通过使用另一个Pandas DataFrame在Pandas DataFrame中填写NA值

Question

import pandas as pd


df1 = pd.DataFrame({
                  'value1': ["a","a","a","b","b","b","c","c"],
                  'value2': [1,2,3,4,4,4,5,5],
                    'value3': [1,2,3, None , None, None, None, None],
                    'value4': [1,2,3,None , None, None, None, None],
                    'value5': [1,2,3,None , None, None, None, None]})

df2 = pd.DataFrame({
                  'value1': ["k","j","l","m","x","y"],
                  'value2': [2, 2, 1, 3, 4, 5],
                  'value3': [2, 2, 2, 3, 4, 5],
                  'value4': [3, 2, 2, 3, 4, 5],
                  'value5': [2, 1, 2, 3, 4, 5]})

df1 = 
  value1  value2  value3  value4  value5
0      a       1     1.0     1.0     1.0
1      a       2     2.0     2.0     2.0
2      a       3     3.0     3.0     3.0
3      b       4     NaN     NaN     NaN
4      b       4     NaN     NaN     NaN
5      b       4     NaN     NaN     NaN
6      c       5     NaN     NaN     NaN
7      c       5     NaN     NaN     NaN

df2 = 
  value1  value2  value3  value4  value5
0      k       2       2       3       2
1      j       2       2       2       1
2      l       1       2       2       2
3      m       3       3       3       3
4      x       4       4       4       4
5      y       5       5       5       5

我想从df2中的值填充df1中的NaN

所以df1的结果看起来像

df1 = 
  value1  value2  value3  value4  value5
0      a       1     1.0     1.0     1.0
1      a       2     2.0     2.0     2.0
2      a       3     3.0     3.0     3.0
3      b       4     2       2       1
4      b       4     2       2       2
5      b       4     3       3       3
6      c       5     4       4       4
7      c       5     5       5       5

我用下面的代码。

tmp1 = df1[df1.value1 == 'b'].iloc[:, 2:]
tmp2 = df2.iloc[1:, 2:]

tmp1 = tmp2可以更新tmp1中的值，但是当我使用以下命令时

df1[df1.value1 == 'b'].iloc[:, 2:]= tmp2

如下所示，它不会更新df1中的值。

  value1  value2  value3  value4  value5
0      a       1     1.0     1.0     1.0
1      a       2     2.0     2.0     2.0
2      a       3     3.0     3.0     3.0
3      b       4     NaN     NaN     NaN
4      b       4     NaN     NaN     NaN
5      b       4     NaN     NaN     NaN
6      c       5     NaN     NaN     NaN
7      c       5     NaN     NaN     NaN

为什么会发生，如何解决这个问题？

谢谢。

Answer 1

该行不执行您认为的操作：

tmp1 = df1[df1.value1 == 'b'].iloc[:, 2:]

方法被顺序地施加，因此df1[df1.value1 == 'b']只保留行3, 4, 5从df1 。 但这不是您想要的，您想要从满足条件的第一个实例开始更新所有行 。

而是，首先找到所需的索引。

idx = df1['value1'].eq('b').values.argmax()

然后，您需要显式分配 df2的最后n行：

df1.iloc[idx:, 2:] = df2.iloc[-(len(df1.index)-idx):, 2:].values

print(df1)

  value1  value2  value3  value4  value5
0      a       1     1.0     1.0     1.0
1      a       2     2.0     2.0     2.0
2      a       3     3.0     3.0     3.0
3      b       4     2.0     2.0     1.0
4      b       4     2.0     2.0     2.0
5      b       4     3.0     3.0     3.0
6      c       5     4.0     4.0     4.0
7      c       5     5.0     5.0     5.0

Answer 2

如果要使用索引对齐替换nan值，请使用pandas fillna

df1.fillna(df2)

如果要更新df1，请添加就位

df1.fillna(df2, inplace=True)

-

• 编辑没有对齐索引的情况：

如果目标值和替换值的索引未对齐，则可以对齐它们，以便可以使用数据框fillna方法。

要对齐索引，请获取要替换的df1中包含nans的行的索引，过滤df2以包括替换值，然后将df1中的替换索引分配为df2的索引。 然后使用fillna将值从df2传输到df1。

# in this case, find index values when df1.value1 is greater than or equal to 'b'
# (alternately could be indexes of rows containing nans)
idx = df1.index[df1.value1 >= 'b']
# get the section of df2 which will provide replacement values
# limit length to length of idx
align_df = df2[1:len(idx) + 1]
# set the index to match the nan rows from df1
align_df.index = idx
# use auto-alignment with fillna to transfer values from align_df(df2) to df1
df1.fillna(align_df)

# or can use df1.combine_first(align_df) because of the matching target and replacement indexes