CODEWITHSUNDEEP
pythonpython-3.x
Keshav M
Keshav M

Reputation: 1349

Count number of occurrences of each unique item in list of lists

I have a list of lists like the following:

listoflist = [["A", "B", "A", "C", "D"], ["Z", "A", "B", "C"], ["D", "D", "X", "Y", "Z"]]

I want to find the number of sublists that each unique value in listoflist occurs in. For example, "A" shows up in two sublists, while "D" shows up in two sublists also, even though it occurs twice in listoflist[3].

How can I get a dataframe which has each unique element in one column and the frequency (number of sublists each unique element shows up in)?

Upvotes: 1

Views: 2299

Answers (3)

Scott Boston
Scott Boston

Reputation: 153540

Another way to do this is to use pandas:

import pandas as pd

df = pd.DataFrame(listoflist)
df.stack().reset_index().groupby(0)['level_0'].nunique().to_dict()

Output:

{'A': 2, 'B': 2, 'C': 2, 'D': 2, 'X': 1, 'Y': 1, 'Z': 2}

Upvotes: 1

Andrey Tyukin
Andrey Tyukin

Reputation: 44992

Essentially, it seems that you want something like

Counter(x for xs in listoflist for x in set(xs))

Each list is converted into a set first, to exclude duplicates. Then the sequence of sets is flatmapped and fed into the Counter.

Full code:

from collections import Counter

listoflist = [["A", "B", "A", "C", "D"], ["Z", "A", "B", "C"], ["D", "D", "X", "Y", "Z"]]

c = Counter(x for xs in listoflist for x in set(xs))

print(c)

Results in:

# output:
# Counter({'B': 2, 'C': 2, 'Z': 2, 'D': 2, 'A': 2, 'Y': 1, 'X': 1})

Upvotes: 2

llllllllll
llllllllll

Reputation: 16434

You can use: itertools.chain together with collections.Counter:

In [94]: import itertools as it

In [95]: from collections import Counter

In [96]: Counter(it.chain(*map(set, listoflist)))
Out[96]: Counter({'A': 2, 'B': 2, 'C': 2, 'D': 2, 'X': 1, 'Y': 1, 'Z': 2})

As mentioned in the comment by @Jean-François Fabre, you can also use:

In [97]: Counter(it.chain.from_iterable(map(set, listoflist)))
Out[97]: Counter({'A': 2, 'B': 2, 'C': 2, 'D': 2, 'X': 1, 'Y': 1, 'Z': 2})

Upvotes: 3

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