How to determine the length of lists in a pandas dataframe column
PythonPandasPython Problem Overview
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?
Python Solutions
Solution 1 - Python
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
Solution 2 - Python
pandas.Series.map(len)andpandas.Series.apply(len)are equivalent in execution time, and slightly faster thanpandas.Series.str.len().- Difference between map, applymap and apply methods in Pandas
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)