您可以使用np.where。如果cond是布尔数组，A并且B是数组，则
C = np.where(cond, A, B)
将C定义为等于A哪里cond为True，B哪里cond为False。
import numpy as np
import pandas as pd
a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])
df['que'] = np.where((df['one'] >= df['two']) & (df['one'] <= df['three'])
, df['one'], np.nan)
产量
one two three que
0 10 1.2 4.2 10
1 15 70 0.03 NaN
2 8 5 0 NaN
如果您有多个条件，则可以使用np.select代替。例如，如果你想df['que']等于df['two']时df['one'] < df['two']，则
conditions = [
(df['one'] >= df['two']) & (df['one'] <= df['three']),
df['one'] < df['two']]
choices = [df['one'], df['two']]
df['que'] = np.select(conditions, choices, default=np.nan)
产量
one two three que
0 10 1.2 4.2 10
1 15 70 0.03 70
2 8 5 0 NaN
如果我们可以假设df['one'] >= df['two']whendf['one'] < df['two']为False，那么条件和选择可以简化为
conditions = [
df['one'] < df['two'],
df['one'] <= df['three']]
choices = [df['two'], df['one']]
（如果包含df['one']或df['two']包含NaN，则该假设可能不正确。）
注意
a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])
用字符串值定义一个DataFrame。由于它们看起来是数字，因此最好将这些字符串转换为浮点数：
df2 = df.astype(float)
但是，这会改变结果，因为字符串会逐个字符地进行比较，而浮点数会进行数字比较。
In [61]: '10' <= '4.2'
Out[61]: True
In [62]: 10 <= 4.2
Out[62]: False
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