Creating a histogram that sorts array values into bins, and shows the the frequency of items in the bins
0
For an array of values between 0 and 1, I want to create a histogram of 5 bins where bin one show the frequency(# of times) numbers between 0-0.2 show up in the array, bin 2 shows the frequency of numbers between 0.2-0.4, bin 3 is 0.4-0.6, bin 4: 0.6-0.8, bin 5 0.8-1.
import numpy as np
arr = np.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = np.histogram(arr, bins=5)
x = range(0,5)
graph = plt.bar(x,height=y)
plt.show()
python arrays numpy matplotlib histogram
edited Nov 21 at 1:10
eyllanesc
72.8k93055
asked Nov 21 at 1:10
Jacob
78324
add a comment |
0
For an array of values between 0 and 1, I want to create a histogram of 5 bins where bin one show the frequency(# of times) numbers between 0-0.2 show up in the array, bin 2 shows the frequency of numbers between 0.2-0.4, bin 3 is 0.4-0.6, bin 4: 0.6-0.8, bin 5 0.8-1.
import numpy as np
arr = np.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = np.histogram(arr, bins=5)
x = range(0,5)
graph = plt.bar(x,height=y)
plt.show()
python arrays numpy matplotlib histogram
edited Nov 21 at 1:10
eyllanesc
72.8k93055
asked Nov 21 at 1:10
Jacob
78324
What is your issue exactly? One issue is If 0 and 1 aren't in the data, it won't be the bins you expect. You'll want to explicitly define your bins (or range) since they will be determined by the min and max of your data.
– busybear
Nov 21 at 1:15
In case you want to use a bar plot, see e.g. this answer.
– ImportanceOfBeingErnest
Nov 21 at 2:05
add a comment |
0
0
0
For an array of values between 0 and 1, I want to create a histogram of 5 bins where bin one show the frequency(# of times) numbers between 0-0.2 show up in the array, bin 2 shows the frequency of numbers between 0.2-0.4, bin 3 is 0.4-0.6, bin 4: 0.6-0.8, bin 5 0.8-1.
import numpy as np
arr = np.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = np.histogram(arr, bins=5)
x = range(0,5)
graph = plt.bar(x,height=y)
plt.show()
python arrays numpy matplotlib histogram
edited Nov 21 at 1:10
eyllanesc
72.8k93055
asked Nov 21 at 1:10
Jacob
78324
For an array of values between 0 and 1, I want to create a histogram of 5 bins where bin one show the frequency(# of times) numbers between 0-0.2 show up in the array, bin 2 shows the frequency of numbers between 0.2-0.4, bin 3 is 0.4-0.6, bin 4: 0.6-0.8, bin 5 0.8-1.
import numpy as np
arr = np.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = np.histogram(arr, bins=5)
x = range(0,5)
graph = plt.bar(x,height=y)
plt.show()
python arrays numpy matplotlib histogram
python arrays numpy matplotlib histogram
edited Nov 21 at 1:10
eyllanesc
72.8k93055
asked Nov 21 at 1:10
Jacob
78324
edited Nov 21 at 1:10
eyllanesc
72.8k93055
asked Nov 21 at 1:10
Jacob
78324
edited Nov 21 at 1:10
eyllanesc
72.8k93055
edited Nov 21 at 1:10
eyllanesc
72.8k93055
edited Nov 21 at 1:10
eyllanesc
72.8k93055
72.8k93055
asked Nov 21 at 1:10
Jacob
78324
asked Nov 21 at 1:10
Jacob
78324
asked Nov 21 at 1:10
Jacob
78324
78324
What is your issue exactly? One issue is If 0 and 1 aren't in the data, it won't be the bins you expect. You'll want to explicitly define your bins (or range) since they will be determined by the min and max of your data.
– busybear
Nov 21 at 1:15
In case you want to use a bar plot, see e.g. this answer.
– ImportanceOfBeingErnest
Nov 21 at 2:05
add a comment |
What is your issue exactly? One issue is If 0 and 1 aren't in the data, it won't be the bins you expect. You'll want to explicitly define your bins (or range) since they will be determined by the min and max of your data.
– busybear
Nov 21 at 1:15
In case you want to use a bar plot, see e.g. this answer.
