Pandas / Python：替换多列中的多个值

Question

All, I have an analytical csv file with 190 columns and 902 rows. I need to recode values in several columns (18 to be exact) from it's current 1-5 Likert scaling to 0-4 Likert scaling.

I've tried using replace:

df.replace({'Job_Performance1': {1:0, 2:1, 3:2, 4:3, 5:4}}, inplace=True)

But that throws a Value Error: "Replacement not allowed with overlapping keys and values"

I can use map:

df['job_perf1'] = df.Job_Performance1.map({1:0, 2:1, 3:2, 4:3, 5:4})

But, I know there has to be a more efficient way to accomplish this since this use case is standard in statistical analysis and statistical software eg SPSS

I've reviewed multiple questions on StackOverFlow but none of them quite fit my use case. eg Pandas - replacing column values , pandas replace multiple values one column , Python pandas: replace values multiple columns matching multiple columns from another dataframe

Suggestions?

Answer 1

You can simply subtract a scalar value from your column which is in effect what you're doing here:

df['job_perf1'] = df['job_perf1'] - 1

Also as you need to do this on 18 cols, then I'd construct a list of the 18 column names and just subtract 1 from all of them at once:

df[col_list] = df[col_list] - 1

Answer 2

No need for a mapping. This can be done as a vector addition, since effectively, what you're doing, is subtracting 1 from each value. This works elegantly:

df['job_perf1'] = df['Job_Performance1'] - numpy.ones(len(df['Job_Performance1']))

Or, without numpy :

df['job_perf1'] = df['Job_Performance1'] - [1] * len(df['Job_Performance1'])