问题描述

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)
@H_502_11@
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)
@H_502_11@
产量
  one  two three  que
0  10  1.2   4.2   10
1  15   70  0.03  NaN
2   8    5     0  NaN
@H_502_11@

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)
@H_502_11@
产量
  one  two three  que
0  10  1.2   4.2   10
1  15   70  0.03   70
2   8    5     0  NaN
@H_502_11@
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']]
@H_502_11@
（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'])
@H_502_11@
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)
@H_502_11@
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
@H_502_11@
解决方法

                    
以此为起点：
a = [['10','1.2','4.2'],['15','70','0.03'],['8','5','0']]
df = pd.DataFrame(a,columns=['one','two','three'])

Out[8]: 
  one  two three
0   10  1.2   4.2
1   15  70   0.03
2    8   5     0

我想if在熊猫中使用类似声明的内容。
if df['one'] >= df['two'] and df['one'] <= df['three']:
    df['que'] = df['one']

基本上，通过if语句检查每一行，然后创建新列。
文档说要使用，.all但没有示例…