– ImportanceOfBeingErnest
Nov 21 at 2:05
What is your issue exactly? One issue is If 0 and 1 aren't in the data, it won't be the bins you expect. You'll want to explicitly define your bins (or range) since they will be determined by the min and max of your data.
– busybear
Nov 21 at 1:15
What is your issue exactly? One issue is If 0 and 1 aren't in the data, it won't be the bins you expect. You'll want to explicitly define your bins (or range) since they will be determined by the min and max of your data.
– busybear
Nov 21 at 1:15
In case you want to use a bar plot, see e.g. this answer.
– ImportanceOfBeingErnest
Nov 21 at 2:05
In case you want to use a bar plot, see e.g. this answer.
– ImportanceOfBeingErnest
Nov 21 at 2:05
add a comment |
2 Answers
2
active
oldest
votes
2
I think you are looking for matplotlib's hist method.
With your sample array the code would look like:
plt.hist(arr, bins=np.linspace(0,1,6), ec='black')
answered Nov 21 at 1:32
user7374610
6981422
I don't want to specify the number of bins because the number of bins is variable, in this example I only have 5 but I could 30, or 40, or 1000. The graph in this picture is an example of what I want, but I might have 5 graphs like this beside eachother, so ideally I want each array index of y to be a bar
– Jacob
Nov 21 at 12:37
Ignore everything after the last comma in my last sentence
– Jacob
Nov 21 at 12:48
add a comment |
-1
Is it what you are after ?
import numpy
from numpy.random import random
import matplotlib.pyplot as plt
arr = random(100)
y, other_stuff = numpy.histogram(arr, bins=5)
x = numpy.linspace(0.1, 0.9, 5)
graph = plt.bar(x, height=y, width=0.2, edgecolor='black')
plt.show()
As pointed out in the comment below, the snippet above does not define the edges of the bins in the call to histogram(). The one below corrects for that.
import numpy
import matplotlib.pyplot as plt
arr = numpy.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = numpy.histogram(arr, bins=numpy.linspace(0, 1, 6))
graph = plt.bar(numpy.linspace(0.1, 0.9, 5), height=y, width=0.2,
edgecolor='black')
plt.show()
answered Nov 21 at 1:33
Patol75
6136
1
Attention! this only looks like it's working correctly. And that only by coincidence. You will need to define the bins as in the other answer.
– ImportanceOfBeingErnest
Nov 21 at 1:53
What do you mean ? The call to histogram() gives back 5 values as per bins=5, and the call to bar() specifies the location of the center of the bins (through x and the default keyword align='center') as well as their total width. I agree hist() does the job way better (and I have upvoted the answer), but I do not see why my answer would be only coincidentally working ? I just used and improved the snippet provided by the OP.
– Patol75
Nov 21 at 4:42
1
If you take 100 random points between 0 and 1 and put them into 5 bins, the likelyhood is pretty high that those bins are close to[0,0.2), [0.2,0.4), etc. However, it is an assumption which is in general not true, and will lead to wrong results if you took e.g. the data provided in the question. Your approach would then give this result, which is obviously different from @user7374610's result and you may easily count the values by hand to see that it's not correctly binning the values.
– ImportanceOfBeingErnest
Nov 21 at 4:57
Ah, yes, makes sense now, thank you.
– Patol75
Nov 21 at 5:00
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
I think you are looking for matplotlib's hist method.
With your sample array the code would look like:
plt.hist(arr, bins=np.linspace(0,1,6), ec='black')
answered Nov 21 at 1:32
user7374610
6981422
I don't want to specify the number of bins because the number of bins is variable, in this example I only have 5 but I could 30, or 40, or 1000. The graph in this picture is an example of what I want, but I might have 5 graphs like this beside eachother, so ideally I want each array index of y to be a bar
– Jacob
Nov 21 at 12:37
Ignore everything after the last comma in my last sentence
– Jacob
Nov 21 at 12:48
add a comment |
2
I think you are looking for matplotlib's hist method.
With your sample array the code would look like:
plt.hist(arr, bins=np.linspace(0,1,6), ec='black')
answered Nov 21 at 1:32
user7374610
6981422
I don't want to specify the number of bins because the number of bins is variable, in this example I only have 5 but I could 30, or 40, or 1000. The graph in this picture is an example of what I want, but I might have 5 graphs like this beside eachother, so ideally I want each array index of y to be a bar
– Jacob
Nov 21 at 12:37
Ignore everything after the last comma in my last sentence
– Jacob
Nov 21 at 12:48
add a comment |
2
2
2
I think you are looking for matplotlib's hist method.
