Python Pandas Count Most Frequent Occurrences
This is my sample data frame with data about orders: import pandas as pd my_dict = { 'status' : ['a', 'b', 'c', 'd', 'a','a', 'd'], 'city' : ['London','Berlin','Paris',
Solution 1:
Here are some more "pandas" ways of doing the same thing:
To get top three components
#Using list comprehension usually faster than .str accessor in pandas
pd.concat([pd.Series(i.split(',')) for i in df.components]).value_counts().head(3)
#OR using "pure" pandas methods
df.components.str.split(',', expand=True).stack().value_counts().head(3)
Output:
e05 6
e04 5
d02 4
dtype: int64
Next find cohorts, 3 components reported together n=3:
from itertools import combinations
n=3
pd.concat([pd.Series(list(combinations(i.split(','), n))) for i in df.components])\
.value_counts().head(3)
Output:
( с43, e04, e05) 4
(a02, e04, e05) 3
( с43, d07, e05) 3
dtype: int64
Solution 2:
@ScottBoston's answer shows vectorized (hence probably faster) ways to achieve this.
Top occurring
from collections import Counter
from itertools import chain
n = 3
individual_components = chain.from_iterable(df['components'].str.split(', '))
counter = Counter(individual_components)
print(counter.most_common(n))
# [('e05', 6), ('e04', 5), ('a02', 4)]
Top-n co-occuring
Note that I'm using n
twice, once for "the size of the co-occurrence" and once for the "top-n" part. Obviously, you can use 2 different variables.
from collections import Counter
from itertools import combinations
n = 3
individual_components = []
for components in df['components']:
order_components = sorted(components.split(', '))
individual_components.extend(combinations(order_components, n))
counter = Counter(individual_components)
print(counter.most_common(n))
# [(('e04', 'e05', 'с43'), 4), (('a02', 'b08', 'e05'), 3), (('a02', 'd07', 'e05'), 3)]
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