如何使用布尔掩码将pandas数据帧中的“任意字符串”替换为nan？

Question

我有一个227x4的数据帧与国家名称和数值清洁（wrangle？）。

以下是数据帧的抽象：

import pandas as pd
import random
import string
import numpy as np
pdn = pd.DataFrame(["".join([random.choice(string.ascii_letters) for i in range(3)]) for j in range (6)], columns =['Country Name'])
measures = pd.DataFrame(np.random.random_integers(10,size=(6,2)), columns=['Measure1','Measure2'])
df = pdn.merge(measures, how= 'inner', left_index=True, right_index =True)

df.iloc[4,1] = 'str'
df.iloc[1,2] = 'stuff'
print(df)

  Country Name Measure1 Measure2
0          tua        6        3
1          MDK        3    stuff
2          RJU        7        2
3          WyB        7        8
4          Nnr      str        3
5          rVN        7        4

如何将所有列中的字符串值替换为np.nan，而不涉及国家名称？

我尝试使用布尔掩码：

mask = df.loc[:,measures.columns].applymap(lambda x: isinstance(x, (int, float))).values
print(mask)

[[ True  True]
 [ True False]
 [ True  True]
 [ True  True]
 [False  True]
 [ True  True]]

# I thought the following would replace by default false with np.nan in place, but it didn't
df.loc[:,measures.columns].where(mask, inplace=True)
print(df)

  Country Name Measure1 Measure2
0          tua        6        3
1          MDK        3    stuff
2          RJU        7        2
3          WyB        7        8
4          Nnr      str        3
5          rVN        7        4


# this give a good output, unfortunately it's missing the country names
print(df.loc[:,measures.columns].where(mask))

  Measure1 Measure2
0        6        3
1        3      NaN
2        7        2
3        7        8
4      NaN        3
5        7        4

我看了几个与我有关的问题（[1]、[2]、[3]、[4]、[5]、[6]、[7]、[8]），但找不到一个回答我所关心的问题。

Answer 1

使用带有错误的数字强制，即

cols = ['Measure1','Measure2']
df[cols] = df[cols].apply(pd.to_numeric,errors='coerce')

 Country Name  Measure1  Measure2
0          PuB       7.0       6.0
1          JHq       2.0       NaN
2          opE       4.0       3.0
3          pxl       3.0       6.0
4          ouP       NaN       4.0
5          qZR       4.0       6.0

Answer 2

cols = ['Measure1','Measure2']
df[cols] = df[cols].applymap(lambda x: x if not isinstance(x, str) else np.nan)

或

df[cols] = df[cols].applymap(lambda x: np.nan if isinstance(x, str) else x)

结果:

In [22]: df
Out[22]:
  Country Name  Measure1  Measure2
0          nBl      10.0       9.0
1          Ayp       8.0       NaN
2          diz       4.0       1.0
3          aad       7.0       3.0
4          JYI       NaN      10.0
5          BJO       9.0       8.0

Answer 3

仅指定感兴趣的列：

cols = ['Measure1','Measure2']
mask = df[cols].applymap(lambda x: isinstance(x, (int, float)))

df[cols] = df[cols].where(mask)
print (df)
  Country Name Measure1 Measure2
0          uFv        7        8
1          vCr        5      NaN
2          qPp        2        6
3          QIC       10       10
4          Suy      NaN        8
5          eFS        6        4

一个元问题，我在这里花3个多小时来制定一个问题（包括研究）是正常的吗？

在我看来，是的，创造好的问题真的很难。