How can I split a column of tuples in a Pandas dataframe?

Question

I have a Pandas dataframe (this is only a little piece)

>>> d1
   y norm test  y norm train  len(y_train)  len(y_test)  \
0    64.904368    116.151232          1645          549
1    70.852681    112.639876          1645          549

                                    SVR RBF  \
0   (35.652207342877873, 22.95533537448393)
1  (39.563683797747622, 27.382483096332511)

                                        LCV  \
0  (19.365430594452338, 13.880062435173587)
1  (19.099614489458364, 14.018867136617146)

                                   RIDGE CV  \
0  (4.2907610988480362, 12.416745648065584)
1    (4.18864306788194, 12.980833914392477)

                                         RF  \
0   (9.9484841581029428, 16.46902345373697)
1  (10.139848213735391, 16.282141345406522)

                                           GB  \
0  (0.012816232716538605, 15.950164822266007)
1  (0.012814519804493328, 15.305745202851712)

                                             ET DATA
0  (0.00034337162272515505, 16.284800366214057)  j2m
1  (0.00024811554516431878, 15.556506191784194)  j2m
>>>

I want to split all the columns that contain tuples. For example, I want to replace the column LCV with the columns LCV-a and LCV-b.

How can I do that?

Solution 1

You can do this by doing pd.DataFrame(col.tolist()) on that column:

In [2]: df = pd.DataFrame({'a':[1,2], 'b':[(1,2), (3,4)]})

In [3]: df
Out[3]:
   a       b
0  1  (1, 2)
1  2  (3, 4)

In [4]: df['b'].tolist()
Out[4]: [(1, 2), (3, 4)]

In [5]: pd.DataFrame(df['b'].tolist(), index=df.index)
Out[5]:
   0  1
0  1  2
1  3  4

In [6]: df[['b1', 'b2']] = pd.DataFrame(df['b'].tolist(), index=df.index)

In [7]: df
Out[7]:
   a       b  b1  b2
0  1  (1, 2)   1   2
1  2  (3, 4)   3   4

Note: in an earlier version, this answer recommended to use df['b'].apply(pd.Series) instead of pd.DataFrame(df['b'].tolist(), index=df.index). That works as well (because it makes a Series of each tuple, which is then seen as a row of a dataframe), but it is slower / uses more memory than the tolist version, as noted by the other answers here (thanks to denfromufa).

Solution 2

The str accessor that is available to pandas.Series objects of dtype == object is actually an iterable.

Assume a pandas.DataFrame df:

df = pd.DataFrame(dict(col=[*zip('abcdefghij', range(10, 101, 10))]))

df

        col
0   (a, 10)
1   (b, 20)
2   (c, 30)
3   (d, 40)
4   (e, 50)
5   (f, 60)
6   (g, 70)
7   (h, 80)
8   (i, 90)
9  (j, 100)

We can test if it is an iterable:

from collections import Iterable

isinstance(df.col.str, Iterable)

True

We can then assign from it like we do other iterables:

var0, var1 = 'xy'
print(var0, var1)

x y

Simplest solution

So in one line we can assign both columns:

df['a'], df['b'] = df.col.str

df

        col  a    b
0   (a, 10)  a   10
1   (b, 20)  b   20
2   (c, 30)  c   30
3   (d, 40)  d   40
4   (e, 50)  e   50
5   (f, 60)  f   60
6   (g, 70)  g   70
7   (h, 80)  h   80
8   (i, 90)  i   90
9  (j, 100)  j  100

---

Faster solution

Only slightly more complicated, we can use zip to create a similar iterable:

df['c'], df['d'] = zip(*df.col)

df

        col  a    b  c    d
0   (a, 10)  a   10  a   10
1   (b, 20)  b   20  b   20
2   (c, 30)  c   30  c   30
3   (d, 40)  d   40  d   40
4   (e, 50)  e   50  e   50
5   (f, 60)  f   60  f   60
6   (g, 70)  g   70  g   70
7   (h, 80)  h   80  h   80
8   (i, 90)  i   90  i   90
9  (j, 100)  j  100  j  100

---

Inline

Meaning, don't mutate existing df.

This works because assign takes keyword arguments where the keywords are the new (or existing) column names and the values will be the values of the new column. You can use a dictionary and unpack it with ** and have it act as the keyword arguments.