With your sample array the code would look like:
plt.hist(arr, bins=np.linspace(0,1,6), ec='black')
answered Nov 21 at 1:32
user7374610
6981422
I think you are looking for matplotlib's hist method.
With your sample array the code would look like:
plt.hist(arr, bins=np.linspace(0,1,6), ec='black')
answered Nov 21 at 1:32
user7374610
6981422
answered Nov 21 at 1:32
user7374610
6981422
answered Nov 21 at 1:32
user7374610
6981422
answered Nov 21 at 1:32
user7374610
6981422
6981422
I don't want to specify the number of bins because the number of bins is variable, in this example I only have 5 but I could 30, or 40, or 1000. The graph in this picture is an example of what I want, but I might have 5 graphs like this beside eachother, so ideally I want each array index of y to be a bar
– Jacob
Nov 21 at 12:37
Ignore everything after the last comma in my last sentence
– Jacob
Nov 21 at 12:48
add a comment |
I don't want to specify the number of bins because the number of bins is variable, in this example I only have 5 but I could 30, or 40, or 1000. The graph in this picture is an example of what I want, but I might have 5 graphs like this beside eachother, so ideally I want each array index of y to be a bar
– Jacob
Nov 21 at 12:37
Ignore everything after the last comma in my last sentence
– Jacob
Nov 21 at 12:48
I don't want to specify the number of bins because the number of bins is variable, in this example I only have 5 but I could 30, or 40, or 1000. The graph in this picture is an example of what I want, but I might have 5 graphs like this beside eachother, so ideally I want each array index of y to be a bar
– Jacob
Nov 21 at 12:37
I don't want to specify the number of bins because the number of bins is variable, in this example I only have 5 but I could 30, or 40, or 1000. The graph in this picture is an example of what I want, but I might have 5 graphs like this beside eachother, so ideally I want each array index of y to be a bar
– Jacob
Nov 21 at 12:37
Ignore everything after the last comma in my last sentence
– Jacob
Nov 21 at 12:48
Ignore everything after the last comma in my last sentence
– Jacob
Nov 21 at 12:48
add a comment |
-1
Is it what you are after ?
import numpy
from numpy.random import random
import matplotlib.pyplot as plt
arr = random(100)
y, other_stuff = numpy.histogram(arr, bins=5)
x = numpy.linspace(0.1, 0.9, 5)
graph = plt.bar(x, height=y, width=0.2, edgecolor='black')
plt.show()
As pointed out in the comment below, the snippet above does not define the edges of the bins in the call to histogram(). The one below corrects for that.
import numpy
import matplotlib.pyplot as plt
arr = numpy.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = numpy.histogram(arr, bins=numpy.linspace(0, 1, 6))
graph = plt.bar(numpy.linspace(0.1, 0.9, 5), height=y, width=0.2,
edgecolor='black')
plt.show()
answered Nov 21 at 1:33
Patol75
6136
1
Attention! this only looks like it's working correctly. And that only by coincidence. You will need to define the bins as in the other answer.
– ImportanceOfBeingErnest
Nov 21 at 1:53
What do you mean ? The call to histogram() gives back 5 values as per bins=5, and the call to bar() specifies the location of the center of the bins (through x and the default keyword align='center') as well as their total width. I agree hist() does the job way better (and I have upvoted the answer), but I do not see why my answer would be only coincidentally working ? I just used and improved the snippet provided by the OP.
– Patol75
Nov 21 at 4:42
1
If you take 100 random points between 0 and 1 and put them into 5 bins, the likelyhood is pretty high that those bins are close to[0,0.2), [0.2,0.4), etc. However, it is an assumption which is in general not true, and will lead to wrong results if you took e.g. the data provided in the question. Your approach would then give this result, which is obviously different from @user7374610's result and you may easily count the values by hand to see that it's not correctly binning the values.
– ImportanceOfBeingErnest
Nov 21 at 4:57
Ah, yes, makes sense now, thank you.
