Pandas pythonic way to extract value and index from pd.Series as tuple

Question

I define a Pandas DataFrame as a = pd.DataFrame({'val':[1,2,3,4,5,6],'id':[2,1,4,3,0,5]}) , such that:

In [0]: import pandas as pd
In [1]: a = pd.DataFrame({'val':[1,2,3,4,5,6],'id':[2,1,4,3,0,5]})
In [2]: a
Out[2]: 
   id  val
0   2    1
1   1    2
2   4    3
3   3    4
4   0    5
5   5    6

Also, I perform some functions over it, like a.sort('id',inplace=True) , such that:

In [3]: a
Out[3]: 
   id  val
4   0    5
1   1    2
0   2    1
3   3    4
2   4    3
5   5    6

Note how the index is unordered.

And after, I want to extract information by doing something like a.val[a.id >=2] , which returns a pd.Series :

In [4]: a.val[a.id >=2]
Out[4]: 
0    1
3    4
2    3
5    6
Name: val, dtype: int64

If I want to extract the first value of the pd.Series I do a.val[a.id >= 2].iloc[0] , and for the index a.val[id >= 2].index[0] , and put them as a tuple.

Is there a way to extract both values as a tuple in one line of code? Is there a pythonic way?

Answer 1

Disclaimer: I'm not a real pandas user (though I like to read pandas questions around here), so there could be a better way...

That said, if pandas doesn't provide a better way, I'd just write a function:

def extract_first(series):
    return series.iloc[0], series.index[0]

extract_first(a.val[a.id >= 2])

I would expect that this would allow you to execute the comparatively expensive indexing only once which could be helpful for larger series.

---

Looking at the docs , it appears that you can also use iteritems :

next(a.val[a.id >= 2].iteritems())