按 N 列块重塑 Pandas 数据框列
Reshape Pandas dataframe columns by block of N columns
我有 1 个数据框,其中的列块需要重新整形为行。
我尝试使用 stack() 和 melt() 但找不到正确的方法。
这是我期望的示例:
data = {'id':['a1', 'a2', 'a3', 'a4'],
'year':[20, 20, 19, 18],
'b_A': [1, 2, 3, 4],
'b_B': [5, 6, 7, 8],
'b_C': [9, 10, 11, 12],
'c_A': [13, 14, 15, 16],
'c_B': [17, 18, 19, 20],
'c_C': [21, 22, 23, 24],
'd_A': [25, 26, 27, 28],
'd_B': [29, 30, 31, 32],
'd_C': [33, 34, 35, 36],
}
df = pd.DataFrame(data)
id year b_A b_B b_C c_A c_B c_C d_A d_B d_C
0 a1 20 1 5 9 13 17 21 25 29 33
1 a2 20 2 6 10 14 18 22 26 30 34
2 a3 19 3 7 11 15 19 23 27 31 35
3 a4 18 4 8 12 16 20 24 28 32 36
预期结果应该是:
id year origin A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
感谢您的宝贵时间和帮助。
您可以将具有 _
的非列名称转换为 DataFrame.set_index
, then splitting columns by Series.str.split
and reshape by DataFrame.stack
的索引:
df1 = df.set_index(['id','year'])
df1.columns = df1.columns.str.split('_', expand=True)
df1 = df1.stack(level=0).reset_index()
print (df1)
id year level_2 A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
如果还需要设置列 origin
可以使用 DataFrame.rename_axis
:
df1 = df.set_index(['id','year'])
df1.columns = df1.columns.str.split('_', expand=True)
df1 = df1.rename_axis(['origin',None], axis=1).stack(0).reset_index()
print (df1)
id year origin A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
或使用 wide_to_long
更改值的顺序 _
例如 A_b
到 b_A
:
df.columns = [f'{"_".join(x[::-1])}' for x in df.columns.str.split('_')]
df1 = pd.wide_to_long(df,
stubnames=['A','B','C'],
i=['id','year'],
j='origin',
sep='_',
suffix=r'\w+').reset_index()
print (df1)
id year origin A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
您也可以使用 pivot_longer function from pyjanitor; at the moment you have to install the latest development version from github:
# install latest dev version
# pip install git+https://github.com/ericmjl/pyjanitor.git
import janitor
df.pivot_longer(index=["id", "year"],
names_to=("origin", ".value"),
names_sep="_")
id year origin A B C
0 a1 20 b 1 5 9
1 a2 20 b 2 6 10
2 a3 19 b 3 7 11
3 a4 18 b 4 8 12
4 a1 20 c 13 17 21
5 a2 20 c 14 18 22
6 a3 19 c 15 19 23
7 a4 18 c 16 20 24
8 a1 20 d 25 29 33
9 a2 20 d 26 30 34
10 a3 19 d 27 31 35
11 a4 18 d 28 32 36
names_sep
值拆分列;与 .value
配对的拆分值保留为列 headers,而其他值集中在 origin
列下方。
如果你想要数据出现的顺序,可以使用sort_by_appearance
参数:
df.pivot_longer(
index=["id", "year"],
names_to=("origin", ".value"),
names_sep="_",
sort_by_appearance=True,
)
id year origin A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
我有 1 个数据框,其中的列块需要重新整形为行。 我尝试使用 stack() 和 melt() 但找不到正确的方法。
这是我期望的示例:
data = {'id':['a1', 'a2', 'a3', 'a4'],
'year':[20, 20, 19, 18],
'b_A': [1, 2, 3, 4],
'b_B': [5, 6, 7, 8],
'b_C': [9, 10, 11, 12],
'c_A': [13, 14, 15, 16],
'c_B': [17, 18, 19, 20],
'c_C': [21, 22, 23, 24],
'd_A': [25, 26, 27, 28],
'd_B': [29, 30, 31, 32],
'd_C': [33, 34, 35, 36],
}
df = pd.DataFrame(data)
id year b_A b_B b_C c_A c_B c_C d_A d_B d_C
0 a1 20 1 5 9 13 17 21 25 29 33
1 a2 20 2 6 10 14 18 22 26 30 34
2 a3 19 3 7 11 15 19 23 27 31 35
3 a4 18 4 8 12 16 20 24 28 32 36
预期结果应该是:
id year origin A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
感谢您的宝贵时间和帮助。
您可以将具有 _
的非列名称转换为 DataFrame.set_index
, then splitting columns by Series.str.split
and reshape by DataFrame.stack
的索引:
df1 = df.set_index(['id','year'])
df1.columns = df1.columns.str.split('_', expand=True)
df1 = df1.stack(level=0).reset_index()
print (df1)
id year level_2 A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
如果还需要设置列 origin
可以使用 DataFrame.rename_axis
:
df1 = df.set_index(['id','year'])
df1.columns = df1.columns.str.split('_', expand=True)
df1 = df1.rename_axis(['origin',None], axis=1).stack(0).reset_index()
print (df1)
id year origin A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
或使用 wide_to_long
更改值的顺序 _
例如 A_b
到 b_A
:
df.columns = [f'{"_".join(x[::-1])}' for x in df.columns.str.split('_')]
df1 = pd.wide_to_long(df,
stubnames=['A','B','C'],
i=['id','year'],
j='origin',
sep='_',
suffix=r'\w+').reset_index()
print (df1)
id year origin A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36
您也可以使用 pivot_longer function from pyjanitor; at the moment you have to install the latest development version from github:
# install latest dev version
# pip install git+https://github.com/ericmjl/pyjanitor.git
import janitor
df.pivot_longer(index=["id", "year"],
names_to=("origin", ".value"),
names_sep="_")
id year origin A B C
0 a1 20 b 1 5 9
1 a2 20 b 2 6 10
2 a3 19 b 3 7 11
3 a4 18 b 4 8 12
4 a1 20 c 13 17 21
5 a2 20 c 14 18 22
6 a3 19 c 15 19 23
7 a4 18 c 16 20 24
8 a1 20 d 25 29 33
9 a2 20 d 26 30 34
10 a3 19 d 27 31 35
11 a4 18 d 28 32 36
names_sep
值拆分列;与 .value
配对的拆分值保留为列 headers,而其他值集中在 origin
列下方。
如果你想要数据出现的顺序,可以使用sort_by_appearance
参数:
df.pivot_longer(
index=["id", "year"],
names_to=("origin", ".value"),
names_sep="_",
sort_by_appearance=True,
)
id year origin A B C
0 a1 20 b 1 5 9
1 a1 20 c 13 17 21
2 a1 20 d 25 29 33
3 a2 20 b 2 6 10
4 a2 20 c 14 18 22
5 a2 20 d 26 30 34
6 a3 19 b 3 7 11
7 a3 19 c 15 19 23
8 a3 19 d 27 31 35
9 a4 18 b 4 8 12
10 a4 18 c 16 20 24
11 a4 18 d 28 32 36