Pandas 如何通过应用函数调试组件
我试图理解我的前同事写的函数Pandas 如何通过应用函数调试组件,pandas,pandas-groupby,pandas-apply,Pandas,Pandas Groupby,Pandas Apply,我试图理解我的前同事写的函数 def generate_df(group): date_str = str(group['CallerLocal_Date'].iloc[-1]) + ' {0}:00:00' # some other functions return something enrich_df = df.groupby(['LeadNumber', 'CallerLocal_Date'], as_index=False).apply(generate_
def generate_df(group):
date_str = str(group['CallerLocal_Date'].iloc[-1]) + ' {0}:00:00'
# some other functions
return something
enrich_df = df.groupby(['LeadNumber', 'CallerLocal_Date'], as_index=False).apply(generate_df).reset_index(drop=True)
我无法完全理解上面的函数,所以我尝试按groupby并实际查看date\u str=str(group['CallerLocal\u date'].iloc[-1])+'{0}:00:00'
line的功能
df
看起来像这样
LeadNumber CallerLocal_Date Caller_TimeZone
0 7-OH4UMXXL5 2017-09-13 America/Chicago
1 7-OL4ZHUF47 2017-09-26 America/Chicago
2 7-OL4ZHUF47 2017-09-26 America/Chicago
3 7-OHMFNFFC2 2017-09-13 America/Chicago
4 7-OHMFNFFC2 2017-09-12 America/Chicago
5 7-OGBMIPIIN 2017-09-11 America/Chicago
6 7-OGBMIPIIN 2017-09-07 America/Chicago
7 7-OETJOA7O6 2017-09-01 America/Chicago
8 7-OETJOA7O6 2017-09-06 America/Chicago
9 7-OILTU4T5O 2017-09-18 America/Chicago
10 7-OGJHKCJFZ 2017-09-07 America/Chicago
所以我定义
group = df.groupby(['LeadNumber', 'CallerLocal_Date'], as_index=False)
打电话
date_str = str(group['CallerLocal_Date'].iloc[-1]) + ' {0}:00:00'
然后我得到了
AttributeError: Cannot access callable attribute 'iloc' of 'DataFrameGroupBy' objects, try using the 'apply' method
有人能告诉我如何在不使用apply
函数的情况下调试groupby对象吗?您可以执行以下操作:
groups = df.groupby(['LeadNumber', 'CallerLocal_Date'], as_index=False)
group = groups.get_group(list(groups.groups)[0])
然后您可以逐行运行代码:
date_str = str(group['CallerLocal_Date'].iloc[-1]) + ' {0}:00:00'
在generate_df函数
print(type(group))
、print(group)
和print('\n')
)中添加三个print语句,这将让您准确地看到传递的内容。感觉pandas
应该有一种更优雅的方式来完成这一操作。可以是类似于df.groupby(..,debug=True,…)