Pandas DataFrame乘以列并创建新列 [英] Pandas DataFrame multiplying columns and creating new columns

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问题描述

我有一些包含11列的数据.我需要将第1-10列乘以第11列,然后用这些结果创建10个新列.为此,我正在使用pandas DataFrame.

I have some data that has 11 columns. I need to multiply columns 1-10 by column 11 and then create 10 new columns with those results. To do this I am using pandas DataFrame.

现在,我了解了如何使用这样的代码分别为每一列执行此操作

Now I understand how to do this for each column individually with a code like this

df['newcolumn1'] = df['column1']*df['column11']
df['newcolumn2'] = df['column2']*df['column11']
df['newcolumn3'] = df['column3']*df['column11']

我假设我可以设置一个函数和一个循环来遍历各列并创建新的列.无论如何,我可以通过引用列索引号而不是列名来做到这一点.

I'm assuming I can set up a function and a loop to iterate through the columns and create the new columns. Is there anyway I can do this by referencing the column index number instead of the column names.

推荐答案

您可以使用

Instead of multiplying individually or explicitly looping, you could use multiply to generate a DataFrame of your new columns, then pd.concat to join things together. Doing so by column number as you would like to may look like

pd.concat([df, 
           (df.iloc[:, :10].multiply(df.iloc[:, -1], axis='rows')
                           .add_prefix('new_'))], 
           axis=1)

最小示例

>>> df
   column1  column2  column3  column4
0        0        1        2        3
1        4        5        6        7
2        8        9       10       11
3       12       13       14       15

>>> pd.concat([df, 
                (df.iloc[:, :3].multiply(df.iloc[:, -1], axis='rows')
                               .add_prefix('new_')], axis=1))], 
                axis=1)

   column1  column2  column3  column4  new_column1  new_column2  new_column3
0        0        1        2        3            0            3            6
1        4        5        6        7           28           35           42
2        8        9       10       11           88           99          110
3       12       13       14       15          180          195          210

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