您可以用途：

target = ['Men', 'Women', 'Non-bi']

# get rid of non relevant column
tmp = df.drop(columns='User').notna()

# keep users having at least one target
m1 = tmp[target].any(axis=1)
# drop rows having another match
m2 = ~tmp.drop(columns=target).any(axis=1)

# count
count = (m1&m2).sum()

输出：3

如果需要计算所有答案的组合，将行转换为frozenset s，删除缺失值并使用Series.value_counts：

s = df.set_index('User').apply(lambda x: frozenset(x.dropna()), axis=1).value_counts()
print (s)
(Asexual)          1
(Women, Non-bi)    1
(Men)              1
(Women, Men)       1
dtype: int64

或者通过DataFrame.stack进行整形，并转换为frozenset s：

s = df.set_index('User').stack().groupby('User').agg(frozenset).value_counts()
print (s)
(Asexual)          1
(Women, Non-bi)    1
(Men)              1
(Women, Men)       1
Name: value, dtype: int64

或者，如果需要通过列表匹配值-每行只计数一次，使用DataFrame.isin，通过Series.any至少生成一个True，并通过sum计数布尔值：

need = ['Men', 'Women', 'Non-bi']

out = df.isin(need).any(axis=1).sum()
print (out)
3

need = ['Men', 'Women', 'Non-bi']

out = df.reindex(need, axis=1).isin(need).any(axis=1).sum()
print (out)
3