– Patol75
Nov 21 at 5:00
add a comment |
-1
Is it what you are after ?
import numpy
from numpy.random import random
import matplotlib.pyplot as plt
arr = random(100)
y, other_stuff = numpy.histogram(arr, bins=5)
x = numpy.linspace(0.1, 0.9, 5)
graph = plt.bar(x, height=y, width=0.2, edgecolor='black')
plt.show()
As pointed out in the comment below, the snippet above does not define the edges of the bins in the call to histogram(). The one below corrects for that.
import numpy
import matplotlib.pyplot as plt
arr = numpy.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = numpy.histogram(arr, bins=numpy.linspace(0, 1, 6))
graph = plt.bar(numpy.linspace(0.1, 0.9, 5), height=y, width=0.2,
edgecolor='black')
plt.show()
answered Nov 21 at 1:33
Patol75
6136
1
Attention! this only looks like it's working correctly. And that only by coincidence. You will need to define the bins as in the other answer.
– ImportanceOfBeingErnest
Nov 21 at 1:53
What do you mean ? The call to histogram() gives back 5 values as per bins=5, and the call to bar() specifies the location of the center of the bins (through x and the default keyword align='center') as well as their total width. I agree hist() does the job way better (and I have upvoted the answer), but I do not see why my answer would be only coincidentally working ? I just used and improved the snippet provided by the OP.
– Patol75
Nov 21 at 4:42
1
If you take 100 random points between 0 and 1 and put them into 5 bins, the likelyhood is pretty high that those bins are close to[0,0.2), [0.2,0.4), etc. However, it is an assumption which is in general not true, and will lead to wrong results if you took e.g. the data provided in the question. Your approach would then give this result, which is obviously different from @user7374610's result and you may easily count the values by hand to see that it's not correctly binning the values.
– ImportanceOfBeingErnest
Nov 21 at 4:57
Ah, yes, makes sense now, thank you.
– Patol75
Nov 21 at 5:00
add a comment |
-1
-1
-1
Is it what you are after ?
import numpy
from numpy.random import random
import matplotlib.pyplot as plt
arr = random(100)
y, other_stuff = numpy.histogram(arr, bins=5)
x = numpy.linspace(0.1, 0.9, 5)
graph = plt.bar(x, height=y, width=0.2, edgecolor='black')
plt.show()
As pointed out in the comment below, the snippet above does not define the edges of the bins in the call to histogram(). The one below corrects for that.
import numpy
import matplotlib.pyplot as plt
arr = numpy.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = numpy.histogram(arr, bins=numpy.linspace(0, 1, 6))
graph = plt.bar(numpy.linspace(0.1, 0.9, 5), height=y, width=0.2,
edgecolor='black')
plt.show()
answered Nov 21 at 1:33
Patol75
6136
Is it what you are after ?
import numpy
from numpy.random import random
import matplotlib.pyplot as plt
arr = random(100)
y, other_stuff = numpy.histogram(arr, bins=5)
x = numpy.linspace(0.1, 0.9, 5)
graph = plt.bar(x, height=y, width=0.2, edgecolor='black')
plt.show()
As pointed out in the comment below, the snippet above does not define the edges of the bins in the call to histogram(). The one below corrects for that.
import numpy
import matplotlib.pyplot as plt
arr = numpy.array([0.5, 0.1, 0.05, 0.67, 0.8, 0.9, 1, 0.22, 0.25])
y, other_stuff = numpy.histogram(arr, bins=numpy.linspace(0, 1, 6))
graph = plt.bar(numpy.linspace(0.1, 0.9, 5), height=y, width=0.2,
edgecolor='black')
plt.show()
answered Nov 21 at 1:33
Patol75
6136
edited Nov 21 at 5:05
answered Nov 21 at 1:33
Patol75
6136
answered Nov 21 at 1:33
Patol75
6136
answered Nov 21 at 1:33
Patol75
6136
6136
1
Attention! this only looks like it's working correctly. And that only by coincidence. You will need to define the bins as in the other answer.
– ImportanceOfBeingErnest
Nov 21 at 1:53
What do you mean ? The call to histogram() gives back 5 values as per bins=5, and the call to bar() specifies the location of the center of the bins (through x and the default keyword align='center') as well as their total width. I agree hist() does the job way better (and I have upvoted the answer), but I do not see why my answer would be only coincidentally working ? I just used and improved the snippet provided by the OP.
– Patol75
Nov 21 at 4:42
1
If you take 100 random points between 0 and 1 and put them into 5 bins, the likelyhood is pretty high that those bins are close to[0,0.2), [0.2,0.4), etc. However, it is an assumption which is in general not true, and will lead to wrong results if you took e.g. the data provided in the question. Your approach would then give this result, which is obviously different from @user7374610's result and you may easily count the values by hand to see that it's not correctly binning the values.
– ImportanceOfBeingErnest
Nov 21 at 4:57
Ah, yes, makes sense now, thank you.
– Patol75
Nov 21 at 5:00
add a comment |
1
Attention! this only looks like it's working correctly. And that only by coincidence. You will need to define the bins as in the other answer.
– ImportanceOfBeingErnest
Nov 21 at 1:53
What do you mean ? The call to histogram() gives back 5 values as per bins=5, and the call to bar() specifies the location of the center of the bins (through x and the default keyword align='center') as well as their total width. I agree hist() does the job way better (and I have upvoted the answer), but I do not see why my answer would be only coincidentally working ? I just used and improved the snippet provided by the OP.
– Patol75
Nov 21 at 4:42
1
If you take 100 random points between 0 and 1 and put them into 5 bins, the likelyhood is pretty high that those bins are close to[0,0.2), [0.2,0.4), etc. However, it is an assumption which is in general not true, and will lead to wrong results if you took e.g. the data provided in the question. Your approach would then give this result, which is obviously different from @user7374610's result and you may easily count the values by hand to see that it's not correctly binning the values.
– ImportanceOfBeingErnest
Nov 21 at 4:57
Ah, yes, makes sense now, thank you.
– Patol75
Nov 21 at 5:00
1
1
Attention! this only looks like it's working correctly. And that only by coincidence. You will need to define the bins as in the other answer.
– ImportanceOfBeingErnest
Nov 21 at 1:53
Attention! this only looks like it's working correctly. And that only by coincidence. You will need to define the bins as in the other answer.
– ImportanceOfBeingErnest
Nov 21 at 1:53
What do you mean ? The call to histogram() gives back 5 values as per bins=5, and the call to bar() specifies the location of the center of the bins (through x and the default keyword align='center') as well as their total width. I agree hist() does the job way better (and I have upvoted the answer), but I do not see why my answer would be only coincidentally working ? I just used and improved the snippet provided by the OP.
– Patol75
Nov 21 at 4:42
What do you mean ? The call to histogram() gives back 5 values as per bins=5, and the call to bar() specifies the location of the center of the bins (through x and the default keyword align='center') as well as their total width. I agree hist() does the job way better (and I have upvoted the answer), but I do not see why my answer would be only coincidentally working ? I just used and improved the snippet provided by the OP.
– Patol75
Nov 21 at 4:42
1
1
If you take 100 random points between 0 and 1 and put them into 5 bins, the likelyhood is pretty high that those bins are close to
[0,0.2), [0.2,0.4), etc. However, it is an assumption which is in general not true, and will lead to wrong results if you took e.g. the data provided in the question. Your approach would then give this result, which is obviously different from @user7374610's result and you may easily count the values by hand to see that it's not correctly binning the values.– ImportanceOfBeingErnest
Nov 21 at 4:57
If you take 100 random points between 0 and 1 and put them into 5 bins, the likelyhood is pretty high that those bins are close to
[0,0.2), [0.2,0.4), etc. However, it is an assumption which is in general not true, and will lead to wrong results if you took e.g. the data provided in the question. Your approach would then give this result, which is obviously different from @user7374610's result and you may easily count the values by hand to see that it's not correctly binning the values.– ImportanceOfBeingErnest
Nov 21 at 4:57
Ah, yes, makes sense now, thank you.
– Patol75
Nov 21 at 5:00
Ah, yes, makes sense now, thank you.
– Patol75
Nov 21 at 5:00
add a comment |
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What is your issue exactly? One issue is If 0 and 1 aren't in the data, it won't be the bins you expect. You'll want to explicitly define your bins (or range) since they will be determined by the min and max of your data.
– busybear
Nov 21 at 1:15
In case you want to use a bar plot, see e.g. this answer.
– ImportanceOfBeingErnest
Nov 21 at 2:05