681818
                          #> 3 5 7 2 3 2 7 4.363636
                          #> 4 7 5 3 3 1 1 4.545455


                          Created on 2018-11-24 by the reprex package (v0.2.1)







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 25 '18 at 2:34









                          Jake KauppJake Kaupp

                          5,64721428




                          5,64721428






























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