在 Pandas 的列上应用 lambda
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Apply lambda on a column in pandas
提问by Shiva Krishna Bavandla
I have the below data frame
我有以下数据框
ipdb> csv_data
country_edited sale_edited date_edited transformation_edited
0 India 403171 21091956 1
1 Bhutan 394096 21091956 2
2 Nepal 361372 21091956 3
3 madhya 355883 21091956 4
4 sudan 262949 21091956 5
and below is my code
下面是我的代码
transfactor_count = 5.6
csv_data["transformation_edited"] = csv_data["transformation_edited"].apply(lambda x: x * transfactor_count)
But the above code was giving me an error
但是上面的代码给了我一个错误
*** NameError: global name 'transfactor_count' is not defined
How to solve this ?
如何解决这个问题？
Actual code
实际代码
for foreign_key in data_mapping_record.csvdatabasecolumnmapping_set.all():
data_type = foreign_key.data_type
trans_factor = foreign_key.tranformation_factor
if data_type == "Decimal":
import ipdb; ipdb.set_trace()
csv_data[foreign_key.table_column_name] = csv_data[foreign_key.table_column_name].apply(lambda x: x * trans_factor )
elif data_type in ["Date", "Datetime"]:
csv_data[foreign_key.table_column_name] = csv_data[foreign_key.table_column_name].apply( lambda d: datetime.strptime(d, dates[date]) )
回答by Jan Zeiseweis
As the error already says, the lambda function can't find the globalvariable. You can try make it global by:
正如错误所说，lambda 函数找不到全局变量。您可以尝试通过以下方式使其成为全球性的：
global transfactor_count
transfactor_count = 5.6
csv_data["transformation_edited"] = csv_data["transformation_edited"].apply(lambda x: x * transfactor_count)
But as jezrael pointed out:
但正如 jezrael 指出的那样：
csv_data["transformation_edited"] = csv_data["transformation_edited"] * transfactor_count
is much more elegant.
更优雅。
you might have to change the datatype to float before:
您可能必须在之前将数据类型更改为浮动：
csv_data["transformation_edited"] = csv_data["transformation_edited"].astype(float) * transfactor_count
回答by sandeep
You can use lambda on dataframe as below:
您可以在数据帧上使用 lambda，如下所示：
transfactor_count = 5.6;
csv_data['transformation_edited']=map(lambda x: x * transfactor_count, csv_data['transformation_edited'])
