以新的列名迭代合并 pandas 列
Iteratively merge panda columns with new column names
假设我在一个循环中迭代合并一个熊猫数据框,但在两到三次迭代后,熊猫重复列名,例如考虑以下示例,我在其中迭代合并列,但为简单起见没有循环:
A= {'Name':['A','B','C'],'GPA':[4.0,3.80,3.70], 'School':['U','U','U'], 'Time':[22,26,30]}
A1 = pd.DataFrame(A)
B= {'Name':['D','E','F'],'GPA':[3.50,3.70,3.60], 'School':['S','S','S'],'Time':[34,44,54]}
B1 = pd.DataFrame(B)
C= {'Name':['G','H','I'],'GPA':[3.70,3.50,3.70], 'School':['C','C','C'],'Time':[76,86,96]}
C1 = pd.DataFrame(C)
L= [A1,B1,C1]
comb = A1
for ii in L[1:]:
comb = pd.concat([comb,ii],ignore_index=True)
comb
B = pd.merge(comb, comb, on=['Name','GPA'])
C = pd.merge(B, comb, on=['Name','GPA'])
D = pd.merge(C, comb, on=['Name','GPA'])
你看熊猫把School_x和School_y的名字重复了两次,有没有办法改成School_x和School_y,School_z和School_t。我不是在谈论之后重命名它,而是强制合并为不相同的列选择新的列名。否则如何区分具有 1000 列的数据框并想象 500 个具有相同的列名。
更新:以上只是一个示例,假设您正在像这样循环合并多个数据帧:
for ii in list:
df = df.merge(A,on = 'some column', how = 'outer')
那么在我看来,每次重复相同的列时,你如何迭代地更改列名,即使有后缀。
尝试将 suffixes
参数更改为 ('_z', '_t')
:
的元组
B = pd.merge(comb, comb, on=['Name','GPA'])
C = pd.merge(B, comb, on=['Name','GPA'])
D = pd.merge(C, comb, on=['Name','GPA'], suffixes=('_z', '_t'))
>>> D
Name GPA School_x Time_x School_y Time_y School_z Time_z School_t Time_t
0 A 4.0 U 22 U 22 U 22 U 22
1 B 3.8 U 26 U 26 U 26 U 26
2 C 3.7 U 30 U 30 U 30 U 30
3 D 3.5 S 34 S 34 S 34 S 34
4 E 3.7 S 44 S 44 S 44 S 44
5 F 3.6 S 54 S 54 S 54 S 54
6 G 3.7 C 76 C 76 C 76 C 76
7 H 3.5 C 86 C 86 C 86 C 86
8 I 3.7 C 96 C 96 C 96 C 96
>>>
如 pd.merge
文档中所述:
Parameters:
...
...
suffixes: list-like, default is (“_x”, “_y”)
A length-2 sequence where each element is optionally a string indicating the suffix to add to overlapping column names in left and right respectively. Pass a value of None instead of a string to indicate that the column name from left or right should be left as-is, with no suffix. At least one of the values must not be None.
...
...
编辑:
对于问题的最新更新,请尝试创建一个迭代器并使用 next
。
functools.reduce
更好:
from functools import reduce
from string import ascii_lowercase
it = iter(ascii_lowercase)
print(reduce(lambda x, y: pd.merge(x, y, on=['Name','GPA'], suffixes=('_' + next(it), '_' + next(it))), [comb for _ in range(4)]))
输出:
Name GPA School_a Time_a School_b Time_b School_e Time_e School_f Time_f
0 A 4.0 U 22 U 22 U 22 U 22
1 B 3.8 U 26 U 26 U 26 U 26
2 C 3.7 U 30 U 30 U 30 U 30
3 D 3.5 S 34 S 34 S 34 S 34
4 E 3.7 S 44 S 44 S 44 S 44
5 F 3.6 S 54 S 54 S 54 S 54
6 G 3.7 C 76 C 76 C 76 C 76
7 H 3.5 C 86 C 86 C 86 C 86
8 I 3.7 C 96 C 96 C 96 C 96
如您所见,我使用 [comb for _ in range(4)]
创建了一个列表理解,它将循环并合并 4 次,要更改次数只需更改数字即 [comb for _ in range(10)]
.
