假设df2中有足够的唯一值来填充df1中的0，提取这些唯一值，并使用布尔索引为它们赋值：

# which rows are 0?
m = df1['test'].eq(0)

# extract values of df2 that are not in df1
vals = df2.loc[~df2['test'].isin(df1['test']), 'test'].tolist()
# ['b', 'e', 'f']

# assign the values in the limit of the needed number
df1.loc[m, 'test'] = vals[:m.sum()]

print(df1)

输出：

test
0    b
1    a
2    e
3    c
4    d
5    f

如果df2中的值不总是足够的，并且您希望填充前几个可能的0：

m = df1['test'].eq(0)

vals = df2.loc[~df2['test'].isin(df1['test']), 'test'].unique()
# ['b', 'e', 'f']

m2 = m.cumsum().le(len(vals))

df1.loc[m&m2, 'test'] = vals[:m.sum()]

print(df1)

解决方案假设：

1. df2中“0”的数量==唯一值
   1.有一个类似“test”的列要操作

# get the unique values in df1
uni = df1['test'].unique()

# get the unique values in df2 which are not in df1
unique_df2 = df2[~df2['test'].isin(uni)]

# get the index of all the '0' in df1 in a list
index_df1_0 = df1.index[df1['test'] == 0].tolist()

# replace the '0' in df1 with unique values from df1 (assumption #1 imp!)
for val_ in range(len(index_df1_0)):
    df1.iloc[index_df1_0[val_]] = unique_df2.iloc[val_]

print(df1)