So this is a clever way of assigning a new column named 'g' that is the first item in the df.col.str iterable and 'h' that is the second item in the df.col.str iterable:

df.assign(**dict(zip('gh', df.col.str)))

        col  g    h
0   (a, 10)  a   10
1   (b, 20)  b   20
2   (c, 30)  c   30
3   (d, 40)  d   40
4   (e, 50)  e   50
5   (f, 60)  f   60
6   (g, 70)  g   70
7   (h, 80)  h   80
8   (i, 90)  i   90
9  (j, 100)  j  100

---

My version of the list approach

With modern list comprehension and variable unpacking. Note: also inline using join

df.join(pd.DataFrame([*df.col], df.index, [*'ef']))

        col  g    h
0   (a, 10)  a   10
1   (b, 20)  b   20
2   (c, 30)  c   30
3   (d, 40)  d   40
4   (e, 50)  e   50
5   (f, 60)  f   60
6   (g, 70)  g   70
7   (h, 80)  h   80
8   (i, 90)  i   90
9  (j, 100)  j  100

The mutating version would be

df[['e', 'f']] = pd.DataFrame([*df.col], df.index)

---

Naive Time Test

Use the one defined above:

%timeit df.assign(**dict(zip('gh', df.col.str)))
%timeit df.assign(**dict(zip('gh', zip(*df.col))))
%timeit df.join(pd.DataFrame([*df.col], df.index, [*'gh']))

1.16 ms ± 21.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
635 µs ± 18.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
795 µs ± 42.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

10^3 times bigger

df = pd.concat([df] * 1000, ignore_index=True)

%timeit df.assign(**dict(zip('gh', df.col.str)))
%timeit df.assign(**dict(zip('gh', zip(*df.col))))
%timeit df.join(pd.DataFrame([*df.col], df.index, [*'gh']))

11.4 ms ± 1.53 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.1 ms ± 41.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.33 ms ± 35.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Solution 3

On much larger datasets, I found that .apply() is few orders of magnitude slower than pd.DataFrame(df['b'].values.tolist(), index=df.index).

This performance issue was closed in GitHub, although I do not agree with this decision:

performance issue - apply with pd.Series vs tuple #11615

It is based on this answer.

Solution 4

I think a simpler way is:

>>> import pandas as pd
>>> df = pd.DataFrame({'a':[1,2], 'b':[(1,2), (3,4)]})
>>> df
   a       b
0  1  (1, 2)
1  2  (3, 4)
>>> df['b_a'] = df['b'].str[0]
>>> df['b_b'] = df['b'].str[1]
>>> df
   a       b  b_a  b_b
0  1  (1, 2)    1    2
1  2  (3, 4)    3    4

Solution 5

A caveat of the second solution,

pd.DataFrame(df['b'].values.tolist())

is that it will explicitly discard the index, and add in a default sequential index, whereas the accepted answer

apply(pd.Series)

will not, since the result of apply will retain the row index. While the order is initially retained from the original array, Pandas will try to match the indices from the two dataframes.

This can be very important if you are trying to set the rows into an numerically indexed array, and Pandas will automatically try to match the index of the new array to the old, and cause some distortion in the ordering.

A better hybrid solution would be to set the index of the original dataframe onto the new, i.e.,

pd.DataFrame(df['b'].values.tolist(), index=df.index)

Which will retain the speed of using the second method while ensuring the order and indexing is retained on the result.

View more solutions

Comments

[Donbeo]

is there a way to automate it due to the large number of columns?

[joris]

Not directly I think. But you can easily write a function for it using the above code (+ removing the original one)

[Swier]

pd.DataFrame(df['b'].tolist()) without the .values seems to work just fine too. (And thanks, your solution is much faster than .apply())

[denfromufa]

I was worried about capturing index, hence explicit usage of .values.

[Axel]

If you have a large number of columns you may want to consider to 'tidy' your data: vita.had.co.nz/papers/tidy-data.html You can do this using the melt function.

[Yury Wallet]

.apply(pd.Series) works fine, but for large datasets consumes a lot of memory and can cause Memory Error

[Yury Wallet]

solution by @denfromufa works super fast df[['b1', 'b2']] =pd.DataFrame(df['b'].values.tolist(), index=df.index) and cause no Memory Error (as compared to .apply(pd.Series))

[ApplePie]

This solutions is indeed much more simpler

[denfromufa]

@jinhuawang it appears this is hack on top of str representation of a pd.Series object. Can you explain how this even works?!

[Jinhua Wang]

I think it is just how the str object works? you can access the array object with str

[mammykins]

What if some of the rows have tuples with a different number of values?

[Natacha]

I think this should be the accepted one. It's more 'pandas-onic'...if that's a thing.

[mirekphd]

Consider adding TL;DR: df['a'], df['b'] = df.col.str :)

[Peter Mortensen]

An explanation would be in order. E.g., what is it that allows it to be simpler? What is the idea/gist? Please respond by editing your answer, not here in comments (without "Edit:", "Update:", or similar - the answer should appear as if it was written today).

[zabop]

df.res.str now raises a FutureWarning, but works very well.