[英]Python3 pandas data frame using fillna(method='bfill') with group by
I am new to python and pandas and stuck with request mentioned below Have data in python pandas data frame as 我是python和pandas的新手,并且卡在下面提到的请求中,因此python pandas数据框中的数据为
time_stamp dish_id table_no order_id
2017-10-05 22:11 122 A1
2017-10-05 22:14 127 A1
2017-10-05 22:17 129 A5
2017-10-05 22:19 122 A1 X_001
2017-10-05 22:17 129 A5 X_002
I am filling in the missing order values with 我要填写缺少的订单值
output_sort[['new_order_id']] = output_sort[['order_id']].fillna(method='bfill')
and this gets me result like 这让我的结果像
time_stamp dish_id table_no order_id
2017-10-05 22:11 122 A1 X_001
2017-10-05 22:14 127 A1 X_001
2017-10-05 22:17 129 A5 X_001
2017-10-05 22:19 122 A1 X_001
2017-10-05 22:17 129 A5 X_002
However i want to get results like 但是我想得到像
time_stamp dish_id table_no order_id
2017-10-05 22:11 122 A1 X_001
2017-10-05 22:14 127 A1 X_001
2017-10-05 22:17 129 A5 X_002
2017-10-05 22:19 122 A1 X_001
2017-10-05 22:17 129 A5 X_002
The order_id's get matched with correct_table no I haven't been able to find a way to do it Any help would be really appreciated order_id已与correct_table匹配,否,我一直无法找到一种方法来做,任何帮助将不胜感激
df.groupby('table_no')['order_id'].apply(lambda x :x.ffill().bfill())
Out[529]:
0 X_001
1 X_001
2 X_002
3 X_001
4 X_002
Name: order_id, dtype: object
df['order_id']=df.groupby('table_no')['order_id'].apply(lambda x :x.ffill().bfill())
df
Out[530]:
time_stamp dish_id table_no order_id
0 2017-10-0522:11 122 A1 X_001
1 2017-10-0522:14 127 A1 X_001
2 2017-10-0522:17 129 A5 X_002
3 2017-10-0522:19 122 A1 X_001
4 2017-10-0522:17 129 A5 X_002
df.assign(order_id=df.groupby('table_no').order_id.bfill())
time_stamp dish_id table_no order_id
0 2017-10-05 22:11 122 A1 X_001
1 2017-10-05 22:14 127 A1 X_001
2 2017-10-05 22:17 129 A5 X_002
3 2017-10-05 22:19 122 A1 X_001
4 2017-10-05 22:17 129 A5 X_002
While not as idiomatic as bfill
, map
should be a pretty good alternative. 虽然不如
bfill
惯用, bfill
map
应该是一个不错的选择。
m = dict(df[['table_no', 'order_id']].dropna().values)
print(m)
{'A1': 'X_001', 'A5': 'X_002'}
df['order_id'] = df.table_no.map(m)
print(df)
time_stamp dish_id table_no order_id
0 2017-10-05 22:11 122 A1 X_001
1 2017-10-05 22:14 127 A1 X_001
2 2017-10-05 22:17 129 A5 X_002
3 2017-10-05 22:19 122 A1 X_001
4 2017-10-05 22:17 129 A5 X_002
You can also do this with df.replace
: 您也可以使用
df.replace
进行此df.replace
:
df['order_id'] = df.table_no.replace(m)
print(df)
time_stamp dish_id table_no order_id
0 2017-10-05 22:11 122 A1 X_001
1 2017-10-05 22:14 127 A1 X_001
2 2017-10-05 22:17 129 A5 X_002
3 2017-10-05 22:19 122 A1 X_001
4 2017-10-05 22:17 129 A5 X_002
Another way of generating m
would be: 生成
m
另一种方法是:
m = df[['table_no', 'order_id']].dropna().set_index('table_no').order_id
print(m)
table_no
A1 X_001
A5 X_002
Name: order_id, dtype: object
series_ = df.table_no.tolist()
def fill_():
order_id_ = []
if table_no == 'A1'
order_id_.append('X_001')
else:
order_id_.append('X_005')
return order_id_
df.order_no = list(map(fill_,series_))
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