Pandas/Python:根据另一列中的值设置一列的值 [英] Pandas/Python: Set value of one column based on value in another column

问题描述

我需要基于Pandas数据框中的另一列的值来设置一列的值.这是逻辑:

I need to set the value of one column based on the value of another in a Pandas dataframe. This is the logic:

if df['c1'] == 'Value':
    df['c2'] = 10
else:
    df['c2'] = df['c3']

我无法做到这一点，这就是简单地创建一个具有新值的列(或更改现有列的值:任一个对我有效).

I am unable to get this to do what I want, which is to simply create a column with new values (or change the value of an existing column: either one works for me).

如果我尝试运行上面的代码，或者将其编写为函数并使用apply方法，则会得到以下信息:

If I try to run the code above or if I write it as a function and use the apply method, I get the following:

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

推荐答案

一种方法是将索引与.loc一起使用.

one way to do this would be to use indexing with .loc.

示例

在没有示例数据框的情况下，我将在此处进行补充:

In the absence of an example dataframe, I'll make one up here:

import numpy as np
import pandas as pd

df = pd.DataFrame({'c1': list('abcdefg')})
df.loc[5, 'c1'] = 'Value'

>>> df
      c1
0      a
1      b
2      c
3      d
4      e
5  Value
6      g

假设您要创建一个新列 c2，等效于c1，但c1是Value的情况除外，在这种情况下，您希望将其分配给10:

Assuming you wanted to create a new column c2, equivalent to c1 except where c1 is Value, in which case, you would like to assign it to 10:

首先，您可以创建新列c2，并使用以下两行之一将它们设置为与c1等价的列(它们本质上执行相同的操作):

First, you could create a new column c2, and set it to equivalent as c1, using one of the following two lines (they essentially do the same thing):

df = df.assign(c2 = df['c1'])
# OR:
df['c2'] = df['c1']

然后，使用.loc查找c1等于'Value'的所有索引，并在c2中的这些索引处分配所需的值:

Then, find all the indices where c1 is equal to 'Value' using .loc, and assign your desired value in c2 at those indices:

df.loc[df['c1'] == 'Value', 'c2'] = 10

最终，您将得到以下结果:

And you end up with this:

>>> df
      c1  c2
0      a   a
1      b   b
2      c   c
3      d   d
4      e   e
5  Value  10
6      g   g

如果按照您在问题中所建议的那样，有时您可能只是想替换现有列中的值，而不是创建新列，然后跳过该列的创建，并执行以下操作:

If, as you suggested in your question, you would perhaps sometimes just want to replace the values in the column you already have, rather than create a new column, then just skip the column creation, and do the following:

df['c1'].loc[df['c1'] == 'Value'] = 10
# or:
df.loc[df['c1'] == 'Value', 'c1'] = 10

给你

>>> df
      c1
0      a
1      b
2      c
3      d
4      e
5     10
6      g

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