python - Pandas 数据框中的多个 if else 条件并派生多列

Question

我有一个如下所示的数据框。

import pandas as pd
import numpy as np
raw_data = {'student':['A','B','C','D','E'],
        'score': [100, 96, 80, 105,156], 
    'height': [7, 4,9,5,3],
    'trigger1' : [84,95,15,78,16],
    'trigger2' : [99,110,30,93,31],
    'trigger3' : [114,125,45,108,46]}

df2 = pd.DataFrame(raw_data, columns = ['student','score', 'height','trigger1','trigger2','trigger3'])

print(df2)

我需要根据多个条件导出 Flag 列。

我需要将分数和高度列与触发器 1 -3 列进行比较。

标记栏:

1. 如果 Score 大于等于触发 1 且高度小于 8 则红色 --

2. 如果 Score 大于等于触发 2 且高度小于 8 则黄色 --

3. 如果 Score 大于等于 trigger 3 并且 height 小于 8 那么 Orange --

4. 如果高度大于8则留空

如何在 pandas dataframe 中编写 if else 条件并派生列？

预期输出

  student  score  height  trigger1  trigger2  trigger3    Flag
0       A    100       7        84        99       114  Yellow
1       B     96       4        95       110       125     Red
2       C     80       9        15        30        45     NaN
3       D    105       5        78        93       108  Yellow
4       E    156       3        16        31        46  Orange

对于我原来问题中的其他列 Text1，我已经尝试过这个，但是整数列在使用 astype(str) 任何其他方法连接时不转换字符串？

def text_df(df):

    if (df['trigger1'] <= df['score'] < df['trigger2']) and (df['height'] < 8):
        return df['student'] + " score " + df['score'].astype(str) + " greater than " + df['trigger1'].astype(str) + " and less than height 5"
    elif (df['trigger2'] <= df['score'] < df['trigger3']) and (df['height'] < 8):
        return df['student'] + " score " + df['score'].astype(str) + " greater than " + df['trigger2'].astype(str) + " and less than height 5"
    elif (df['trigger3'] <= df['score']) and (df['height'] < 8):
        return df['student'] + " score " + df['score'].astype(str) + " greater than " + df['trigger3'].astype(str) + " and less than height 5"
    elif (df['height'] > 8):
        return np.nan

Answer 1

您需要使用上限和下限进行链式比较

def flag_df(df):
    
    if (df['trigger1'] <= df['score'] < df['trigger2']) and (df['height'] < 8):
        return 'Red'
    elif (df['trigger2'] <= df['score'] < df['trigger3']) and (df['height'] < 8):
        return 'Yellow'
    elif (df['trigger3'] <= df['score']) and (df['height'] < 8):
        return 'Orange'
    elif (df['height'] > 8):
        return np.nan
    
df2['Flag'] = df2.apply(flag_df, axis = 1)

    student score   height  trigger1    trigger2    trigger3    Flag
0   A       100     7       84          99          114         Yellow
1   B       96      4       95          110         125         Red
2   C       80      9       15          30          45          NaN
3   D       105     5       78          93          108         Yellow
4   E       156     3       16          31          46          Orange

注意:您可以使用非常嵌套的 np.where 来执行此操作，但我更喜欢为多个 if-else 应用一个函数

编辑:回答@Cecilia 的问题

1. 返回的对象不是字符串而是一些计算，比如第一个条件，我们要返回df['height']*2

不确定您尝试了什么，但您可以使用返回派生值而不是字符串

def flag_df(df):

    if (df['trigger1'] <= df['score'] < df['trigger2']) and (df['height'] < 8):
        return df['height']*2
    elif (df['trigger2'] <= df['score'] < df['trigger3']) and (df['height'] < 8):
        return df['height']*3
    elif (df['trigger3'] <= df['score']) and (df['height'] < 8):
        return df['height']*4
    elif (df['height'] > 8):
        return np.nan

1. 如果 osome 列中有“NaN”值并且我想使用 df['xxx'] is None 作为条件怎么办，代码似乎无法正常工作

再次不确定您尝试了什么代码，但使用 pandas isnull 可以解决问题

def flag_df(df):

    if pd.isnull(df['height']):
        return df['height']
    elif (df['trigger1'] <= df['score'] < df['trigger2']) and (df['height'] < 8):
        return df['height']*2
    elif (df['trigger2'] <= df['score'] < df['trigger3']) and (df['height'] < 8):
        return df['height']*3
    elif (df['trigger3'] <= df['score']) and (df['height'] < 8):
        return df['height']*4
    elif (df['height'] > 8):
        return np.nan