Get total values_count from a dataframe with Python Pandas

If performance is an issue try:

from collections import Counter

Counter(df.values.ravel())
#Counter({'string1': 3, 'string2': 2, 'string3': 1})

Or stack it into one Series then use value_counts

df.stack().value_counts()
#string1    3
#string2    2
#string3    1
#dtype: int64

For larger (long) DataFrames with a small number of columns, looping may be faster than stacking:

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

Also, there's a numpy solution:

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

Small 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)

Long 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)