如何为数据框中的每一行分配一个值到不同的列？

Question

I have a dataframe dat that looks like this:

        p1    p2    type  replace
1       0     1     1     1
2       1     0     1     1
3       0     0     2     1
...

I want do something like dat['p + str(type)'] = replace to get:

        p1    p2    type  replace 
1       1     1     1     1
2       1     0     1     1
3       0     1     2     1
...

How can I do this? Of course I can't assign in a loop using something like iterrows...

Answer 1

Maybe there is some one liner to do this, but if performance is not really an issue, you can easily do this with a simple for loop:

In [134]: df
Out[134]: 
   p1  p2  type  replace
0   0   1     1        1
1   1   0     1        1
2   0   0     2        1

In [135]: for i in df.index:
     ...:     df.loc[i, 'p'+str(df.loc[i, 'type'])] = df.loc[i, 'replace']

In [136]: df
Out[136]: 
   p1  p2  type  replace
0   1   1     1        1
1   1   0     1        1
2   0   1     2        1

If you have much more rows than columns, this will be much faster and is actually easier (and if necessary you can loop over 1, 2, ..):

df["p1"][df["type"]==1] = df["replace"][df["type"]==1]
df["p2"][df["type"]==2] = df["replace"][df["type"]==2]

Answer 2

In [47]: df['p1'].where(~(df['type'] == 1), df['replace'], inplace=True)

In [48]: df['p2'].where(~(df['type'] == 2), df['replace'], inplace=True)

In [49]: df
Out[49]: 
   p1  p2  type  replace
1   1   1     1        1
2   1   0     1        1
3   0   1     2        1

Answer 3

Just for completeness, I ended up doing the following, which may or may not be the same as what Dan Allan suggested:

for i in range(2):
    df.loc[df['type'] == i + 1, 'p' + str(i + 1)] = df.loc[df['type'] == i + 1, 'replace']

I have a much larger problem than the example I gave (with something like 30 types and thousands of rows in the dataframe), and this solution seems very fast. Thanks to all for your help in thinking about this problem!