python - 在精确文本匹配时重新索引数据框

Question

当 text 时，我想创建一个数据帧索引与另一个数据帧索引的映射。列(给定)匹配。两个数据帧的长度相等，并且总是会完全匹配。

df_original = pd.DataFrame(dict(text=['The cat sat on the table', 'There is a kind of hush', 'The boy kicked the ball', 'He shot the elephant', 'I want to eat right now!']))
df = pd.DataFrame(dict(text=['He shot the elephant', 'The boy kicked the ball', 'The cat sat on the table', 'I want to eat right now!', 'There is a kind of hush']))

df_original好像:

0   The cat sat on the table
1   There is a kind of hush
2   The boy kicked the ball
3   He shot the elephant
4   I want to eat right now!

df好像:

0   He shot the elephant
1   The boy kicked the ball
2   The cat sat on the table
3   I want to eat right now!
4   There is a kind of hush

我想得到字典映射，就像这样，

d = {2: 0, 4: 1, 1: 2, 0: 3, 3: 4}

例如: df 的第二个索引与 df_original 的第 0 个索引匹配.所以它们必须映射在一起等等。
如果可能，我更喜欢矢量化操作，并且正在寻找一种。
我试着做:

d = {}
for i1, r1 in df_original.iterrows():
    for i2, r2 in df.iterrows():
        if r1[0] == r2[0]: 
            d[i2] = i1
print(d)
# {2: 0, 4: 1, 1: 2, 0: 3, 3: 4}

但这非常慢，因为我有数百万行的数据帧。

Answer 1

试试 merge :

(df_original.reset_index()
   .merge(df.reset_index(), on='text')
   .set_index('index_y')['index_x'].to_dict()
)

出去:

{2: 0, 4: 1, 1: 2, 0: 3, 3: 4}