将多个列值更改为二进制值 [英] Changing multiple column values to binary values

问题描述

我之前曾问过这个问题，但是我得到的答案并没有像我想的那样得出来，所以我就在这里.

I've asked this question before but the answer I got didn't quite work out as I thought it had, so that here I am.

上一个问题:我正在尝试定义一个函数，在该函数中将使用一个数据框并更改列中的值以创建多个新数据框.

I am trying to define a function where it will take a dataframe and change values in a column to create multiple new dataframes.

以df1为例，如下所示:

As an example, from df1 looking like:

  df1:

  class    colB    colC
0   1      1b      1c
1   2      2b      2c
2   3      3b      3c
3   1      4b      4c
4   2      5b      5c

我正在尝试创建多个二进制类来实现一对一分类.因此该函数将创建...

I am trying to create multiple binary classes to implement one-vs-all classification. So the function would create...

df2:
  class    colB    colC
0   1      1b      1c
1   -1      2b      2c
2   -1      3b      3c
3   1      4b      4c
4   -1      5b      5c

df3:
  class    colB    colC
0   -1      1b      1c
1   1      2b      2c
2   -1      3b      3c
3   -1      4b      4c
4   1      5b      5c

df4:
  class    colB    colC
0   -1      1b      1c
1   -1      2b      2c
2    1      3b      3c
3   -1      4b      4c
4   -1      5b      5c

，依此类推.所有唯一值都是1到120之间的增量值.

and so on. All the unique values are an incremental value ranging from 1 to 120.

以前的答案给出的问题(np.identity)是它创建了将每个单个值都设为1或-1的数据帧，而不是将相同的值归为同一类.

The problem with the previous answer give (np.identity) was that it created dataframes taking every single value as either 1 or -1 instead of categorizing identical values as the same class accordingly.

谢谢

推荐答案

使用np.where和unique的类似想法(再次重命名class列，因此它不会覆盖内置名称):

A similar idea using np.where and unique (again renaming your class column so it doesn't override a builtin name):

dfs = [
    df1.assign(class_=np.where(df1['class_'].eq(i), 1, -1)) for i in df1['class_'].unique()
]

for d in dfs:
    print(d, end='\n\n')

   class_ colB colC
0       1   1b   1c
1      -1   2b   2c
2      -1   3b   3c
3       1   4b   4c
4      -1   5b   5c

   class_ colB colC
0      -1   1b   1c
1       1   2b   2c
2      -1   3b   3c
3      -1   4b   4c
4       1   5b   5c

   class_ colB colC
0      -1   1b   1c
1      -1   2b   2c
2       1   3b   3c
3      -1   4b   4c
4      -1   5b   5c

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