如何从dataframe[duplicate]获取3个布尔值

Question

假设我有整数0到100的数据帧。我想把这些值分为3部分，低、中、高，低小于33，高大于66，中介于33和66之间。所以我用

df['low'] = df['int'] <= 33
df['mid'] = 33  < df['int'] < 66
df['high'] = df['int'] >= 66 

我得到了一个错误

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_12080/1299746928.py in <module>
      1 df['low'] = df['int'] <= 33
----> 2 df['mid'] = 33  < df['int'] < 66
      3 df['high'] = df['int'] >= 66

c:\program files\python37\lib\site-packages\pandas\core\generic.py in __nonzero__(self)
   1536     def __nonzero__(self):
   1537         raise ValueError(
-> 1538             f"The truth value of a {type(self).__name__} is ambiguous. "
   1539             "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
   1540         )

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

我已经尝试过if-else语句和和以及其他操作符。低和高起作用，但中间不起作用。

请问我能知道附近的路吗？

Answer 1

当你收到评论时，你可以在（33,66）

df['mid'] = df['int'].between(33, 66)

但你也可以把它写成

df['mid'] = (33 < df['int']) & (df['int'] < 66)

您还可以使用（不低）和（不高）

df['mid'] = ~df['low'] & ~df['high']

最小工作示例

import pandas as pd

data = {
    'int': range(0, 100, 10), 
}

df = pd.DataFrame(data)
 
df['low'] = df['int'] <= 33
df['high'] = df['int'] >= 66

df['mid1'] = (33 < df['int']) & (df['int'] < 66)

df['mid2'] = ~df['low'] & ~df['high']

df['mid-between'] = df['int'].between(33, 66)

print(df)

结果:

   int    low   high   mid1   mid2  mid-between
0    0   True  False  False  False        False
1   10   True  False  False  False        False
2   20   True  False  False  False        False
3   30   True  False  False  False        False
4   40  False  False   True   True         True
5   50  False  False   True   True         True
6   60  False  False   True   True         True
7   70  False   True  False  False        False
8   80  False   True  False  False        False
9   90  False   True  False  False        False

顺便说一句：

如果你使用pandas.cut（）

bins =  pd.cut(df['int'], [-1, 33, 66, 100], labels=['low', 'mid', 'high']) 

print( bins )

然后你就可以

0     low
1     low
2     low
3     low
4     mid
5     mid
6     mid
7    high
8    high
9    high

import pandas as pd

data = {
    'int': range(0, 100, 10), 
}

df = pd.DataFrame(data)

min_value = min(df['int']) - 1
max_value = max(df['int']) + 1

bins = pd.cut(df['int'], [min_value, 33, 66, max_value], labels=['low', 'mid', 'high'])

print(bins)

但它更像是开闭范围（在数学中（a，b））：

min < x <= 33
 33 < x <= 66
 66 < x <= max