对于函数:
from functools import reduce
from string import ascii_lowercase
def cumulative_merge(df, n):
it = iter(ascii_lowercase)
return reduce(lambda x, y: pd.merge(x, y, on=['Name','GPA'], suffixes=('_' + next(it), '_' + next(it))), [comb for _ in range(n)])
执行:
print(cumulative_merge(df, 4))
输出:
Name GPA School_a Time_a School_b Time_b School_e Time_e School_f Time_f
0 A 4.0 U 22 U 22 U 22 U 22
1 B 3.8 U 26 U 26 U 26 U 26
2 C 3.7 U 30 U 30 U 30 U 30
3 D 3.5 S 34 S 34 S 34 S 34
4 E 3.7 S 44 S 44 S 44 S 44
5 F 3.6 S 54 S 54 S 54 S 54
6 G 3.7 C 76 C 76 C 76 C 76
7 H 3.5 C 86 C 86 C 86 C 86
8 I 3.7 C 96 C 96 C 96 C 96
假设我在一个循环中迭代合并一个熊猫数据框,但在两到三次迭代后,熊猫重复列名,例如考虑以下示例,我在其中迭代合并列,但为简单起见没有循环:
A= {'Name':['A','B','C'],'GPA':[4.0,3.80,3.70], 'School':['U','U','U'], 'Time':[22,26,30]}
A1 = pd.DataFrame(A)
B= {'Name':['D','E','F'],'GPA':[3.50,3.70,3.60], 'School':['S','S','S'],'Time':[34,44,54]}
B1 = pd.DataFrame(B)
C= {'Name':['G','H','I'],'GPA':[3.70,3.50,3.70], 'School':['C','C','C'],'Time':[76,86,96]}
C1 = pd.DataFrame(C)
L= [A1,B1,C1]
comb = A1
for ii in L[1:]:
comb = pd.concat([comb,ii],ignore_index=True)
comb
B = pd.merge(comb, comb, on=['Name','GPA'])
C = pd.merge(B, comb, on=['Name','GPA'])
D = pd.merge(C, comb, on=['Name','GPA'])
你看熊猫把School_x和School_y的名字重复了两次,有没有办法改成School_x和School_y,School_z和School_t。我不是在谈论之后重命名它,而是强制合并为不相同的列选择新的列名。否则如何区分具有 1000 列的数据框并想象 500 个具有相同的列名。
更新:以上只是一个示例,假设您正在像这样循环合并多个数据帧:
for ii in list:
df = df.merge(A,on = 'some column', how = 'outer')
那么在我看来,每次重复相同的列时,你如何迭代地更改列名,即使有后缀。
尝试将 suffixes
参数更改为 ('_z', '_t')
:
B = pd.merge(comb, comb, on=['Name','GPA'])
C = pd.merge(B, comb, on=['Name','GPA'])
D = pd.merge(C, comb, on=['Name','GPA'], suffixes=('_z', '_t'))
>>> D
Name GPA School_x Time_x School_y Time_y School_z Time_z School_t Time_t
0 A 4.0 U 22 U 22 U 22 U 22
1 B 3.8 U 26 U 26 U 26 U 26
2 C 3.7 U 30 U 30 U 30 U 30
3 D 3.5 S 34 S 34 S 34 S 34
4 E 3.7 S 44 S 44 S 44 S 44
5 F 3.6 S 54 S 54 S 54 S 54
6 G 3.7 C 76 C 76 C 76 C 76
7 H 3.5 C 86 C 86 C 86 C 86
8 I 3.7 C 96 C 96 C 96 C 96
>>>
如 pd.merge
文档中所述:
Parameters:
...
...suffixes: list-like, default is (“_x”, “_y”)
A length-2 sequence where each element is optionally a string indicating the suffix to add to overlapping column names in left and right respectively. Pass a value of None instead of a string to indicate that the column name from left or right should be left as-is, with no suffix. At least one of the values must not be None.
... ...
编辑:
对于问题的最新更新,请尝试创建一个迭代器并使用 next
。
functools.reduce
更好:
from functools import reduce
from string import ascii_lowercase
it = iter(ascii_lowercase)
print(reduce(lambda x, y: pd.merge(x, y, on=['Name','GPA'], suffixes=('_' + next(it), '_' + next(it))), [comb for _ in range(4)]))
输出:
Name GPA School_a Time_a School_b Time_b School_e Time_e School_f Time_f
0 A 4.0 U 22 U 22 U 22 U 22
1 B 3.8 U 26 U 26 U 26 U 26
2 C 3.7 U 30 U 30 U 30 U 30
3 D 3.5 S 34 S 34 S 34 S 34
4 E 3.7 S 44 S 44 S 44 S 44
5 F 3.6 S 54 S 54 S 54 S 54
6 G 3.7 C 76 C 76 C 76 C 76
7 H 3.5 C 86 C 86 C 86 C 86
8 I 3.7 C 96 C 96 C 96 C 96
如您所见,我使用 [comb for _ in range(4)]
创建了一个列表理解,它将循环并合并 4 次,要更改次数只需更改数字即 [comb for _ in range(10)]
.
对于函数:
from functools import reduce
from string import ascii_lowercase
def cumulative_merge(df, n):
it = iter(ascii_lowercase)
return reduce(lambda x, y: pd.merge(x, y, on=['Name','GPA'], suffixes=('_' + next(it), '_' + next(it))), [comb for _ in range(n)])
执行:
print(cumulative_merge(df, 4))
输出:
Name GPA School_a Time_a School_b Time_b School_e Time_e School_f Time_f
0 A 4.0 U 22 U 22 U 22 U 22
1 B 3.8 U 26 U 26 U 26 U 26
2 C 3.7 U 30 U 30 U 30 U 30
3 D 3.5 S 34 S 34 S 34 S 34
4 E 3.7 S 44 S 44 S 44 S 44
5 F 3.6 S 54 S 54 S 54 S 54
6 G 3.7 C 76 C 76 C 76 C 76
7 H 3.5 C 86 C 86 C 86 C 86
8 I 3.7 C 96 C 96 C 96 C 96