具有 Nan 支持的 Pandas Lambda 函数
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Pandas Lambda Function with Nan Support
提问by Tyler Russell
I am trying to write a lambda function in Pandas that checks to see if Col1 is a Nan and if so, uses another column's data. I have having trouble getting code (below) to compile/execute correctly.
我正在尝试在 Pandas 中编写一个 lambda 函数来检查 Col1 是否为 Nan，如果是，则使用另一列的数据。我在获取代码（如下）以正确编译/执行时遇到问题。
import pandas as pd
import numpy as np
df=pd.DataFrame({ 'Col1' : [1,2,3,np.NaN], 'Col2': [7, 8, 9, 10]})
df2=df.apply(lambda x: x['Col2'] if x['Col1'].isnull() else x['Col1'], axis=1)
Does anyone have any good idea on how to write a solution like this with a lambda function or have I exceeded the abilities of lambda? If not, do you have another solution? Thanks.
有没有人对如何使用 lambda 函数编写这样的解决方案有什么好主意，或者我是否超出了 lambda 的能力？如果没有，您有其他解决方案吗？谢谢。
回答by jezrael
You need pandas.isnullfor check if scalar is NaN:
您需要pandas.isnull检查标量是否为NaN：
df = pd.DataFrame({ 'Col1' : [1,2,3,np.NaN],
'Col2' : [8,9,7,10]})
df2 = df.apply(lambda x: x['Col2'] if pd.isnull(x['Col1']) else x['Col1'], axis=1)
print (df)
Col1 Col2
0 1.0 8
1 2.0 9
2 3.0 7
3 NaN 10
print (df2)
0 1.0
1 2.0
2 3.0
3 10.0
dtype: float64
But better is use Series.combine_first:
但更好的是使用Series.combine_first：
df['Col1'] = df['Col1'].combine_first(df['Col2'])
print (df)
Col1 Col2
0 1.0 8
1 2.0 9
2 3.0 7
3 10.0 10
Another solution with Series.update:
另一个解决方案Series.update：
df['Col1'].update(df['Col2'])
print (df)
Col1 Col2
0 8.0 8
1 9.0 9
2 7.0 7
3 10.0 10
回答by Gerges
Assuming that you do have a second column, that is:
假设您确实有第二列，即：
df = pd.DataFrame({ 'Col1' : [1,2,3,np.NaN], 'Col2': [1,2,3,4]})
df = pd.DataFrame({ 'Col1' : [1,2,3,np.NaN], 'Col2': [1,2,3,4]})
The correct solution to this problem would be:
这个问题的正确解决方案是：
df['Col1'].fillna(df['Col2'], inplace=True)
回答by Allen
You need to use np.nan()
你需要使用 np.nan()
#import numpy as np
df2=df.apply(lambda x: 2 if np.isnan(x['Col1']) else 1, axis=1)
df2
Out[1307]:
0 1
1 1
2 1
3 2
dtype: int64
回答by jiahe
Within pandas 0.24.2, I use
在Pandas 0.24.2 中，我使用
df.apply(lambda x: x['col_name'] if x[col1] is np.nan else expressions_another, axis=1)
because pd.isnull() doesn't work.
因为 pd.isnull() 不起作用。
in my work,I found the following phenomenon,
在我的工作中，我发现了以下现象，
No running results:
没有运行结果：
df['prop'] = df.apply(lambda x: (x['buynumpday'] / x['cnumpday']) if pd.isnull(x['cnumpday']) else np.nan, axis=1)
Results exist:
结果存在：
df['prop'] = df.apply(lambda x: (x['buynumpday'] / x['cnumpday']) if x['cnumpday'] is not np.nan else np.nan, axis=1)
