有没有办法对熊猫/numpy中计数项目的共现进行矢量化？

Question

I frequently need to generate network graphs based on the co-occurences of items in a column. I start of with something like this:

           letters
0  [b, a, e, f, c]
1        [a, c, d]
2        [c, b, j]

In the following example, I want a to make a table of all pairs of letters, and then have a "weight" column, which describes how many times each two letter pair appeared in the same row together (see bottom for example).

I am currently doing large parts of it using a for loop, and I was wondering if there is a way for me to vectorize it, as I am often dealing with enormous datasets that take an extremely long time to process in this way. I am also concerned about keeping things within memory limits. This is my code right now:

import pandas as pd

# Make some data
df = pd.DataFrame({'letters': [['b','a','e','f','c'],['a','c','d'],['c','b','j']]})

# I make a list of sets, which contain pairs of all the elements
# that co-occur in the data in the same list
sets = []
for lst in df['letters']:
    for i, a in enumerate(lst):
        for b in lst[i:]:
            if not a == b:
                sets.append({a, b})

# Sets now looks like:
# [{'a', 'b'},
#  {'b', 'e'},
#  {'b', 'f'},...

# Dataframe with one column containing the sets
df = pd.DataFrame({'weight': sets})

# We count how many times each pair occurs together
df = df['weight'].value_counts().reset_index()

# Split the sets into two seperate columns
split = pd.DataFrame(df['index'].values.tolist()) \
          .rename(columns = lambda x: f'Node{x+1}') \
          .fillna('-')

# Merge the 'weight' column back onto the dataframe
df = pd.concat([df['weight'], split], axis = 1)

print(df.head)

# Output:
   weight Node1 Node2
0       2     c     b
1       2     a     c
2       1     f     e
3       1     d     c
4       1     j     b

Answer 1

Notes:

As suggested in the other answers, make use of collections.Counter for the counting. Since it behaves like a dict though, it needs hashable types. {a,b} is not hashable, because it's a set. Replacing it with a tuple fixes the hashability problem, but introduces possible duplicates (eg, ('a', 'b') and ('b', 'a') ). To fix this issue, just sort the tuple.

since sorted returns a list , we need to turn that back into a tuple: tuple(sorted((a,b))) . A bit cumbersome, but convenient in combination with Counter .

Quick and easy speedup: Comprehensions instead of loops

When rearranged, your nested loops can be replaced with the following comprehension:

sets = [ sorted((a,b)) for lst in df['letters'] for i,a in enumerate(lst) for b in lst[i:] if not a == b ]

Python has optimizations in place for comprehension execution, so this will already bring some speedup.

Bonus: If you combine it with Counter , you don't even need the result as a list, but can instead use a generator expression (almost no extra memory is used instead of storing all pairs):

Counter( tuple(sorted((a, b))) for lst in lists for i,a in enumerate(lst) for b in lst[i:] if not a == b ) # note the lack of [ ] around the comprehension

---

Evaluation: What is the faster approach?

As usual, when dealing with performance, the final answer must come from testing different approaches and choosing the best one. Here I compare the (IMO very elegant and readable) itertools -based approach by @yatu, the original nested-for and the comprehension. All tests run on the same sample data, randomly generated to look like the given example.

from timeit import timeit

setup = '''
import numpy as np
import random
from collections import Counter
from itertools import combinations, chain
random.seed(42)
np.random.seed(42)

DF_SIZE = 50000 # make it big
MAX_LEN = 6
list_lengths = np.random.randint(1, 7, DF_SIZE)

letters = 'abcdefghijklmnopqrstuvwxyz'

lists = [ random.sample(letters, ln) for ln in list_lengths ] # roughly equivalent to df.letters.tolist()
'''

#################

comprehension = '''Counter( tuple(sorted((a, b))) for lst in lists for i,a in enumerate(lst) for b in lst[i:] if not a == b )'''
itertools = '''Counter(chain.from_iterable(combinations(sorted(i), r=2) for i in lists))'''
original_for_loop = '''
sets = []
for lst in lists:
    for i, a in enumerate(lst):
        for b in lst[i:]:
            if not a == b:
                sets.append(tuple(sorted((a, b))))
Counter(sets)
'''

print(f'Comprehension: {timeit(setup=setup, stmt=comprehension, number=10)}')
print(f'itertools: {timeit(setup=setup, stmt=itertools, number=10)}')
print(f'nested for: {timeit(setup=setup, stmt=original_for_loop, number=10)}')

Running the code above on my machine (python 3.7) prints:

Comprehension: 1.6664735930098686
itertools: 0.5829475829959847
nested for: 1.751666523006861

So, both suggested approaches improve over the nested for loops, but itertools is indeed faster in this case.

Answer 2

For a performance improvement you could use itertooos.combinations in order to get all length 2 combinations from the inner lists, and Counter to count the pairs in a flattened list.

Note that in addition to obtaining all combinations from each sublist, sorting is a necessary step since it will ensure that all pairs of tuples will appear in the same order:

from itertools import combinations, chain
from collections import Counter

l = df.letters.tolist()
t = chain.from_iterable(combinations(sorted(i), r=2) for i in l)

print(Counter(t))

Counter({('a', 'b'): 1,
         ('a', 'c'): 2,
         ('a', 'e'): 1,
         ('a', 'f'): 1,
         ('b', 'c'): 2,
         ('b', 'e'): 1,
         ('b', 'f'): 1,
         ('c', 'e'): 1,
         ('c', 'f'): 1,
         ('e', 'f'): 1,
         ('a', 'd'): 1,
         ('c', 'd'): 1,
         ('b', 'j'): 1,
         ('c', 'j'): 1})

Answer 3

A numpy/scipy solution using sparse incidence matrices:

from itertools import chain
import numpy as np
from scipy import sparse
from simple_benchmark import BenchmarkBuilder, MultiArgument

B = BenchmarkBuilder()

@B.add_function()
def pp(L):
    SZS = np.fromiter(chain((0,),map(len,L)),int,len(L)+1).cumsum()
    unq,idx = np.unique(np.concatenate(L),return_inverse=True)
    S = sparse.csr_matrix((np.ones(idx.size,int),idx,SZS),(len(L),len(unq)))
    SS = (S.T@S).tocoo()
    idx = (SS.col>SS.row).nonzero()
    return unq[SS.row[idx]],unq[SS.col[idx]],SS.data[idx] # left, right, count


from collections import Counter
from itertools import combinations

@B.add_function()
def yatu(L):
    return Counter(chain.from_iterable(combinations(sorted(i),r=2) for i in L))

@B.add_function()
def feature_engineer(L):
    Counter((min(nodes), max(nodes))
            for row in L for nodes in combinations(row, 2))

from string import ascii_lowercase as ltrs

ltrs = np.array([*ltrs])

@B.add_arguments('array size')
def argument_provider():
    for exp in range(4, 30):
        n = int(1.4**exp)
        L = [ltrs[np.maximum(0,np.random.randint(-2,2,26)).astype(bool).tolist()] for _ in range(n)]
        yield n,L

r = B.run()
r.plot()

We see that the method presented here ( pp ) comes with the typical numpy constant overhead, but from ~100 sublists it starts winning.

OPs example:

import pandas as pd

df = pd.DataFrame({'letters': [['b','a','e','f','c'],['a','c','d'],['c','b','j']]})
pd.DataFrame(dict(zip(["left", "right", "count"],pp(df['letters']))))

Prints:

   left right  count
0     a     b      1
1     a     c      2
2     b     c      2
3     c     d      1
4     a     d      1
5     c     e      1
6     a     e      1
7     b     e      1
8     c     f      1
9     e     f      1
10    a     f      1
11    b     f      1
12    b     j      1
13    c     j      1

Answer 4

Notes to improve efficiency:

1. Instead of storing the pairs in sets, which are memory hogs and require expensive computation for adding elements, use a tuple where the first element is the smallest.

2. To calculate the combinations quickly, use itertools.combinations.

3. To count the combinations use collections.Counter

4. optionally, convert the count to a DataFrame.

Here's an example implementation:

from collections import Counter
from itertools import combinations

data = df.letters.tolist()

#    data = [['b', 'a', 'e', 'f', 'c'],
#            ['a', 'c', 'd'],
#            ['c', 'b', 'j']]

counts = Counter((min(nodes), max(nodes)) for row in data for nodes in combinations(row, 2))