如何匹配和合并具有完全不同值的两个数据帧，除了数据帧列中的数字？

Question

有一个有价值的数据框ABC

      id         |     price                          |   type
0     easdca     | Rs.1,599.00 was trasn by you       | unknown
1     vbbngy     | txn of INR 191.00 using            | unknown
2     awerfa     | Rs.190.78 credits was used by you  | unknown
3     zxcmo5     | DLR.2000 credits was used by you   | unknown

和其他有价值的XYZ

         price          |   type
0      190.78           | food
1      191.00           | movie
2      2,000            | football
3      1,599.00         | basketball

如何使用ABC映射XYZ，以便ABC中的类型使用XYZ价格中的值（数字）更新xyz中的类型。

输出我需要

       id         |     price                          |   type
0     easdca     | Rs.1,599.00 was trasn by you        | basketball
1     vbbngy     | txn of INR 191.00 using             | movie
2     awerfa     | Rs.190.78 credits was used by you   | food
3     zxcmo5     | DLR.2,000 credits was used by you| football

用过这个

d = dict(zip(XYZ['PRICE'],XYZ['TYPE']))

pat = (r'({})'.format('|'.join(d.keys())))

ABC['TYPE']=ABC['PRICE'].str.extract(pat,expand=False).map(d)

但是像190.78和191.00这样的价值观正在变得不匹配。例如，当处理大量数据时，190.78应该与食物值匹配，例如190.77与食物不匹配，其中它具有分配给它的其他值。 198.78也与其他一些与食物相匹配的产品不匹配

Answer 1

DF

        id                price                                type
0       easdca        Rs.1,599.00 was trasn by you          unknown
1       vbbngy        txn of INR 191.00 using               unknown
2       awerfa        Rs.190.78 credits was used by you     unknown
3       zxcmo5        DLR.2000 credits was used by you      unknown

DF2

           price                   type
0        190.78                    food
1        191.00                   movie
2        2,000                 football
3        1,599.00            basketball

使用re

df['price_'] = df['price'].apply(lambda x: re.findall(r'(?<=[\.\s])[\d\.]+',x.replace(',',''))[0])
df2.columns = ['price_','type']
df2['price_'] = df2['price_'].str.repalce(',','')

将类型更改为float

df2['price_']  = df2['price_'].astype(float)
df['price_']  = df['price_'] .astype(float)

使用pd.merge

df = df.merge(df2, on='price_')
df.drop('type_x', axis=1)

产量

                id                                 price   price_       type_y
0      easdca        Rs.1,599.00 was trasn by you         1599.00   basketball
1      vbbngy        txn of INR 191.00 using               191.00        movie
2      awerfa        Rs.190.78 credits was used by you     190.78         food
3      zxcmo5        DLR.2000 credits was used by you        2000     football

Answer 2

您可以执行以下操作：

'''
First we make a artificial key column to be able to merge
We basically just substract the floating numbers from the string
And convert it to type float
'''

df1['price_key'] = df1['price'].str.replace(',', '').str.extract('(\d+\.\d+)').astype(float)

# After that we do a merge on price and price_key and drop the columns which we dont need
df_final = pd.merge(df1, df2, left_on='price_key', right_on='price', suffixes=['', '_2'])
df_final = df_final.drop(['type', 'price_key', 'price_2'], axis='columns')

产量

    id      price                               type_2
0   easdca  Rs.1,599.00 was trasn by you        basketball
1   vbbngy  txn of INR 191.00 using             movie
2   awerfa  Rs.190.78 credits was used by you   food
3   zxcmo5  DLR.2000.78 credits was used by you football

我假设你在xyz表中输了一个错字，第三个价格应该是2000.78而不是2000。