您可以使用python库计数器

counts = [[Counter(C)[word] for C in (C1, C2, C3, C4)] for word in l]
res_df = pd.DataFrame(counts, columns=['C1', 'C2', 'C3', 'C4'], index=l)

输出

C1  C2  C3  C4
good   2   1   0   1
blue   0   0   1   0

使用.loc的另一种替代方法：

df = pd.DataFrame({'C1': Counter(C1), 'C2': Counter(C2), 'C3': Counter(C3), 'C4': Counter(C4)}).loc[l].fillna(0).astype('int')

示例如下：

from collections import Counter
import pandas as pd

C1 = ['hello','good','good','desk']
C2 = ['nice','good','desk','paper']
C3 = ['red','blue','green']
C4 = ['good']

l= ['good','blue']

df = pd.DataFrame({'C1': Counter(C1), 'C2': Counter(C2), 'C3': Counter(C3), 'C4': Counter(C4)}).loc[l].fillna(0).astype('int')

print(df)

输出：

C1  C2  C3  C4
good   2   1   0   1
blue   0   0   1   0

一个想法是通过列表过滤值，列表转换为集合，然后按Counter计数，最后传递给DataFrame，并添加0和整数：

from collections import Counter

d = {'C1':C1, 'C2':C2, 'C3':C3, 'C4':C4}

s = set(l)     

df = (pd.DataFrame({k:Counter([y for y in v if y in s]) for k, v in d.items()})
        .fillna(0).astype(int))
print (df)
      C1  C2  C3  C4
good   2   1   0   1
blue   0   0   1   0

如果可能，列表中不存在值：

from collections import Counter

l= ['good','blue','non']

d = {'C1':C1, 'C2':C2, 'C3':C3, 'C4':C4}

s = set(l)     

df = (pd.DataFrame({k:Counter([y for y in v if y in s]) for k, v in d.items()})
        .fillna(0)
        .astype(int)
        .reindex(l, fill_value=0))
print (df)
    
      C1  C2  C3  C4
good   2   1   0   1
blue   0   0   1   0
non    0   0   0   0