如何使用python重新排列数据框行？答案

【问题标题】：How to rearrange row of data frame using python?如何使用python重新排列数据框行？
【发布时间】：2020-08-16 19:06:37
【问题描述】：

我有一个数据框，其中数据位于另一列中，我想从该列中获取这些日期并创建一个日期列并存储它们。这是我的示例数据。

df=[['Monday, 13 January 2020','',''],['Task 1',13588,'Jack'],['','','Address 1'],['','','City 1'],['Task 2',13589,'Ammie'],['','','Address 2'],['','','City'],['Task 3',13589,'Amanda'],['','','Address 3'],['','','City 3'],['Tuesday, 14 January 2020','',''],['Task 4',13587,'Chelsea'],['','','Address 4'],['','','City 4'],['Task 5','13586','Ibrahim'],['','','Address 5'],['','','City 5'],['Task 6',13585,'Kate'],['','','Address 6'],['','','City 6']]

df=pd.DataFrame(df)
df.columns = ['Task','ID','Supervisor']
df=df.replace(np.nan,'')
df

    Task    ID  Supervisor
0   Monday, 13 January 2020     
1   Task 1  13588   Jack
2           Address 1
3           City 1
4   Task 2  13589   Ammie
5           Address 2
6           City
7   Task 3  13589   Amanda
8           Address 3
9           City 3
10  Tuesday, 14 January 2020        
11  Task 4  13587   Chelsea
12          Address 4
13          City 4
14  Task 5  13586   Ibrahim
15          Address 5
16          City 5
17  Task 6  13585   Kate
18          Address 6
19          City 6

我想得到以下输出。

    Date                    Task    ID      Supervisor
0 Monday, 13 January 2020   Task 1  13588   Jack Address 1 City 1
1 Monday, 13 January 2020   Task 2  13589   Ammie Address 2 City
2 Monday, 13 January 2020   Task 3  13589   Amanda Address 3 City 3
3 Tuesday, 14 January 2020  Task 4  13587   Chelsea Address 4 City 4
4 Tuesday, 14 January 2020  Task 5  13586   Ibrahim Address 5 City 5
5 Tuesday, 14 January 2020  Task 6  13585   Kate Address 6 City 6

这是我的尝试。

def rowMerger(a,b):
    try:
        rule1 = lambda x: x not in ['']
        u = a.loc[a.iloc[:,0].apply(rule1) & a.iloc[:,1].apply(rule1) & a.iloc[:,2].apply(rule1)].index
        print(u)
        findMergerindexs = list(u)
        findMergerindexs.sort()
        a = pd.DataFrame(a)
        tabcolumns = pd.DataFrame(a.columns)
        totalcolumns = len(tabcolumns)
        b = pd.DataFrame(columns = list(tabcolumns))
        if (len(findMergerindexs) > 0):
            for m in range(len(findMergerindexs))
                if not (m == (len(findMergerindexs)-1)): 
                    startLoop = findMergerindexs[m]
                    endLoop = findMergerindexs[m+1]
                else:
                    startLoop = findMergerindexs[m]
                    endLoop = len(a)
                listValues = []
                for i in range(totalcolumns):
                    value = ' '
                    for n in range(startLoop,endLoop):
                        value = value + ' ' + str(a.iloc[n,i])
                    listValues.insert(i,(value.strip()))
                b = b.append(pd.Series(listValues),ignore_index = True)
        else:
            print("File is not having a row for merging instances - Please check the file manually for instance - ")
        return b
    except: 
        print("Error - While merging the rows")
    return b

这段代码给了我下面的输出。

rowMerger(df,0)
       0       1             2
0   Task 1  13588   Jack Address 1 City 1
1   Task 2  13589   Ammie Address 2 City
2   Task 3 Tuesday, 14 January 2020 13589   Amanda Address 3 City 3
3   Task 4  13587   Chelsea Address 4 City 4
4   Task 5  13586   Ibrahim Address 5 City 5
5   Task 6  13585   Kate Address 6 City 6

但问题是这段代码只会合并行。不确定如何在所需输出中显示的各行中复制日期并将其放在不同的列中。谁能帮助我实现这一目标？

【问题讨论】：

如何在您的第一个代码块的示例中使用 rowMerger(a,b)。 b是什么？
b 是函数返回的数据框。该函数有两个参数，数据框和 0。rowMerger(df,0) 将为我们提供我上面显示的结果。

