Question

我有一个简单的数据框df，其中包含一列列表lists。我想基于lists生成另外一列。

df如下：

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看起来像这样：

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，我想对滚动的2个列表进行“求和” / append。注意第1行，因为我只有1个列表1，所以rolllists是该列表。但是在第2行中，我有2个列表要附加。然后对于第三行，追加df[2].lists和df[3].lists等。之前我已经做过类似的工作，请参考：Pandas Dataframe, Column of lists, Create column of sets of cumulative lists, and record by record differences。
另外，如果我们可以在上面获得这一部分，那么我想在groupby中进行操作（因此，下面的示例将是1组，例如，df可能看起来像这样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df.lists.rolling（2）.sum（）之类的各种方法，但出现此错误：

TypeError: cannot handle this type -> object

在pandas 0.24.1中为

，在pandas 0.22.0中为unfortunatley，该命令不会出错，而是返回与lists中相同的值。如此看来，较新版本的Pandas无法汇总列表？这是次要的问题。

爱任何帮助！玩得开心！

Answer 1

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]

Answer 2

怎么样？

rolllists = [df.lists[1].copy()]
for row in df.iterrows():
    index, values = row
    if index > 1:  # or > 0 if zero-indexed
        rolllists.append(df.loc[index - 1, 'lists'] + values['lists'])
df['rolllists'] = rolllists

或者作为一个稍微扩展一些的功能：

lists={1:[[1]],2:[[1,2,3]],3:[[2,9,7,9]],4:[[2,7,3,5]]}
df=pd.DataFrame.from_dict(lists,orient='index')
df=df.rename(columns={0:'lists'})

def rolling_lists(df, roll_period=2):
    new_roll, rolllists = [], [df.lists[1].copy()] * (roll_period - 1)
    for row in df.iterrows():
        index, values = row
        if index > roll_period - 1:  # or -2 if zero-indexed
            res = []
            for i in range(index - roll_period, index):
                res.append(df.loc[i + 1, 'lists'])  # or i if 0-indexed
            rolllists.append(res)
    for li in rolllists:
        while isinstance(li[0], list):
            li = [item for sublist in li for item in sublist]  # flatten nested list
        new_roll.append(li)
    df['rolllists'] = new_roll
    return df

也可以轻松扩展到groupby，只需将其包装在函数中并使用df.apply(rolling_lists)。您可以提供任意数量的滚动行以用作roll_period。希望这会有所帮助！

Python Pandas滚动聚合列表的一列

2 个答案: