How to form tuple column from two columns in Pandas

Question

I've got a Pandas DataFrame and I want to combine the 'lat' and 'long' columns to form a tuple.

<class 'pandas.core.frame.DataFrame'>
Int64Index: 205482 entries, 0 to 209018
Data columns:
Month           205482  non-null values
Reported by     205482  non-null values
Falls within    205482  non-null values
Easting         205482  non-null values
Northing        205482  non-null values
Location        205482  non-null values
Crime type      205482  non-null values
long            205482  non-null values
lat             205482  non-null values
dtypes: float64(4), object(5)

The code I tried to use was:

def merge_two_cols(series): 
    return (series['lat'], series['long'])

sample['lat_long'] = sample.apply(merge_two_cols, axis=1)

However, this returned the following error:

---------------------------------------------------------------------------
 AssertionError                            Traceback (most recent call last)
<ipython-input-261-e752e52a96e6> in <module>()
      2     return (series['lat'], series['long'])
      3 
----> 4 sample['lat_long'] = sample.apply(merge_two_cols, axis=1)
      5

...

AssertionError: Block shape incompatible with manager 

How can I solve this problem?

Solution 1

Get comfortable with zip. It comes in handy when dealing with column data.

df['new_col'] = list(zip(df.lat, df.long))

It's less complicated and faster than using apply or map. Something like np.dstack is twice as fast as zip, but wouldn't give you tuples.

Solution 2

In [10]: df
Out[10]:
          A         B       lat      long
0  1.428987  0.614405  0.484370 -0.628298
1 -0.485747  0.275096  0.497116  1.047605
2  0.822527  0.340689  2.120676 -2.436831
3  0.384719 -0.042070  1.426703 -0.634355
4 -0.937442  2.520756 -1.662615 -1.377490
5 -0.154816  0.617671 -0.090484 -0.191906
6 -0.705177 -1.086138 -0.629708  1.332853
7  0.637496 -0.643773 -0.492668 -0.777344
8  1.109497 -0.610165  0.260325  2.533383
9 -1.224584  0.117668  1.304369 -0.152561

In [11]: df['lat_long'] = df[['lat', 'long']].apply(tuple, axis=1)

In [12]: df
Out[12]:
          A         B       lat      long                             lat_long
0  1.428987  0.614405  0.484370 -0.628298      (0.484370195967, -0.6282975278)
1 -0.485747  0.275096  0.497116  1.047605      (0.497115615839, 1.04760475074)
2  0.822527  0.340689  2.120676 -2.436831      (2.12067574274, -2.43683074367)
3  0.384719 -0.042070  1.426703 -0.634355      (1.42670326172, -0.63435462504)
4 -0.937442  2.520756 -1.662615 -1.377490     (-1.66261469102, -1.37749004179)
5 -0.154816  0.617671 -0.090484 -0.191906  (-0.0904840623396, -0.191905582481)
6 -0.705177 -1.086138 -0.629708  1.332853     (-0.629707821728, 1.33285348929)
7  0.637496 -0.643773 -0.492668 -0.777344   (-0.492667604075, -0.777344111021)
8  1.109497 -0.610165  0.260325  2.533383        (0.26032456699, 2.5333825651)
9 -1.224584  0.117668  1.304369 -0.152561     (1.30436900612, -0.152560909725)

Solution 3

Pandas has the itertuples method to do exactly this:

list(df[['lat', 'long']].itertuples(index=False, name=None))

Solution 4

You should try using pd.to_records(index=False):

import pandas as pd
df = pd.DataFrame({'language': ['en', 'ar', 'es'], 'greeting': ['Hi', 'اهلا', 'Hola']})
df

   language  greeting
0       en    Hi
1       ar    اهلا
2       es   Hola

df['list_of_tuples'] = list(df[['language', 'greeting']].to_records(index=False))
df['list_of_tuples']

0    [en, Hi]
1    [ar, اهلا]
2    [es, Hola]

enjoy!

Solution 5

I'd like to add df.values.tolist(). (as long as you don't mind to get a column of lists rather than tuples)

import pandas as pd
import numpy as np

size = int(1e+07)
df = pd.DataFrame({'a': np.random.rand(size), 'b': np.random.rand(size)}) 

%timeit df.values.tolist()
1.47 s ± 38.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit list(zip(df.a,df.b))
1.92 s ± 131 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

View more solutions

Comments

[elksie5000]

That's brilliant. Thank you. Clearly need to get my head around lambda functions.

[Wouter Overmeire]

Did this work on your data? If so, can you share your pandas version and the data? I wonder why your code did not work, it should.

[elksie5000]

The version is 0.10.1_20130131. Excuse my ignorance, but what's the best way of uploading a section of the data for you? (Still a relative newbie).

[Wouter Overmeire]

I failed to reproduce on 0.10.1. Best way of uploading? You can either create code that generates a frame holding random data, that has the same issue and share that code or pickle the frame above (sample) and transfer it via a free big file transfer service. How to pickle (in two lines, without ","): import pickle, with open('sample.pickle', 'w') as file: pickle.dump(sample, file)

[Balzer82]

I need exactly this, but in the opposite direction. I have a column with the lat_long tuple and need two columns with lat and long. How to unpack the tuple?

[Wouter Overmeire]

There are several ways, see e.g stackoverflow.com/questions/22799300/…

[imrek]

I have df[["year", "month", "day"]].apply(tuple, axis=1) where "year", "month", "day" are just integers, and this fails to do anything. EDIT: This works for floats only, what a ***y language.

[paulwasit]

in python3, you have to use list. This should work: df['new_col'] = list(zip(df.lat, df.long))

[Dale]

@paulwasit ah yes, my love hate relationship with python 3's lazy behavior. thanks.

[Pengju Zhao]

This method list(zip(df.lat, df.long)) in 124ms is much more efficient than df[['lat', 'long']].apply(tuple, axis=1) in 14.2 s for 900k rows. The ratio is more than 100.

[seeiespi]

I am trying to use this with a longer list of columns df['new_col'] = list(zip(df[cols_to_keep])) but keep getting an error: Length of values does not match length of index any advice?

[rishi jain]

I have upvoted this as I need to zip 10 columns and don't want to give dataframe name 10 times. Just want to give Column names.

[Peter Hansen]

zip(df[cols_to_keep]) will iterate over the DataFrame, creating a list of columns instead of a list of Series. you need zip( [df[c] for c in cols_to_keep])

[ChaimG]

When you have more than just these two columns: %timeit df[['a', 'b']].values.tolist(). It's still much faster.

[jedge]

@PeterHansen's answer helped me but think it may have been missing an * to unpack the list first - i.e. df['new_col'] = list(zip(*[df[c] for c in cols_to_keep])

[Zizzipupp]

This one fails for me with: TypeError: only integer scalar arrays can be converted to a scalar index.

[ThatNewGuy]

It's faster to create it, but any operations on that column will be faster in tuple form. For example, try calling .value_counts() on a column of lists vs a column of tuples.