标签： python pandas dataframe date conda

【解决方案1】：

所以本质上我们使用 lambda 将日期与任务编号分开，pd.Series.fillna(method='ffill') 填写最后一个有效日期。

所以我们要添加这些行：

# Split Date column
df['Date'] = df.apply(lambda x: " ".join(x[0].split(' ')[2:]) if len(x[0].split(' ')) > 2 else np.nan,axis=1).fillna(method='ffill')

# Clean Task column
df['Task'] = df.apply(lambda x: " ".join(x[0].split(' ')[:2]) if len(x[0].split(' ')) > 1 else x[0],axis=1)

# Rename and reorder remaining columns
df['ID'] = df[1]
df['Supervisor'] = df[2]
df = df[['Date','Task','ID','Supervisor']]

【讨论】：

【解决方案2】：

您可以尝试以下方法：

task_mask = df.Task.str.match("Task\s+\d")
df.assign(Task = df.Task[task_mask],
          Date = pd.Series(np.where(~task_mask, df["Task"], np.NaN)).shift()) \
        .replace("", np.NaN) \
        .dropna(how='all') \
        .ffill() \
        .groupby(["Task", "ID", "Date"]).agg({"Supervisor": lambda x: " ".join(x)}) \
        .reset_index()

输出

#      Task     ID                      Date                Supervisor
# 0  Task 1  13588   Monday, 13 January 2020     Jack Address 1 City 1
# 1  Task 2  13589   Monday, 13 January 2020      Ammie Address 2 City
# 2  Task 3  13589   Monday, 13 January 2020   Amanda Address 3 City 3
# 3  Task 4  13587  Tuesday, 14 January 2020  Chelsea Address 4 City 4
# 4  Task 5  13586  Tuesday, 14 January 2020  Ibrahim Address 5 City 5
# 5  Task 6  13585  Tuesday, 14 January 2020     Kate Address 6 City 6

解释：

过滤Task 列：dates 和task id。
- 一种解决方案是使用正则表达式匹配task id。 pandas.Series.str.match 完成这项工作。使用的正则表达式非常简单："Task\s+\d" 表示 Task + 任意空格 + 数字。

task_mask = df.Task.str.match("Task\s+\d")

从这个掩码中，我们可以提取Date 和Tasks。通过task_mask 和df.Task[task_mask] 可以轻松访问这些任务
Dates 更难提取。
- 我们使用np.where 设置Task 值或NaN。
- 然后，我们将这个array 转换成pd.Series
- 最后，我们使用shift 将所有值移1。移动行让我们可以在第 5 步中轻松删除 NaN 行。

pd.Series(np.where(~task_mask, df["Task"], np.NaN)).shift()

使用replace将所有空字符串替换为NaN
使用dropna 和how="all" 删除空行（例如，只有Date 的旧行）
使用ffill 将所有NaN 值填充为以前的非NaN 值
按"Task", "ID"a and "Date" 分组并使用agg 聚合行。聚合函数基于str.join:lambda x: " ".join(x)
使用reset_index从groupby重置索引。

希望这很清楚！

代码+插图

# Create dataframe
data = [['Monday, 13 January 2020', '', ''], ['Task 1', 13588, 'Jack'], ['', '', 'Address 1'], ['', '', 'City 1'], ['Task 2', 13589, 'Ammie'], ['', '', 'Address 2'], ['', '', 'City'], ['Task 3', 13589, 'Amanda'], ['', '', 'Address 3'], ['', '', 'City 3'], [
    'Tuesday, 14 January 2020', '', ''], ['Task 4', 13587, 'Chelsea'], ['', '', 'Address 4'], ['', '', 'City 4'], ['Task 5', '13586', 'Ibrahim'], ['', '', 'Address 5'], ['', '', 'City 5'], ['Task 6', 13585, 'Kate'], ['', '', 'Address 6'], ['', '', 'City 6']]
df = pd.DataFrame(data)
df.columns = ['Task', 'ID', 'Supervisor']
print(df)

