小编典典

如果性能是一个问题，请尝试：

from collections import Counter

Counter(df.values.ravel())

#Counter({'string1': 3, 'string2': 2, 'string3': 1})

或者stack它变成一个Series再使用value_counts

df.stack().value_counts()

#string1 3

#string2 2

#string3 1

#dtype: int64

对于具有少量列的较大(长)DataFrame，循环可能比堆栈快：

s = pd.Series()

for col in df.columns:

s = s.add(df[col].value_counts(), fill_value=0)

#string1 3.0

#string2 2.0

#string3 1.0

#dtype: float64

此外，还有一个numpy解决方案：

import numpy as np

np.unique(df.to_numpy(), return_counts=True)

#(array(['string1', 'string2', 'string3'], dtype=object),

# array([3, 2, 1], dtype=int64))

df = pd.DataFrame({'row1': ['string1', 'string1', 'string2'],

'row2': ['string3', 'string1', 'string2']})

def vc_from_loop(df):

s = pd.Series()

for col in df.columns:

s = s.add(df[col].value_counts(), fill_value=0)

return s

小 DataFrame

%timeit Counter(df.values.ravel())

#11.1 µs ± 56.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit df.stack().value_counts()

#835 µs ± 5.46 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit vc_from_loop(df)

#2.15 ms ± 34.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit np.unique(df.to_numpy(), return_counts=True)

#23.8 µs ± 241 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

长 DataFrame

df = pd.concat([df]*300000, ignore_index=True)

%timeit Counter(df.values.ravel())

#124 ms ± 1.85 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit df.stack().value_counts()

#337 ms ± 3.59 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit vc_from_loop(df)

#182 ms ± 1.58 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit np.unique(df.to_numpy(), return_counts=True)

#1.16 s ± 1.09 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

2020-12-20