Pythonic计算字符串列表中出现次数的方法

Question

What's the best way to find the count of occurrences of strings from a list in a target string? Specifically, I have a list :

string_list = [
    "foo",
    "bar",
    "baz"
]

target_string = "foo bar baz bar"

# Trying to write this function!
count = occurrence_counter(target_string) # should return 4

I'd like to optimize to minimize speed and memory usage, if that makes a difference. In terms of size, I would expect that string_list may end up containing several hundred substrings.

Answer 1

Another way using collelctions.Counter :

from collections import Counter
word_counts = Counter(target_string.split(' '))
total = sum(word_counts.get(w, 0)) for w in string_list)

Answer 2

This works!

def occurrence_counter(target_string):
    return sum(map(lambda x: x in string_list, target_string.split(' ')))

The string gets split into tokens, then each token gets transformed into a 1 if it is in the list, a 0 otherwise. The sum function, at last, sums those values.

EDIT: also:

def occurrence_counter(target_string):
    return len(list(filter(lambda x: x in string_list, target_string.split(' '))))

Answer 3

This Python3 should work:

In [4]: string_list = [
   ...:     "foo",
   ...:     "bar",
   ...:     "baz"
   ...: ]
   ...: 
   ...: set_of_counted_word = set(string_list)
   ...: 
   ...: def occurrence_counter(target_str, words_to_count=set_of_counted_word):
   ...:     return sum(1 for word in target_str.strip().split()
   ...:                if word in words_to_count)
   ...: 
   ...: 
   ...: for target_string in ("foo bar baz bar", " bip foo bap foo dib baz   "):
   ...:     print("Input: %r -> Count: %i" % (target_string, occurrence_counter(target_string)))
   ...: 
   ...: 
Input: 'foo bar baz bar' -> Count: 4
Input: ' bip foo bap foo dib baz   ' -> Count: 3

In [5]:

Answer 4

You could use a variable to store a running count is you iterate through the list like so:

def occurence_counter(x):
    count = 0
    for y in x:
        count +=1
    return count

Answer 5

Another solution:

def occurrence_counter(target_string, string_list):
    target_list = target_string.split(' ')
    return len([w for w in target_list if w in string_list])

Answer 6

Combo of sum and string.count :

def counter(s, lst)
    return sum(s.count(sub) for sub in lst)

This will not count overlapping occurrences of the same pattern.

Answer 7

You could use a Trie to convert your substrings to a regex pattern (eg (?:ba[rz]|foo) ) and parse your target_string :

import re
from trie import Trie

trie = Trie()

substrings = [
    "foo",
    "bar",
    "baz"
]
for substring in substrings:
    trie.add(substring)
print(trie.pattern())
# (?:ba[rz]|foo)

target_string = "foo bar baz bar"
print(len(re.findall(trie.pattern(), target_string)))
# 4

The required library is here : trie.py

It should be much faster than parsing the whole target_string for each substring , but it might not return the desired result for overlapping substrings. It returns 2 for ["foo", "bar", "foobar"] and "foobar" .

A related question was : " Speed up millions of regex replacements in Python 3 " : here's an answer with sets and one with a trie regex .

Answer 8

I am not sure this is the most pythonic way, but you can try it:

string_list_B = target_string.split(" ")
commonalities = set(string_list) - (set(string_list) - set(string_list_B))