# Step 1
task_mask = df.Task.str.match("Task\s+\d")
print(task_mask)
# 0     False
# 1      True
# 2     False
# 3     False
# 4      True
# 5     False
# 6     False
# 7      True
# 8     False
# 9     False
# 10    False
# 11     True
# 12    False
# 13    False
# 14     True
# 15    False
# 16    False
# 17     True
# 18    False
# 19    False
# Name: Task, dtype: bool

# Step 2
print(df.Task[task_mask])
# 1     Task 1
# 4     Task 2
# 7     Task 3
# 11    Task 4
# 14    Task 5
# 17    Task 6
# Name: Task, dtype: object

# Step 3
print(pd.Series(np.where(~task_mask, df["Task"], np.NaN)).shift())
# 0                          NaN
# 1      Monday, 13 January 2020
# 2                          NaN
# 3
# 4
# 5                          NaN
# 6
# 7
# 8                          NaN
# 9
# 10
# 11    Tuesday, 14 January 2020
# 12                         NaN
# 13
# 14
# 15                         NaN
# 16
# 17
# 18                         NaN
# 19
# dtype: object

# Step 4
print(df.assign(Task=df.Task[task_mask],
                Date=pd.Series(np.where(~task_mask, df["Task"], np.NaN)).shift())
        .replace("", np.NaN))
#       Task     ID Supervisor                      Date
# 0      NaN    NaN        NaN                       NaN
# 1   Task 1  13588       Jack   Monday, 13 January 2020
# 2      NaN    NaN  Address 1                       NaN
# 3      NaN    NaN     City 1                       NaN
# 4   Task 2  13589      Ammie                       NaN
# 5      NaN    NaN  Address 2                       NaN
# 6      NaN    NaN       City                       NaN
# 7   Task 3  13589     Amanda                       NaN
# 8      NaN    NaN  Address 3                       NaN
# 9      NaN    NaN     City 3                       NaN
# 10     NaN    NaN        NaN                       NaN
# 11  Task 4  13587    Chelsea  Tuesday, 14 January 2020
# 12     NaN    NaN  Address 4                       NaN
# 13     NaN    NaN     City 4                       NaN
# 14  Task 5  13586    Ibrahim                       NaN
# 15     NaN    NaN  Address 5                       NaN
# 16     NaN    NaN     City 5                       NaN
# 17  Task 6  13585       Kate                       NaN
# 18     NaN    NaN  Address 6                       NaN
# 19     NaN    NaN     City 6                       NaN

# Step 5:
print(df.assign(Task = df.Task[task_mask],
                Date = pd.Series(np.where(~task_mask, df["Task"], np.NaN)).shift()) \
        .replace("", np.NaN) \
        .dropna(how='all'))
#       Task     ID Supervisor                      Date
# 1   Task 1  13588       Jack   Monday, 13 January 2020
# 2      NaN    NaN  Address 1                       NaN
# 3      NaN    NaN     City 1                       NaN
# 4   Task 2  13589      Ammie                       NaN
# 5      NaN    NaN  Address 2                       NaN
# 6      NaN    NaN       City                       NaN
# 7   Task 3  13589     Amanda                       NaN
# 8      NaN    NaN  Address 3                       NaN
# 9      NaN    NaN     City 3                       NaN
# 11  Task 4  13587    Chelsea  Tuesday, 14 January 2020
# 12     NaN    NaN  Address 4                       NaN
# 13     NaN    NaN     City 4                       NaN
# 14  Task 5  13586    Ibrahim                       NaN
# 15     NaN    NaN  Address 5                       NaN
# 16     NaN    NaN     City 5                       NaN
# 17  Task 6  13585       Kate                       NaN
# 18     NaN    NaN  Address 6                       NaN
# 19     NaN    NaN     City 6                       NaN

# Step 6:
print(df.assign(Task = df.Task[task_mask],
                 Date = pd.Series(np.where(~task_mask, df["Task"], np.NaN)).shift()) \
            .replace("", np.NaN) \
            .dropna(how='all') \
            .ffill())
#       Task     ID Supervisor                      Date
# 1   Task 1  13588       Jack   Monday, 13 January 2020
# 2   Task 1  13588  Address 1   Monday, 13 January 2020
# 3   Task 1  13588     City 1   Monday, 13 January 2020
# 4   Task 2  13589      Ammie   Monday, 13 January 2020
# 5   Task 2  13589  Address 2   Monday, 13 January 2020
# 6   Task 2  13589       City   Monday, 13 January 2020
# 7   Task 3  13589     Amanda   Monday, 13 January 2020
# 8   Task 3  13589  Address 3   Monday, 13 January 2020
# 9   Task 3  13589     City 3   Monday, 13 January 2020
# 11  Task 4  13587    Chelsea  Tuesday, 14 January 2020
# 12  Task 4  13587  Address 4  Tuesday, 14 January 2020
# 13  Task 4  13587     City 4  Tuesday, 14 January 2020
# 14  Task 5  13586    Ibrahim  Tuesday, 14 January 2020
# 15  Task 5  13586  Address 5  Tuesday, 14 January 2020
# 16  Task 5  13586     City 5  Tuesday, 14 January 2020
# 17  Task 6  13585       Kate  Tuesday, 14 January 2020
# 18  Task 6  13585  Address 6  Tuesday, 14 January 2020
# 19  Task 6  13585     City 6  Tuesday, 14 January 2020

# Step 7
print(df.assign(Task = df.Task[task_mask],
                Date = pd.Series(np.where(~task_mask, df["Task"], np.NaN)).shift()) \
        .replace("", np.NaN) \
        .dropna(how='all') \
        .ffill() \
        .groupby(["Task", "ID", "Date"]).agg({"Supervisor": lambda x: " ".join(x)}))
#                                                      Supervisor
# Task   ID    Date
# Task 1 13588 Monday, 13 January 2020      Jack Address 1 City 1
# Task 2 13589 Monday, 13 January 2020       Ammie Address 2 City
# Task 3 13589 Monday, 13 January 2020    Amanda Address 3 City 3
# Task 4 13587 Tuesday, 14 January 2020  Chelsea Address 4 City 4
# Task 5 13586 Tuesday, 14 January 2020  Ibrahim Address 5 City 5
# Task 6 13585 Tuesday, 14 January 2020     Kate Address 6 City 6

# Step 8
df = df.assign(Task = df.Task[task_mask],
               Date = pd.Series(np.where(~task_mask, df["Task"], np.NaN)).shift()) \
        .replace("", np.NaN) \
        .dropna(how='all') \
        .ffill() \
        .groupby(["Task", "ID", "Date"]).agg({"Supervisor": lambda x: " ".join(x)}) \
        .reset_index()
print(df)

#      Task     ID                      Date                Supervisor
# 0  Task 1  13588   Monday, 13 January 2020     Jack Address 1 City 1
# 1  Task 2  13589   Monday, 13 January 2020      Ammie Address 2 City
# 2  Task 3  13589   Monday, 13 January 2020   Amanda Address 3 City 3
# 3  Task 4  13587  Tuesday, 14 January 2020  Chelsea Address 4 City 4
# 4  Task 5  13586  Tuesday, 14 January 2020  Ibrahim Address 5 City 5
# 5  Task 6  13585  Tuesday, 14 January 2020     Kate Address 6 City 6

【讨论】：

【解决方案3】：

@Alexandre 的回答很棒 - 这是一个替代方案，我避免了正则表达式提取和移位：

#convert empty cells to null
(df.replace("",np.nan)
 #create a new column containing only the dates
 #we'll use the null cells in Supervisor column to pick out the dates
 .assign(Date = lambda x: x.loc[x.Supervisor.isna(),'Task'])
 .ffill()
 .dropna(subset=['ID'])
 #drop Dates found in the Task column
 .query('Task != Date')
 .groupby(['Date','Task','ID'],as_index=False)
 .Supervisor.agg(' '.join)
)

        Date                    Task      ID      Supervisor
0   Monday, 13 January 2020     Task 1  13588   Jack Address 1 City 1
1   Monday, 13 January 2020     Task 2  13589   Ammie Address 2 City
2   Monday, 13 January 2020     Task 3  13589   Amanda Address 3 City 3
3   Tuesday, 14 January 2020    Task 4  13587   Chelsea Address 4 City 4
4   Tuesday, 14 January 2020    Task 5  13586   Ibrahim Address 5 City 5
5   Tuesday, 14 January 2020    Task 6  13585   Kate Address 6 City 6

【讨论】：