熊猫几行数据融合

Question

print(ScimEn.head(20))
    Rank             Country
0      1               China
1      2       United States
2      3               Japan
3      4      United Kingdom
4      5  Russian Federation
5      6              Canada
6      7             Germany
7      8               India
8      9              France
9     10         South Korea
10    11               Italy
11    12               Spain
12    13                Iran
13    14           Australia
14    15              Brazil
15    16              Taiwan
16    17              Turkey
17    18              Norway
18    19         Netherlands
19    20              Sweden

我想将这个DF合并到另一个行，但是只有排名0:15的行。如果我经常这样做：

dfs = [ScimEn, energy, GDP[year].reset_index()]
newdf = reduce(lambda left, right: pd.merge(left, right, on='Country'), dfs)
newdf = newdf.set_index('Country')
print(len(newdf))

我的DF值为140项，但我希望上面提到的每个条目的DF值为15项。我也试过：

 dfs = [ScimEn.where(ScimEn['Rank'] < 15), energy, GDP[year].reset_index()]
newdf = reduce(lambda left, right: pd.merge(left, right, on='Country'), dfs)
newdf = newdf.set_index('Country')
print(len(newdf))

但它只返回了4个条目。

Answer 1

我想你可以用query过滤

dfs = [ScimEn.query("Rank < 15"), energy, GDP[year].reset_index()]
newdf = reduce(lambda left, right: pd.merge(left, right, on='Country'), dfs)
newdf = newdf.set_index('Country')
print(len(newdf))

或者像提到的那样由boolean indexing执行，Phong Phung comment

dfs = [ScimEn[ScimEn['Rank'] < 15], energy, GDP[year].reset_index()]
newdf = reduce(lambda left, right: pd.merge(left, right, on='Country'), dfs)
newdf = newdf.set_index('Country')
print(len(newdf))