我的追求:

我有一个简单的问题 DataFrame 我希望通过搜索提取视图的位置 list searches . 例子:

import pandas as pd
data = {k: [v+str(i) for i in range(10)] for k, v in zip(('OrderNo','Name', 'Useless','Description'),('1000','Product ', 'Junk ','Short Desc '))}
df = pd.DataFrame(data)
df.loc[2:6, ('Useless',)] = pd.np.nan
# to mock some nan data in my real one.

因而发生的 df :

  OrderNo       Name Useless   Description
0   10000  Product 0  Junk 0  Short Desc 0
1   10001  Product 1  Junk 1  Short Desc 1
2   10002  Product 2     Nan  Short Desc 2
3   10003  Product 3     Nan  Short Desc 3
4   10004  Product 4     Nan  Short Desc 4
5   10005  Product 5     Nan  Short Desc 5
6   10006  Product 6     Nan  Short Desc 6
7   10007  Product 7  Junk 7  Short Desc 7
8   10008  Product 8  Junk 8  Short Desc 8
9   10009  Product 9  Junk 9  Short Desc 9

现在我想找一个 列表 的 OrderNos 像这样:

searches = ['10005','10009','10003','10000']

我正试图看到这样的景象:

  OrderNo       Name Useless   Description
5   10005  Product 5     Nan  Short Desc 5
9   10009  Product 9  Junk 9  Short Desc 9
3   10003  Product 3     Nan  Short Desc 3
0   10000  Product 0  Junk 0  Short Desc 0

因此,我最终可以将视图转换为以下内容(注意,我删除了一些无用的列):

                        0             1             2             3
OrderNo             10005         10009         10003         10000
Name            Product 5     Product 9     Product 3     Product 0
Description  Short Desc 5  Short Desc 9  Short Desc 3  Short Desc 0

我所尝试的:

This great question/answer 搜查 ,但返回的视图不符合我的顺序:

found = df.loc[df['OrderNo'].isin(searches)]

  OrderNo       Name Useless   Description
0   10000  Product 0  Junk 0  Short Desc 0
3   10003  Product 3     Nan  Short Desc 3
5   10005  Product 5     Nan  Short Desc 5
9   10009  Product 9  Junk 9  Short Desc 9

['my_sort'] 到 found 因此,我可以根据列表重新排序:

found['my_sort'] = found['OrderNo'].apply(lambda x: searches.index(x))
found.sort_values(by='my_sort', inplace=True)
# For now assume index will always be matched and ValueError will be handled.
# This detail is not critical

而这 有点 pandas SettingWithCopyWarning 到处都是,叫我用 .loc[row_indexer,col_indexer] = ... 仍然 建立 抛出相同的,所以我怀疑问题来自搜索。最后我把它包装成了一个新的 要不再看到警告,请执行以下操作:

found = pd.DataFrame(df.loc[df['OrderNo'].isin(searches)])
found['my_sort'] = found['OrderNo'].apply(lambda x: searches.index(x))
found = found[columns].T

虽然这是可行的,但我还是忍不住觉得这很复杂,效率也不高,因为我不得不引入一个新的列来进行排序,然后再次删除。我研究了一些相关的函数,如 reindex where 和 dropna nan 对象),但它们似乎都无法实现我的目标。