python数据框中的正则表达式：计算模式的出现次数

Question

I want to count how often a regex-expression (prior and ensuing characters are needed to identify the pattern) occurs in multiple dataframe columns. I found a solution which seems a litte slow. Is there a more sophisticated way?

column_A	column_B	column_C
Test • test abc	winter • sun	snow rain blank
blabla • summer abc	break • Data	test letter • stop.

So far I created a solution which is slow:

print(df["column_A"].str.count("(?<=[A-Za-z]) • (?=[A-Za-z])").sum() + df["column_B"].str.count("(?<=[A-Za-z]) • (?=[A-Za-z])").sum() + df["column_C"].str.count("(?<=[A-Za-z]) • (?=[A-Za-z])").sum())

Answer 1

You can use list comprehension and re.search . You can reduce 938 µs to 26.7 µs . (make sure don't create list and use generator )

res = sum(sum(True for item in df[col] if re.search("(?<=[A-Za-z]) • (?=[A-Za-z])", item))
       for col in ['column_A', 'column_B','column_C'])
print(res)
# 5

Benchmark:

%%timeit 
sum(sum(True for item in df[col] if re.search("(?<=[A-Za-z]) • (?=[A-Za-z])", item)) for col in ['column_A', 'column_B','column_C'])
# 26 µs ± 2.2 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

---

%%timeit 
df["column_A"].str.count("(?<=[A-Za-z]) • (?=[A-Za-z])").sum() + df["column_B"].str.count("(?<=[A-Za-z]) • (?=[A-Za-z])").sum() + df["column_C"].str.count("(?<=[A-Za-z]) • (?=[A-Za-z])").sum()
# 938 µs ± 149 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
# --------------------------------------------------------------------#

Answer 2

The str.count should be able to apply to the whole dataframe without hard coding this way. Try

sum(df.apply(lambda x: x.str.count("(?<=[A-Za-z]) • (?=[A-Za-z])").sum()))

I have tried with 1000 * 1000 dataframes. Here is a benchmark for your reference.

%timeit sum(df.apply(lambda x: x.str.count("(?<=[A-Za-z]) • (?=[A-Za-z])").sum()))
1.97 s ± 54.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)