Get Row Position instead of Row Index from iterrows() in Pandas

Question

I'm new to stackoverflow and I have research but have not find a satisfying answer.

I understand that I can get a row index by using df.iterrows() to iterate through a df. But what if I want to get a row position instead of row idx. What method can I use?

Example code that I'm working on is below:

df = pd.DataFrame({'month': ['Jan', 'Feb', 'March', 'April'],
               'year': [2012, 2014, 2013, 2014],
               'sale':[55, 40, 84, 31]})

df = df.set_index('month')

for idx, value in df.iterrows():
    print(idx)

How can I get an output of:

0
1
2
3

Thanks!

Answer 1

If you need row number instead of index, you should:

1. Use enumerate for a counter within a loop.
2. Don't extract the index, see options below.

Option 1

In most situations, for performance reasons you should try and use df.itertuples instead of df.iterrows. You can specify index=False so that the first element is not the index.

for idx, row in enumerate(df.itertuples(index=False)):
    # do something

df.itertuples returns a namedtuple for each row.

Option 2

Use df.iterrows. This is more cumbersome, as you need to separate out an unused variable. In addition, this is inefficient vs itertuples.

for idx, (_, row) in enumerate(df.iterrows()):
    # do something