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]
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。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！
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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