根据每行中的值获取列标题

Question

I have a pandas dataframe, something like below (just an illustration):

import datetime
todays_date = datetime.datetime.now().date()   
index = pd.date_range(todays_date-datetime.timedelta(10), periods=2, freq='D')
columnheader=['US', 'Canada', 'UK', 'Japan']
data=np.array([[3,4,2,1],[1,4,3,2]])
df = pd.DataFrame(data, index=index, columns=columnheader)

Which results in:

            US  Canada  UK  Japan
2015-07-26   3       4   2      1
2015-07-27   1       4   3      2

I need to find the column header whose value is 1 and 2 for each row.

so I should get

['Japan', 'UK']
['US', 'Japan']

Answer 1

You can do the following, this tests each row for membership of 1,2 using isin and if so this generates a boolean series, you can use this to index into the columns by calling apply again, we convert this to a list because the dimensions won't align if you don't do this:

In [191]:
df.apply(lambda x: x.isin([1,2]), axis=1).apply(lambda x: list(df.columns[x]), axis=1)

Out[191]:
2015-07-26    [UK, Japan]
2015-07-27    [US, Japan]
Freq: D, dtype: object

output from inner apply :

In [192]:
df.apply(lambda x: x.isin([1,2]), axis=1)

Out[192]:
               US Canada     UK Japan
2015-07-26  False  False   True  True
2015-07-27   True  False  False  True

EDIT

If you want to maintain order then you can define a func to test each value and return this as a series:

In [209]:
filter_vals=[1,2]
def func(x):
    l=[]
    for val in filter_vals:
        for col in df:
            if x[col] == val:
                l.append(col)
​
    return pd.Series(l)
df.apply(func, axis=1)

Out[209]:
                0      1
2015-07-26  Japan     UK
2015-07-27     US  Japan