Question

我有以下数据框。我想要实现的是，对于每个“已完成日期”的工作，“时间已花费”列中的小时总和应等于7。例如，在6/10/2019，小时总和已为7，因此不需要进行调整。在2019年6月12日，小时总和为4.25，因此我需要插入带有“ Tab_description”“ Difference”的行，该行在“ Time_spent”下的差异为2.75。 6/13/2019和6/14/2019已经达到7，因此在那里无需执行任何操作。对于6/19/2019，我需要执行与6/12/2019相同的操作，插入总和为6的行，以使总和为7。

Date_worked Tab_description Time_spent
    0   6/10/2019   Perform planning procedures 7.0
    1   6/11/2019   Perform planning procedures 7.0
    2   6/12/2019   Time off (away from the office) 2.25
    3   6/12/2019   Staff meeting   1.0
    4   6/12/2019   Accounting & Risk Management Luncheon   1.0
    5   6/13/2019   Perform planning procedures 7.0
    6   6/14/2019   Time off (away from the office) 2.0
    7   6/14/2019   Review policies and procedures  5.0
    8   6/17/2019   Time off (away from the office) 7.0
    9   6/18/2019   Perform planning procedures 7.0
    10  6/19/2019   Staff meeting   1.0
    11  6/20/2019   Time off (away from the office) 2.0
    12  6/21/2019   Time off (away from the office) 1.0
    13  6/24/2019   Staff meeting (FY 20 planning)  7.0
    14  6/25/2019   FCR Kick-off meeting    1.0
    15  6/26/2019   Time off (away from the office) 1.5
    16  6/26/2019   Staff meeting   1.0
    17  6/28/2019   Time off (away from the office) 1.0

Answer 1

有很多方法可以做到这一点，我将使用groupby和concat向您展示。

首先让我们计算总时间和差异，

print(df)
      Date_worked                           Tab_description  Time_spent
0    6/10/2019              Perform planning procedures         7.00
1    6/11/2019              Perform planning procedures         7.00
2    6/12/2019          Time off (away from the office)         0.25
3    6/12/2019                          Staff meeting           1.00
4    6/12/2019  Accounting & Risk Management Luncheon           1.00
5    6/13/2019              Perform planning procedures         7.00
6    6/14/2019          Time off (away from the office)         2.00
7    6/14/2019          Review policies and procedures          5.00
8    6/17/2019          Time off (away from the office)         7.00
9    6/18/2019              Perform planning procedures         7.00
10   6/19/2019                          Staff meeting           1.00
11   6/20/2019          Time off (away from the office)         2.00
12   6/21/2019          Time off (away from the office)         1.00
13   6/24/2019                        Staff meeting (FY         7.00
14   6/25/2019                  FCR Kick-off meeting            1.00
15   6/26/2019          Time off (away from the office)         1.50
16   6/26/2019                          Staff meeting           1.00
17   6/28/2019          Time off (away from the office)         1.00

我们从groupby和一个简单的差和开始，将其分配给一个名为df2的新变量。

df2 = df.groupby('Date_worked')['Time_spent'].sum().reset_index()
df2['variance'] = df2['Time_spent'] - 7.00

我们现在创建您的标签栏并创建您要求的描述，

df2.loc[df2['variance'] != 0, 'Tab_description'] = 'Difference'

然后，我们删除所有NaN行，删除'Time_spent'列，并将“方差”列重命名为concat中的时间。

pd.concat(
    [
        df,
        df2.dropna()
        .drop("Time_spent", axis=1)
        .rename(columns={"variance": "Time_spent"}),
    ],
    sort=False,
)
print(df)


  Date_worked                           Tab_description  Time_spent
0    6/10/2019              Perform planning procedures         7.00
1    6/11/2019              Perform planning procedures         7.00
2    6/12/2019          Time off (away from the office)         0.25
3    6/12/2019                          Staff meeting           1.00
4    6/12/2019  Accounting & Risk Management Luncheon           1.00
5    6/13/2019              Perform planning procedures         7.00
6    6/14/2019          Time off (away from the office)         2.00
7    6/14/2019          Review policies and procedures          5.00
8    6/17/2019          Time off (away from the office)         7.00
9    6/18/2019              Perform planning procedures         7.00
10   6/19/2019                          Staff meeting           1.00
11   6/20/2019          Time off (away from the office)         2.00
12   6/21/2019          Time off (away from the office)         1.00
13   6/24/2019                        Staff meeting (FY         7.00
14   6/25/2019                  FCR Kick-off meeting            1.00
15   6/26/2019          Time off (away from the office)         1.50
16   6/26/2019                          Staff meeting           1.00
17   6/28/2019          Time off (away from the office)         1.00
2    6/12/2019                                Difference       -4.75
7    6/19/2019                                Difference       -6.00
8    6/20/2019                                Difference       -5.00
9    6/21/2019                                Difference       -6.00
11   6/25/2019                                Difference       -6.00
12   6/26/2019                                Difference       -4.50
13   6/28/2019                                Difference       -6.00

计算一组时间戳中的小时总和与固定数量之间的差

1 个答案: