Question

我有一个包含索引，功能和时间数据的数据框，但时间数据在一列中，如下所示：

ID日期功能

1 date1 feature1

2 date2 feature2

1 date2 feature3

我想将其转化为：

ID日期功能

1 date1 feature1 date2 feature3

2 date2 feature2 NaN NaN

通过明确定义数据帧，查询和连接已经做到了这一点，但未能找到动态方式。我写的：

df = pd.read_excel('some path')

import pandas as pd

list1 = []
list2 = []
list3 = []

def placeholder_lists():
    for i in range(7):
        if len(str(i)) == 1:
            if i not in [8,9]:
                i = "0"+str(i+3)
            else:
                i = str(i+3)
        else:
            i = str(i+3)
        list1.append(i)

    for l in range(7):
        if len(str(l)) == 1:
            if l not in [10,9]:
                l = "0"+str(l+2)
            else:
                l = str(l+2)
        else:
            l = str(l+2)
        list2.append(l)

    for g in range(7):
        if len(str(g)) == 1:
            if g not in [9,8]:
                g = "0"+str(g+1)
            else:
                g = str(g+1)
        else:
            g = str(g+1)
        list3.append(g)

placeholder_lists()

for m,n,u in zip(list1,list2, list3):
    df01 = df.query('dw_creation_date == "01-AUG-17" ')
    e = str(u)+"-AUG-17"

    currentdf = df.query('dw_creation_date == "%s"' % e)

    if 1 == "01":
        currentdf = df01
    first = "df"+m
    second = "df"+n
    listie = range(50)
    first = second.join(currentdf.set_index('unique_identifier'), on='unique_identifier', lsuffix = listie[n])

......我得到的错误：

first = second.join(currentdf.set_index('unique_identifier'), lsuffix = listie[n])

TypeError：join（）不带关键字参数

有什么想法吗？

Answer 1

cols = ['id','date','feature']
df = pd.DataFrame({'date': {0: 'date1', 1: 'date2', 2: 'date2'}, 
                   'id': {0: 1, 1: 2, 2: 1}, 
                  'feature': {0: 'feature1', 1: 'feature2', 2: 'feature3'}}, columns=cols)

print (df)
   id   date   feature
0   1  date1  feature1
1   2  date2  feature2
2   1  date2  feature3

您可以id df并申请新的Multiindex。然后按groupby重新整形，并按Multiindex的第二级unstack对列进行排序。

列中的最后展开df = df.groupby('id')['date','feature'] \ .apply(lambda x: pd.DataFrame(x.values, columns=['feature','date'])) \ .unstack() \ .sort_index(1, level=1) print (df) feature date feature date 0 0 1 1 id 1 date1 feature1 date2 feature3 2 date2 feature2 None None df.columns = ['{0[0]}_{0[1]}'.format(x) for x in df.columns] df = df.reset_index() print (df) id feature_0 date_0 feature_1 date_1 0 1 date1 feature1 date2 feature3 1 2 date2 feature2 None None和sort_index。

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      th:action="@{ + ${controllerName} + '/deleteAssignment'}" 
      name="assignmentDeleteForm"
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将pandas dataframe列分解为单个df中的多个列

1 个答案: