Python Pandas-根据单元格值查找列值 [英] Python Pandas - Lookup a column value based on a cell value

问题描述

我正在尝试使用一个单元格的值在另一列中查找一个单元格的值.第一个单元格值指示要查找的列.

I'm trying to use the value of one cell to find the value of a cell in another column. The first cell value dictates which column to lookup.

import pandas as pd

df = pd.DataFrame({'A': ['John', 'Andrew', 'Bob', 'Fred'], 'B': [
                  'Fred', 'Simon', 'Andrew', 'Andrew'], 'source': ['A', 'B', 'A', 'B']}, )

print(df)


        A       B source
0    John    Fred      A
1  Andrew   Simon      B
2     Bob  Andrew      A
3    Fred  Andrew      B

我在输出"列中所需的输出值是对源"的查找

My required output value in the 'output' column is a lookup of the 'source'

        A       B source  output
0    John    Fred      A    John
1  Andrew   Simon      B   Simon
2     Bob  Andrew      A     Bob
3    Fred  Andrew      B  Andrew

尝试失败

df['output'] = df[df['source']]

这将导致 ValueError:错误传递了4个项目，放置的含义为1 ，因为 df ['source'] 是通过Series而不是字符串传递的.我尝试使用以下方法转换为字符串:

This results in a ValueError: Wrong number of items passed 4, placement implies 1 because the df['source'] passes in a Series, not a string. I tried converting to a string using:

df['output'] = df[df['source'].convertDTypes(convert_string=True)]

给出错误 AttributeError:系列"对象没有属性"convertDTypes" .

工作解决方案

我发现一种解决方案可能是使用以下方法来遍历行:

I found a solution might by iterating through the rows using:

for index, row in df.iterrows():
    column = df.loc[index, 'source']
    df.at[index, 'output'] = df.loc[index, column]

但是，这篇文章提示，迭代是一个坏主意.代码也不是很优雅.

However, this post suggests iterating is a bad idea. The code doesn't seem very elegant, either.

我觉得我在这里错过了一些基本的东西；这真的不应该那么难.

I feel I've missed something basic here; this really should not be that hard.

推荐答案

让我们以 numpy 的方式进行操作，因为 lookup 在以后的版本中将不再起作用

Let us do numpy way since lookup will not longer work in the future version

df['new'] = df.values[df.index,df.columns.get_indexer(df.source)]
df
Out[339]: 
        A       B source     new
0    John    Fred      A    John
1  Andrew   Simon      B   Simon
2     Bob  Andrew      A     Bob
3    Fred  Andrew      B  Andrew

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