Python for loop over two lists for all the pairs

Question

I have a dataframe with date, id - I need to pull out each date and id combination and create a new dataframe.

  date           id
2016-05-13       abc
2016-05-13       pqr
2016-05-14       abc
2016-05-14       pqr

ids = list(sorted(set(df['id'])))
Out: ['abc','pqr']

dates = list(sorted(set(df[df.id == ids[i]]['date'])))
Out: ['2016-05-13','2016-05-14']

for i in range(0,len(ids)):
    df2 = df[(df.date == dates[i]) & (df.id == id[i])]

The above code is resulting the output (df2) for relative index values only (First date, First Id & Second date, Second Id), but I need the output for all the pairs. Please let me know what to change in the loop?

piRSquared · Accepted Answer · 2016-09-16 21:20:32Z

0

Create a new dataframe with each id in columns and each date in rows. You can fill it in later.

pd.DataFrame([], set(df.date), set(df.id))

if you just want the list of combinations

pd.MultiIndex.from_product([set(df.id), set(df.date)]).tolist()

[('pqr', '2016-05-14'),
 ('pqr', '2016-05-13'),
 ('abc', '2016-05-14'),
 ('abc', '2016-05-13')]

answered Sep 16, 2016 at 21:20

piRSquared

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Comments

ewcz · Accepted Answer · 2016-09-16 20:06:42Z

-2

to get all the pairs ids vs. dates, you could use itertools as

import itertools

for iid, ddate in itertools.product(ids, dates):
    df2 = df[(df.date == ddate) & (df.id == iid)]

answered Sep 16, 2016 at 20:06

ewcz

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2 Comments

piRSquared Over a year ago

itertools is a fine answer, but this snippet doesn't even work. You just reassign some slice of df to the same variable df2 over and over again.

ewcz Over a year ago

@piRSquared from what I understood, the OP wanted to generate such slice for each pair of id/date - slice which is presumably subject to further processing...

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