Question

想象一下两个数据帧：

X = pd.DataFrame([[1,2],[3,4],[5,6]], columns=["a", "b"])
Y = pd.DataFrame([10,20,30], columns=["a"])

>>> X
   a  b
0  1  2
1  3  4
2  5  6
>>> Y
   a
0  10
1  20
2  30

总的来说，我希望我的最终输出是这样的：

   a_X  b_X  a_Y b_Y sum_a sum_b
0    1  2    10  NaN  11      2
1    3  4    20  NaN  23      4
2    5  6    30  NaN  35      6

我试图通过以下方式实现：

merged = X.join(Y, lsuffix="_X", rsuffix="_Y")
merged['sum_a'] = merged['a_X'] + merged['a_Y'] # works
merged['sum_b'] = merged['b_X'] + merged['b_Y'] # doesn't work

显然sum_b列会失败，因为Y集中没有b列。它可能存在，但它没有，我的数据集没有任何保证。看起来我不能使用内置连接在那里添加“NaN”列。

Answer 1

与pd.concat -

连接

k = ['X', 'Y']

df = pd.concat([X, Y], keys=k, axis=1)
df

   X      Y
   a  b   a
0  1  2  10
1  3  4  20
2  5  6  30

生成MultiIndex并使用它来重新索引 -

idx = pd.MultiIndex.from_product([k, df.columns.levels[1].unique()])
df = df.reindex(columns=idx)
df

   X      Y    
   a  b   a   b
0  1  2  10 NaN
1  3  4  20 NaN
2  5  6  30 NaN

重新设置列名称 -

df.columns = df.columns.map('_'.join)
df

   X_a  X_b  Y_a  Y_b
0    1    2   10  NaN
1    3    4   20  NaN
2    5    6   30  NaN

现在，您可以分组后缀并查找总和 -

v = df.groupby(by=lambda x: x.split('_')[1], axis=1).sum().add_prefix('sum_')
v

   sum_a  sum_b
0   11.0    2.0
1   23.0    4.0
2   35.0    6.0

将其与原文连接：

pd.concat([df, v], 1)

   X_a  X_b  Y_a  Y_b  sum_a  sum_b
0    1    2   10  NaN   11.0    2.0
1    3    4   20  NaN   23.0    4.0
2    5    6   30  NaN   35.0    6.0

Answer 2

df=pd.concat([X,Y.reindex(columns=X.columns)],keys=['x','y'],axis=1)

x=df.groupby(level=1,axis=1).sum().add_prefix('sum_')

df.columns=df.columns.map('{0[1]}{0[0]}'.format)


pd.concat([df,x],1)
Out[58]: 
   ax  bx  ay  by  sum_a  sum_b
0   1   2  10 NaN   11.0    2.0
1   3   4  20 NaN   23.0    4.0
2   5   6  30 NaN   35.0    6.0

Answer 3

你可以这样做：

import numpy as np

Y['b'] = np.nan
merged = X.join(Y, lsuffix="_X", rsuffix="_Y")
merged['sum_a'] = merged['a_X'] + merged['a_Y']
merged['sum_b'] = merged['b_X'] + merged.fillna(0)['b_Y']

#>>> merged
#   a_X  b_X  a_Y  b_Y  sum_a  sum_b
#0    1    2   10  NaN     11    2.0
#1    3    4   20  NaN     23    4.0
#2    5    6   30  NaN     35    6.0

Answer 4

更接近你正在做的事情。由于Y不必与X具有相同的列，您可以将reindex用于Y，然后使用fill_value选项执行操作：

Y = Y.reindex(columns=X.columns)
>>> Y
#    a    b
#0  10  NaN
#1  20  NaN  
#2  30  NaN

merged = X.join(Y, lsuffix="_X", rsuffix="_Y")
merged['sum_a'] = merged['a_X'].add(merged['a_Y'], fill_value=0)
merged['sum_b'] = merged['b_X'].add(merged['b_Y'], fill_value=0)

Pandas - 以缺少的列加入NaN

4 个答案: