Python 在Pivot和x27之间添加计算字段;列';基于';价值观';数据

Python 在Pivot和x27之间添加计算字段;列';基于';价值观';数据,python,pandas,pivot-table,Python,Pandas,Pivot Table,我有一份报告,我正在工作,以显示两个季度之间的差异。我有一个SQL查询,我正在读取到一个数据帧中,然后进行数据透视 这是我的代码: df = pd.read_sql_query(mtd_query, cnxn, params=[report_start, end_mtd, report_start, end_mtd, whse]) ##(m-1)//3 + 1 Determine which Quarter each month is ## Create the "Pe

我有一份报告,我正在工作,以显示两个季度之间的差异。我有一个SQL查询,我正在读取到一个数据帧中,然后进行数据透视

这是我的代码:

    df = pd.read_sql_query(mtd_query, cnxn, params=[report_start, end_mtd, report_start, end_mtd, whse])
    ##(m-1)//3 + 1  Determine which Quarter each month is
    ## Create the "Period" column by combining the Quater and the Month
    df['QUARTER'] = (df['INV_MONTH'].astype(int) - 1)//3 + 1
    df['PERIOD'] = df['INV_YEAR'].astype(str) + 'Q' + df['QUARTER'].astype(int).astype(str)
    df['MARGIN'] = (df['PROFIT'].astype(float) / df['SALES'].astype(float))

    df = df.drop('INV_MONTH', axis=1)
    df = df.drop('INV_YEAR', axis=1)
    df = pd.pivot_table(df, index=['REP', 'REP_NAME', 'CUST_NO', 'CUST_NAME', 'TOTALSALES'], columns=['PERIOD'], values=['SALES', 'PROFIT', 'MARGIN'], fill_value=0)
    df = df.reorder_levels([1, 0], axis=1).sort_index(axis=1, ascending=False)
    df = df.sortlevel(level=0, ascending=True)
我试图确定“期间”和“保证金”列之间的差异。我一直找不到任何方法来实现这一点。如有任何建议,我们将不胜感激

当前输出显示:

PERIOD                                                                                            2017Q4                                 2017Q3                                 2017Q2                                 2017Q1                                 2016Q4                        
                                                                                                   SALES        PROFIT    MARGIN          SALES        PROFIT    MARGIN          SALES        PROFIT    MARGIN          SALES        PROFIT    MARGIN          SALES        PROFIT    MARGIN
REP    REP_NAME                       CUST_NO  CUST_NAME                      TOTALSALES                                                                                                                                                                                                    
1.0    Greensboro - House             245.0    TE CONNECTIVITY CORPORATION    103361.05         0.000000      0.000000  0.000000     434.500000     69.520000  0.160000   20391.666667   3262.666667  0.160000       0.000000      0.000000  0.000000       0.000000      0.000000  0.000000
                                      1789.0   GOOD HOUSEKEEPER               50108.47        678.508182     80.170909  0.145883     585.301429     64.180476  0.121915     718.685000     92.033125  0.130453     720.729333     97.955333  0.134821    1237.308333     88.210000  0.099450
所需的输出如下所示:

PERIOD                                                                                            2017Q4                                 2017Q3                                 2017Q2                                 2017Q1                                 2016Q4                        
                                                                                                   SALES        PROFIT    MARGIN   VARIANCE          SALES        PROFIT    MARGIN    VARIANCE          SALES        PROFIT    MARGIN    VARIANCE          SALES        PROFIT    MARGIN    VARIANCE          SALES        PROFIT    MARGIN
REP    REP_NAME                       CUST_NO  CUST_NAME                      TOTALSALES                                                                                                                                                                                                    
1.0    Greensboro - House             245.0    TE CONNECTIVITY CORPORATION    103361.05         0.000000      0.000000  0.000000    -.16         434.500000     69.520000  0.160000    0           20391.666667   3262.666667  0.160000    .16           0.000000      0.000000  0.000000      0            0.000000      0.000000  0.000000
                                      1789.0   GOOD HOUSEKEEPER               50108.47        678.508182     80.170909  0.145883    .023968     585.301429     64.180476  0.121915    -0.008537     718.685000     92.033125  0.130453    -.004368     720.729333     97.955333  0.134821     .035372       1237.308333     88.210000  0.099450
如下所示:

IIUC:

资料来源:

In [60]: df
Out[60]:
  2016Q4                     2017Q1                  2017Q2               \
  MARGIN PROFIT        SALES MARGIN     PROFIT SALES MARGIN       PROFIT
0    0.0   0.00     0.000000    0.0   0.000000   0.0   0.16  3262.666667
1    NaN  88.21  1237.308333    NaN  97.955333   NaN    NaN          NaN

                   2017Q3                   2017Q4
          SALES    MARGIN     PROFIT  SALES MARGIN PROFIT SALES
0  20391.666667  0.160000  69.520000  434.5    0.0    0.0   0.0
1    718.685000  0.121915  64.180476    NaN    NaN    NaN   NaN
解决方案:

In [61]: tmp = (df.loc[:, pd.IndexSlice[:, 'MARGIN']]
    ...:          .fillna(0)
    ...:          .diff(axis=1)
    ...:          .rename(columns=lambda x: 'VARIANCE' if x=='MARGIN' else x))
    ...:

In [62]: pd.concat([df, tmp], axis=1).sort_index(axis=1)
Out[62]:
  2016Q4                              2017Q1                           2017Q2  \
  MARGIN PROFIT        SALES VARIANCE MARGIN     PROFIT SALES VARIANCE MARGIN
0    0.0   0.00     0.000000      NaN    0.0   0.000000   0.0      0.0   0.16
1    NaN  88.21  1237.308333      NaN    NaN  97.955333   NaN      0.0    NaN

                                         2017Q3                              \
        PROFIT         SALES VARIANCE    MARGIN     PROFIT  SALES  VARIANCE
0  3262.666667  20391.666667     0.16  0.160000  69.520000  434.5  0.000000
1          NaN    718.685000     0.00  0.121915  64.180476    NaN  0.121915

  2017Q4
  MARGIN PROFIT SALES  VARIANCE
0    0.0    0.0   0.0 -0.160000
1    NaN    NaN   NaN -0.121915

我无法使用上述解决方案。
所以我这样做了:

df['MARGIN'] = (df['PROFIT'].astype(float) / df['SALES'].astype(float))
df['MARGIN'] = df['MARGIN'].astype(float)
df['PREV_MARGIN'] = df['MARGIN'].shift(-1)
df['VARIANCE'] = df['MARGIN'] - df['PREV_MARGIN']
df = df.drop('PREV_MARGIN', axis=1)

这为我提供了完成工作所需的数据

@MaxU Done。。。很抱歉,我需要手动修改:)请您为您的源(数据透视)df发布一个
print(df.to_dict('r'))
输出,因为我们需要相当长的时间来复制这个多索引、多列df…@MaxU done。再次感谢!它不喜欢df.loc,所以我尝试了df.iloc,现在我在切片上遇到了错误:
TypeError:unorderable types:Slice()>=str()
,我不知道如何解决这个问题。尝试在边距列上执行
.asype(float)
,但没有成功。
df['MARGIN'] = (df['PROFIT'].astype(float) / df['SALES'].astype(float))
df['MARGIN'] = df['MARGIN'].astype(float)
df['PREV_MARGIN'] = df['MARGIN'].shift(-1)
df['VARIANCE'] = df['MARGIN'] - df['PREV_MARGIN']
df = df.drop('PREV_MARGIN', axis=1)