比较两个列,一个按浮点,另一个按字符串,以获得匹配的值

loc 和 isin 如下图所示

df = df1.copy()

df['Matching'] = np.nan
df.loc[(df.Number.isin(df2.Number)) & (df.Item.isin(df2.Item)), 'Matching'] = 'Matched'
df.Matching.fillna('No Match', inplace=True)

Number    Item      Matching

1.0   Phone     No Match
3.0   Watch     Matched
4.0   Pen       Matched
5.0   Pencil    No Match
8.0   Pencil    Matched
12.0  toolkit   Matched
32.0  box       Matched
58.0  fork      Matched

1000 转换成整数然后 merge 对于left join,因为匹配应该有问题,所以两列中的baciuse float精度应该不同:

df1['Number1'] = df1['Number'].mul(1000).astype(int)
df2['Number1'] = df2['Number'].mul(1000).astype(int)

df = pd.merge(df1, df2.drop('Number', 1), how='left', on=['Item','Number1'], indicator=True)
df['Matching'] = df['_merge'].map({'left_only':'No Match', 'both':'Match'})

df = df.drop(['Number1','_merge'], axis=1)
print (df)

   Number     Item  Matching
0     1.0    Phone  No Match
1     3.0    Watch     Match
2     4.0      Pen     Match
3     5.0   Pencil  No Match
4     8.0   Pencil     Match
5    12.0  toolkit  No Match
6    32.0      box     Match
7    58.0     fork     Match

indicator=True

res = pd.merge(df1, df2, how='left', indicator=True)

print(res)

      Item  Number     _merge
0    Phone     1.0  left_only
1    Watch     3.0       both
2      Pen     4.0       both
3   Pencil     5.0  left_only
4   Pencil     8.0       both
5  toolkit    12.0  left_only
6      box    32.0       both
7     fork    58.0       both

一般来说,避免显式的 for 当有专门构建的方法可用时循环,因为这些方法通常针对性能进行了优化。如果愿意,可以通过字典映射替换字符串:

d = {'left_only': 'No Match', 'both': 'Matched'}
df['_merge'] = df['_merge'].map(d)