112

How can the length of the lists in the column be determine without iteration?

I have a dataframe like this:

                                                    CreationDate
2013-12-22 15:25:02                  [ubuntu, mac-osx, syslinux]
2009-12-14 14:29:32  [ubuntu, mod-rewrite, laconica, apache-2.2]
2013-12-22 15:42:00               [ubuntu, nat, squid, mikrotik]

I am calculating the length of lists in the CreationDate column and making a new Length column like this:

df['Length'] = df.CreationDate.apply(lambda x: len(x))

Which gives me this:

                                                    CreationDate  Length
2013-12-22 15:25:02                  [ubuntu, mac-osx, syslinux]       3
2009-12-14 14:29:32  [ubuntu, mod-rewrite, laconica, apache-2.2]       4
2013-12-22 15:42:00               [ubuntu, nat, squid, mikrotik]       4

Is there a more pythonic way to do this?

cottontail
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Mohammad Yusuf
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3 Answers3

166

You can use the str accessor for some list operations as well. In this example,

df['CreationDate'].str.len()

returns the length of each list. See the docs for str.len.

df['Length'] = df['CreationDate'].str.len()
df
Out: 
                                                    CreationDate  Length
2013-12-22 15:25:02                  [ubuntu, mac-osx, syslinux]       3
2009-12-14 14:29:32  [ubuntu, mod-rewrite, laconica, apache-2.2]       4
2013-12-22 15:42:00               [ubuntu, nat, squid, mikrotik]       4

For these operations, vanilla Python is generally faster. pandas handles NaNs though. Here are timings:

ser = pd.Series([random.sample(string.ascii_letters, 
                               random.randint(1, 20)) for _ in range(10**6)])

%timeit ser.apply(lambda x: len(x))
1 loop, best of 3: 425 ms per loop

%timeit ser.str.len()
1 loop, best of 3: 248 ms per loop

%timeit [len(x) for x in ser]
10 loops, best of 3: 84 ms per loop

%timeit pd.Series([len(x) for x in ser], index=ser.index)
1 loop, best of 3: 236 ms per loop
ayhan
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20
import pandas as pd

data = {'os': [['ubuntu', 'mac-osx', 'syslinux'], ['ubuntu', 'mod-rewrite', 'laconica', 'apache-2.2'], ['ubuntu', 'nat', 'squid', 'mikrotik']]}
index = ['2013-12-22 15:25:02', '2009-12-14 14:29:32', '2013-12-22 15:42:00']

df = pd.DataFrame(data, index)

# create Length column
df['Length'] = df.os.map(len)

# display(df)
                                                              os  Length
2013-12-22 15:25:02                  [ubuntu, mac-osx, syslinux]       3
2009-12-14 14:29:32  [ubuntu, mod-rewrite, laconica, apache-2.2]       4
2013-12-22 15:42:00               [ubuntu, nat, squid, mikrotik]       4

%timeit

import pandas as pd
import random
import string

random.seed(365)

ser = pd.Series([random.sample(string.ascii_letters, random.randint(1, 20)) for _ in range(10**6)])

%timeit ser.str.len()
252 ms ± 12.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit ser.map(len)
220 ms ± 7.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit ser.apply(len)
222 ms ± 8.31 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Trenton McKinney
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1
Convert to list and map a function

Pandas dataframe columns are not meant to store collections such as lists, tuples etc. because virtually none of the optimized methods work on these columns, so when a dataframe contains such items, it's usually more efficient to convert the column into a Python list and manipulate the list.

Also, if a function (especially a built-in one like len()) needs to be called on each item in a list, it's usually faster to map this function, rather than calling it in a loop.

mylist = df['CreationDate'].tolist()
df['Length'] = list(map(len, mylist))
Handle NaNs

Nice thing about str.len() is that it handles NaNs but a custom function with try-except should fill that gap.

def nanlen(x):
    try:
        return len(x)
    except TypeError:
        return float('nan')
    
df['Length'] = list(map(nanlen, mylist))
Runtime benchmarks

Essentially, mapping len over lists is approx. 2.5 times faster than looping over a Series, which in turn is 2.5 times faster than pd.Series.str.len for large frames.

res

Code used to produce the plot above:

import pandas as pd
import random, string, perfplot
random.seed(365)

perfplot.plot(
    setup=lambda n: pd.Series([random.sample(string.ascii_letters, random.randint(1, 20)) for _ in range(n)]),
    kernels=[lambda ser: ser.str.len(), lambda ser: ser.map(len), lambda ser: list(map(len, ser.tolist())), lambda ser: [len(x) for x in ser]],
    labels=["ser.str.len()", "ser.map(len)", "list(map(len, ser.tolist()))", "[len(x) for x in ser]"],
    n_range=[2**k for k in range(21)],
    xlabel='Length of dataframe',
    equality_check=lambda x,y: x.eq(y).all()
)
cottontail
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