【发布时间】:2017-01-03 03:54:03
【问题描述】:
我正在尝试复制一个“正确填充”类似 excel 的函数,该函数填充值直到下一个值不是 null/nan/empty。仅当紧随其后的行中的值不为空或“nan”时才进行此“右填充”练习。此外,这必须针对每个组进行。我有以下熊猫数据框数据集。我当前的输入表是“有”。我的输出表是“想要的”。
我只是python的初学者。所以任何帮助将不胜感激。 对于那些希望在分组操作中进行此操作的人,数据如下: 表“have”如下,分组字段“groups”:
import pandas as pd
have = pd.DataFrame({ \
"groups": pd.Series(["group1","group1","group1","group2","group2","group2"]) \
,"0": pd.Series(["abc","1","something here","abc2","1","something here"]) \
,"1": pd.Series(["","2","something here","","","something here"]) \
,"2": pd.Series(["","3","something here","","3","something here"]) \
,"3": pd.Series(["something","1","something here","something","1","something here"]) \
,"4": pd.Series(["","2","something here","","2","something here"]) \
,"5": pd.Series(["","","something here","","","something here"]) \
,"6": pd.Series(["","","something here","","","something here"]) \
,"7": pd.Series(["cdf","5","something here","mnop","5","something here"]) \
,"8": pd.Series(["","6","something here","","6","something here"]) \
,"9": pd.Series(["xyz","1","something here","xyz","1","something here"]) \
})
带有分组字段“groups”的表“want”:
import pandas as pd
want = pd.DataFrame({ \
"groups": pd.Series(["group1","group1","group1","group2","group2","group2"]) \
,"0": pd.Series(["abc","1","something here","anything","1","something here"]) \
,"1": pd.Series(["abc","2","something here"," anything ","2","something here"]) \
,"2": pd.Series(["abc","3","something here"," anything ","3","something here"]) \
,"3": pd.Series(["something","1","something here","","","something here"]) \
,"4": pd.Series(["something ","2","something here","","","something here"]) \
,"5": pd.Series(["","","something here","","","something here"]) \
,"6": pd.Series(["","","something here","","","something here"]) \
,"7": pd.Series(["cdf","5","something here","mnop","5","something here"]) \
,"8": pd.Series(["cdf ","6","something here"," mnop ","6","something here"]) \
,"9": pd.Series(["xyz","1","something here","xyz","1","something here"]) \
})
我尝试使用此代码,但我仍在尝试熟悉 groupby 和 apply 语句:
grouped=have.groupby('groups')
have.groupby('groups').apply(lambda g: have.loc[g].isnull() )
#cond = have.loc[1].isnull() | have.loc[1].ne('')
want.loc[0, cond] = want.loc[0, cond].str.strip().replace('', None)
want
【问题讨论】: