iterrows pandas get next rows value

Question

I have a df in pandas

import pandas as pd df = pd.DataFrame(['AA', 'BB', 'CC'], columns = ['value']) 

I want to iterate over rows in df. For each row i want rows value and next rows value Something like(it does not work):

for i, row in df.iterrows():      print row['value']      i1, row1 = next(df.iterrows())      print row1['value'] 

As a result I want

'AA' 'BB' 'BB' 'CC' 'CC' *Wrong index error here   

At this point i have mess way to solve this

for i in range(0, df.shape[0])    print df.irow(i)['value']    print df.irow(i+1)['value'] 

Is there more efficient way to solve this issue?

Answer 1

Firstly, your "messy way" is ok, there's nothing wrong with using indices into the dataframe, and this will not be too slow. iterrows() itself isn't terribly fast.

A version of your first idea that would work would be:

row_iterator = df.iterrows() _, last = row_iterator.next()  # take first item from row_iterator for i, row in row_iterator:     print(row['value'])     print(last['value'])     last = row 

The second method could do something similar, to save one index into the dataframe:

last = df.irow(0) for i in range(1, df.shape[0]):     print(last)     print(df.irow(i))     last = df.irow(i) 

When speed is critical you can always try both and time the code.

Answer 2

There is a pairwise() function example in the itertools document:

from itertools import tee, izip def pairwise(iterable):     "s -> (s0,s1), (s1,s2), (s2, s3), ..."     a, b = tee(iterable)     next(b, None)     return izip(a, b)  import pandas as pd df = pd.DataFrame(['AA', 'BB', 'CC'], columns = ['value'])  for (i1, row1), (i2, row2) in pairwise(df.iterrows()):     print i1, i2, row1["value"], row2["value"] 

Here is the output:

0 1 AA BB 1 2 BB CC 

But, I think iter rows in a DataFrame is slow, if you can explain what's the problem you want to solve, maybe I can suggest some better method.