Python Pandas滚动聚合一列列表
问题描述
我有一个简单的dataframe df,其中有一列列表lists
。我想根据lists
生成一个附加列。
df
如下所示:
import pandas as pd
lists={1:[[1]],2:[[1,2,3]],3:[[2,9,7,9]],4:[[2,7,3,5]]}
#create test dataframe
df=pd.DataFrame.from_dict(lists,orient='index')
df=df.rename(columns={0:'lists'})
df
lists
1 [1]
2 [1, 2, 3]
3 [2, 9, 7, 9]
4 [2, 7, 3, 5]
我希望df
如下所示:
df
Out[9]:
lists rolllists
1 [1] [1]
2 [1, 2, 3] [1, 1, 2, 3]
3 [2, 9, 7, 9] [1, 2, 3, 2, 9, 7, 9]
4 [2, 7, 3, 5] [2, 9, 7, 9, 2, 7, 3, 5]
基本上我想‘求和’/append
滚动的2个列表。请注意,第1行,因为我只有1个列表1,所以滚动列表就是那个列表。但是在第2行,我有2个我想要追加的列表。然后第三行,加上df[2].lists
和df[3].lists
等,我以前做过类似的事情,参考这个:Pandas Dataframe, Column of lists, Create column of sets of cumulative lists, and record by record differences。此外,如果我们可以获得上面的这一部分,那么我想在
groupby
中执行此操作(例如,下面的示例将是1组,因此,例如df
中的df
可能类似于groupby
):
Group lists rolllists
1 A [1] [1]
2 A [1, 2, 3] [1, 1, 2, 3]
3 A [2, 9, 7, 9] [1, 2, 3, 2, 9, 7, 9]
4 A [2, 7, 3, 5] [2, 9, 7, 9, 2, 7, 3, 5]
5 B [1] [1]
6 B [1, 2, 3] [1, 1, 2, 3]
7 B [2, 9, 7, 9] [1, 2, 3, 2, 9, 7, 9]
8 B [2, 7, 3, 5] [2, 9, 7, 9, 2, 7, 3, 5]
我尝试了各种方法,如df.lists.Rolling(2).sum(),得到以下错误:
TypeError: cannot handle this type -> object
在Pandas 0.24.1中和在Pandas 0.22.0中不走运,该命令不会出错,而是返回与lists
中完全相同的值。所以看起来新版本的 pandas 不能把名单加起来?那是次要问题。
谢谢您的帮助!玩得开心!
解决方案
您可以从
开始import pandas as pd
mylists={1:[[1]],2:[[1,2,3]],3:[[2,9,7,9]],4:[[2,7,3,5]]}
mydf=pd.DataFrame.from_dict(mylists,orient='index')
mydf=mydf.rename(columns={0:'lists'})
mydf = pd.concat([mydf, mydf], axis=0, ignore_index=True)
mydf['group'] = ['A']*4 + ['B']*4
# initialize your new series
mydf['newseries'] = mydf['lists']
# define the function that appends lists overs rows
def append_row_lists(data):
for i in data.index:
try: data.loc[i+1, 'newseries'] = data.loc[i, 'lists'] + data.loc[i+1, 'lists']
except: pass
return data
# loop over your groups
for gp in mydf.group.unique():
condition = mydf.group == gp
mydf[condition] = append_row_lists(mydf[condition])
输出
lists Group newseries
0 [1] A [1]
1 [1, 2, 3] A [1, 1, 2, 3]
2 [2, 9, 7, 9] A [1, 2, 3, 2, 9, 7, 9]
3 [2, 7, 3, 5] A [2, 9, 7, 9, 2, 7, 3, 5]
4 [1] B [1]
5 [1, 2, 3] B [1, 1, 2, 3]
6 [2, 9, 7, 9] B [1, 2, 3, 2, 9, 7, 9]
7 [2, 7, 3, 5] B [2, 9, 7, 9, 2, 7, 3, 5]
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