Question

我被困住了。在下面的代码中，我可以成功创建要查找的小计数据透视表，但不能产生总计。 [以下代码利用了arcgis模块；只需将表（在本例中为MSSQL表）转换为NumPy数组]

import numpy as np
import pandas as pd
import arcpy

table = "\\\\filserver\\MAP_PROJECTS\\LV_WEB\\SDE_CONNECTIONS\\LV_NEXUS.sde\\LV_NEXUS.DATAOWNER.NORTHEAST\\LV_NEXUS.DATAOWNER.NE_HARVEST_OPS"
HUID = "669-NMTC-139"
whereClause = """ "LV_HARVEST_UNIT_ID" = '{0}' """.format(HUID)
tableArray = arcpy.da.TableToNumPyArray(table, ['STAND_NUMB', 'SUPER_TYPE','STRATA', 'OS_TYPE', 'SILV_PRES', 'ACRES'], where_clause = whereClause)
df = pd.DataFrame(tableArray)
report = df.groupby(['SUPER_TYPE']).apply(lambda sub_df:  sub_df.pivot_table(index=['STRATA', 'OS_TYPE', 'STAND_NUMB', 'SILV_PRES'], values=['ACRES'],aggfunc=np.sum, margins=True,margins_name= 'TOTAL'))
np.round(report,1)

这为每个“ SUPER_TYPE”组提供了总计，但是我无法创建总计。我尝试了以下方法：

grandtotal = np.round(np.sum(report),1)
grandtotal.name = 'Grand Total'
report.append(grandtotal)

那只会造成可怕的混乱。它附加了总计，但破坏了我的数据框的格式。

粘贴在下面的数据框不确定如何保留格式

   STAND_NUMB SUPER_TYPE   STRATA OS_TYPE    SILV_PRES      ACRES
0        3113         SH     SH3B    SH3B  OSR/2SS/SCC   0.612748
1        3608         HW     H3AB     H3B  OSR/2SS/SCC  12.936038
2        3105         HW     H3AB     H3B  OSR/2SS/SCC  35.199887
3        3607         HS     HS3B    HS3B  OSR/2SS/SCC  27.683348
4        3601         HW     H3AB     H3A  OSR/2SS/SCC  11.941338
5        3603         HW      H4B     H4B  OSR/2SS/SCC  25.307238
6        3092         HS     HS3B    HS3B  OSR/2SS/SCC  17.331220
7        3600         HW      H4B     H4B  OSR/2SS/SCC  13.443112
8        3596         HW     H3AB     H3B  OSR/2SS/SCC  12.375962
9        3597         HW     H3AB     H3A  OSR/2SS/SCC  41.639072
10       3591         SW     S4BC     S4A  OSR/2SS/SCC  11.355869
11       3594         HS     HS3B    HS3B  OSR/2SS/SCC  31.747874
12       3586         HW     H3AB     H3B  OSR/2SS/SCC  19.437834
13       3588         HW     H3AB     H3A  OSR/2SS/SCC  18.129702
14       3587         HS     HS3B    HS3B  OSR/2SS/SCC  13.788853
15       3585         HW     H3AB     H3A  OSR/2SS/SCC  25.775322
16       3582         SH     SH3B    SH3B  OSR/2SS/SCC  11.026199
17       3581         HS     HS3B    HS3B  OSR/2SS/SCC  16.634195
18       3589         HW     H3AB     H3A  OSR/2SS/SCC  54.684222
19       3579         SH     SH3B    SH3B  OSR/2SS/SCC  17.313354
20       3578         HW  H4C_H2C     H4C  OSR/2SS/SCC  30.255013
21       3576         HW      H3C     H3C  OSR/2SS/SCC  11.310230
22       3573         HW     H3AB     H3A  OSR/2SS/SCC  30.369559
23       3575         HW  H4C_H2C     H4C  OSR/2SS/SCC  53.088547
24       3569         HW      H4A     H4A  OSR/2SS/SCC  12.809001
25       3567         HW      H4B     H4B  OSR/2SS/SCC  24.026682
26       3568         HW     H3AB     H3B  OSR/2SS/SCC  57.934207
27       3565         HW      H4B     H4B  OSR/2SS/SCC  33.545768
28       3605         HW     H3AB     H3B  OSR/2SS/SCC  74.424945
29       3580         HS     HS3B    HS3B  OSR/2SS/SCC   8.062028
30       3571         HW     H3AB     H3A  OSR/2SS/SCC  30.718121
31       3562         HW     H3AB     H3B  OSR/2SS/SCC  22.774026
32       3110         SW      S3C     S3C  OSR/2SS/SCC   2.240600
33     3120.1         SH     SH3B    SH3B  OSR/2SS/SCC   3.726728

Answer 1

根据您发布的示例：

def integration(func, a, n, r, n_steps):
    z = r * np.exp(2j * np.pi * np.arange(0, 1, 1. / n_steps))
    return math.factorial(n) * np.mean(func(a + z) / z**n)

np.allclose(1., integration(np.sin, a=0., n=1, r=1., n_steps=100))

True

将总计添加到Pandas数据透视表

1 个答案: