CODEWITHSUNDEEP
pythonpandascountfrequencyfrequency-analysis
Pinky the mouse
Pinky the mouse

Reputation: 197

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", "Berlin", "Boston", "Paris", "Boston"],
     'components': ["a01, a02, b01, b07, b08, с03, d07, e05, e06", 
                    "a01, b02, b35, b68, с43, d02, d07, e04, e05, e08", 
                    "a02, a05, b08, с03, d02, d06, e04, e05, e06", 
                    "a03, a26, a28, a53, b08, с03, d02, f01, f24", 
                    "a01, a28, a46, b37, с43, d06, e04, e05, f02", 
                    "a02, a05, b35, b68, с43, d02, d07, e04, e05, e08", 
                    "a02, a03, b08, b68, с43, d06, d07, e04, e05, e08"]
}
df = pd.DataFrame(my_dict)
df

enter image description here

I need to count most frequent:

  1. Top-n co-occurring components in orders
  2. Top-n most frequent components (regardless of co-occurrence)

What will be the best way to do it?

I can see the relation to market basket analysis problem as well, but not sure how to do it.

Upvotes: 0

Views: 1265

Answers (2)

DeepSpace
DeepSpace

Reputation: 81594

@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)]

Upvotes: 2

Scott Boston
Scott Boston

Reputation: 153460

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

Upvotes: 3

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