How to select several continue rows with interval in pandas?
I want to select 3 rows every 5 rows. For example, the first 5 rows, I want to remain the last 3 rows.
Input:
import pandas as pd
df = pd.DataFrame({'a': np.arange(16)})
print(df)
Output:
a
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
Expected:
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Hopefully for help!
python pandas dataframe
asked Nov 24 '18 at 10:54
rosefunrosefun
415210
add a comment |
I want to select 3 rows every 5 rows. For example, the first 5 rows, I want to remain the last 3 rows.
Input:
import pandas as pd
df = pd.DataFrame({'a': np.arange(16)})
print(df)
Output:
a
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
Expected:
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Hopefully for help!
python pandas dataframe
asked Nov 24 '18 at 10:54
rosefunrosefun
415210
add a comment |
I want to select 3 rows every 5 rows. For example, the first 5 rows, I want to remain the last 3 rows.
Input:
import pandas as pd
df = pd.DataFrame({'a': np.arange(16)})
print(df)
Output:
a
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
Expected:
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Hopefully for help!
python pandas dataframe
asked Nov 24 '18 at 10:54
rosefunrosefun
415210
I want to select 3 rows every 5 rows. For example, the first 5 rows, I want to remain the last 3 rows.
Input:
import pandas as pd
df = pd.DataFrame({'a': np.arange(16)})
print(df)
Output:
a
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
Expected:
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Hopefully for help!
python pandas dataframe
python pandas dataframe
asked Nov 24 '18 at 10:54
rosefunrosefun
415210
asked Nov 24 '18 at 10:54
rosefunrosefun
415210
asked Nov 24 '18 at 10:54
rosefunrosefun
415210
asked Nov 24 '18 at 10:54
rosefunrosefun
415210
asked Nov 24 '18 at 10:54
rosefunrosefun
415210
415210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Use groupby with index with floor division and get last 3 rows by tail:
df = df.groupby(df.index // 5).tail(3)
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
15 15 <- last group have only one value, so tail select it
Another idea is get index values of each rows by np.arange with reshape to 2d array, select last 'columns' and flatten by ravel, get intersection with real index values and select by loc:
N = 5
M = 3
pos = np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel()
idx = np.intersect1d(df.index, pos)
df = df.loc[idx]
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Detail:
print(np.arange((len(df) // N + 1) * N).reshape(-1, N))
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]
[15 16 17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:])
[[ 2 3 4]
[ 7 8 9]
[12 13 14]
[17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel())
[ 2 3 4 7 8 9 12 13 14 17 18 19]
print(np.intersect1d(df.index, pos))
[ 2 3 4 7 8 9 12 13 14]
answered Nov 24 '18 at 10:56
jezraeljezrael
336k25281357
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Use groupby with index with floor division and get last 3 rows by tail:
df = df.groupby(df.index // 5).tail(3)
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
15 15 <- last group have only one value, so tail select it
Another idea is get index values of each rows by np.arange with reshape to 2d array, select last 'columns' and flatten by ravel, get intersection with real index values and select by loc:
N = 5
M = 3
pos = np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel()
idx = np.intersect1d(df.index, pos)
df = df.loc[idx]
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Detail:
print(np.arange((len(df) // N + 1) * N).reshape(-1, N))
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]
[15 16 17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:])
[[ 2 3 4]
[ 7 8 9]
[12 13 14]
[17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel())
[ 2 3 4 7 8 9 12 13 14 17 18 19]
print(np.intersect1d(df.index, pos))
[ 2 3 4 7 8 9 12 13 14]
answered Nov 24 '18 at 10:56
jezraeljezrael
336k25281357
add a comment |
Use groupby with index with floor division and get last 3 rows by tail:
df = df.groupby(df.index // 5).tail(3)
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
15 15 <- last group have only one value, so tail select it
Another idea is get index values of each rows by np.arange with reshape to 2d array, select last 'columns' and flatten by ravel, get intersection with real index values and select by loc:
N = 5
M = 3
pos = np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel()
idx = np.intersect1d(df.index, pos)
df = df.loc[idx]
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Detail:
print(np.arange((len(df) // N + 1) * N).reshape(-1, N))
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]
[15 16 17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:])
[[ 2 3 4]
[ 7 8 9]
[12 13 14]
[17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel())
[ 2 3 4 7 8 9 12 13 14 17 18 19]
print(np.intersect1d(df.index, pos))
[ 2 3 4 7 8 9 12 13 14]
answered Nov 24 '18 at 10:56
jezraeljezrael
336k25281357
add a comment |
Use groupby with index with floor division and get last 3 rows by tail:
df = df.groupby(df.index // 5).tail(3)
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
15 15 <- last group have only one value, so tail select it
Another idea is get index values of each rows by np.arange with reshape to 2d array, select last 'columns' and flatten by ravel, get intersection with real index values and select by loc:
N = 5
M = 3
pos = np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel()
idx = np.intersect1d(df.index, pos)
df = df.loc[idx]
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Detail:
print(np.arange((len(df) // N + 1) * N).reshape(-1, N))
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]
[15 16 17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:])
[[ 2 3 4]
[ 7 8 9]
[12 13 14]
[17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel())
[ 2 3 4 7 8 9 12 13 14 17 18 19]
print(np.intersect1d(df.index, pos))
[ 2 3 4 7 8 9 12 13 14]
answered Nov 24 '18 at 10:56
jezraeljezrael
336k25281357
Use groupby with index with floor division and get last 3 rows by tail:
df = df.groupby(df.index // 5).tail(3)
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
15 15 <- last group have only one value, so tail select it
Another idea is get index values of each rows by np.arange with reshape to 2d array, select last 'columns' and flatten by ravel, get intersection with real index values and select by loc:
N = 5
M = 3
pos = np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel()
idx = np.intersect1d(df.index, pos)
df = df.loc[idx]
print(df)
a
2 2
3 3
4 4
7 7
8 8
9 9
12 12
13 13
14 14
Detail:
print(np.arange((len(df) // N + 1) * N).reshape(-1, N))
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]
[15 16 17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:])
[[ 2 3 4]
[ 7 8 9]
[12 13 14]
[17 18 19]]
print (np.arange((len(df) // N + 1) * N).reshape(-1, N)[:, -M:].ravel())
[ 2 3 4 7 8 9 12 13 14 17 18 19]
print(np.intersect1d(df.index, pos))
[ 2 3 4 7 8 9 12 13 14]
answered Nov 24 '18 at 10:56
jezraeljezrael
336k25281357
edited Nov 24 '18 at 11:24
answered Nov 24 '18 at 10:56
jezraeljezrael
336k25281357
answered Nov 24 '18 at 10:56
jezraeljezrael
336k25281357
answered Nov 24 '18 at 10:56
jezraeljezrael
336k25281357
336k25281357
add a comment |
add a comment |
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