Python上对多列、行级和非列级的求和
Python上对多列、行级和非列级的求和
提问于 2022-10-08 14:50:23
我想找个办法：
COUNTIF(Col2,Col4,Col6,Col8,Col10,Col12,Col14,Col16,Col18,">=0.05")
SUMIF(Col2,Col4,Col6,Col8,Col10,Col12,Col14,Col16,Col18,">=0.05")
我的尝试：
import pandas as pd
df=pd.read_excel(r'C:\\Users\\Downloads\\Prepped.xls') #Please use: https://github.com/BeboGhattas/temp-repo/blob/main/Prepped.xls
df.iloc[:, [2,4,6,8,10,12,14,16,18]].astype(float) #changing dtype to float
#unconditional sum
df['sum']=df.iloc[:, [2,4,6,8,10,12,14,16,18]].astype(float).sum(axis=1)下面的一切都是行不通的
#sum if
df['greater-than-0.05']=df.iloc[:, [2,4,6,8,10,12,14,16,18]].astype(float).sum([c for c in col if c >= 0.05])| | # | word | B64684807 | B64684807Measure | B649845471 | B649845471Measure | B83344143 | B83344143Measure | B67400624 | B67400624Measure | B85229235 | B85229235Measure | B85630406 | B85630406Measure | B82615898 | B82615898Measure | B87558236 | B87558236Measure | B00000009 | B00000009Measure | 有效竞品数 | 关键词抓取时间 | 搜索量排名 | 月搜索量 | 在售商品数 | 竞争度 |
|---:|----:|:--------|------------:|:-------------------|-------------:|:-------------------------|------------:|:-------------------------|------------:|:-------------------|------------:|:-------------------|------------:|:-------------------|------------:|:-------------------|------------:|-------------------:|------------:|:-------------------|-------------:|:--------------------|-------------:|-----------:|-------------:|---------:|
| 0 | 1 | word 1 | 0.055639 | [主要流量词] | 0.049416 | nan | 0.072298 | [精准流量词, 主要流量词] | 0.00211 | nan | 0.004251 | nan | 0.007254 | nan | 0.074409 | [主要流量词] | 0.033597 | nan | 0.000892 | nan | 9 | 2022-10-06 00:53:56 | 5726 | 326188 | 3810 | 0.01 |
| 1 | 2 | word 2 | 0.045098 | nan | 0.005472 | nan | 0.010791 | nan | 0.072859 | [主要流量词] | 0.003423 | nan | 0.012464 | nan | 0.027396 | nan | 0.002825 | nan | 0.060989 | [主要流量词] | 9 | 2022-10-07 01:16:21 | 9280 | 213477 | 40187 | 0.19 |
| 2 | 3 | word 3 | 0.02186 | nan | 0.05039 | [主要流量词] | 0.007842 | nan | 0.028832 | nan | 0.044385 | [精准流量词] | 0.001135 | nan | 0.003866 | nan | 0.021035 | nan | 0.017202 | nan | 9 | 2022-10-07 00:28:31 | 24024 | 81991 | 2275 | 0.03 |
| 3 | 4 | word 4 | 0.000699 | nan | 0.01038 | nan | 0.001536 | nan | 0.021512 | nan | 0.007658 | nan | 5e-05 | nan | 0.048682 | nan | 0.001524 | nan | 0.000118 | nan | 9 | 2022-10-07 00:52:12 | 34975 | 53291 | 30970 | 0.58 |
| 4 | 5 | word 5 | 0.00984 | nan | 0.030248 | nan | 0.003006 | nan | 0.014027 | nan | 0.00904 | [精准流量词] | 0.000348 | nan | 0.000414 | nan | 0.006721 | nan | 0.00153 | nan | 9 | 2022-10-07 02:36:05 | 43075 | 41336 | 2230 | 0.05 |
| 5 | 6 | word 6 | 0.010029 | [精准流量词] | 0.120739 | [精准流量词, 主要流量词] | 0.014359 | nan | 0.002796 | nan | 0.002883 | nan | 0.028747 | [精准流量词] | 0.007022 | nan | 0.017803 | nan | 0.001998 | nan | 9 | 2022-10-07 00:44:51 | 49361 | 34791 | 517 | 0.01 |
| 6 | 7 | word 7 | 0.002735 | nan | 0.002005 | nan | 0.005355 | nan | 6.3e-05 | nan | 0.000772 | nan | 0.000237 | nan | 0.015149 | nan | 2.1e-05 | nan | 2.3e-05 | nan | 9 | 2022-10-07 09:48:20 | 53703 | 31188 | 511 | 0.02 |
| 7 | 8 | word 8 | 0.003286 | [精准流量词] | 0.058161 | [主要流量词] | 0.013681 | [精准流量词] | 0.000748 | [精准流量词] | 0.002684 | [精准流量词] | 0.013916 | [精准流量词] | 0.029376 | nan | 0.019792 | nan | 0.005602 | nan | 9 | 2022-10-06 01:51:53 | 58664 | 27751 | 625 | 0.02 |
| 8 | 9 | word 9 | 0.004273 | [精准流量词] | 0.025581 | [精准流量词] | 0.014784 | [精准流量词] | 0.00321 | [精准流量词] | 0.000892 | nan | 0.00223 | nan | 0.005315 | nan | 0.02211 | nan | 0.027008 | [精准流量词] | 9 | 2022-10-07 01:34:28 | 73640 | 20326 | 279 | 0.01 |
| 9 | 10 | word 10 | 0.002341 | [精准流量词] | 0.029604 | nan | 0.007817 | [精准流量词] | 0.000515 | [精准流量词] | 0.001865 | [精准流量词] | 0.010128 | [精准流量词] | 0.015378 | nan | 0.019677 | nan | 0.003673 | nan | 9 | 2022-10-07 01:17:44 | 80919 | 17779 | 207 | 0.01 |因此，我的问题是，我如何才能对确切的表(应该使用col2，col4.等等，因为每个文件都有相同的格式，但是头不同，所以使用df['B64684807']是没有帮助的)
示例文件可以在：https://github.com/BeboGhattas/temp-repo/blob/main/Prepped.xls中找到
回答 1
Stack Overflow用户
回答已采纳
发布于 2022-10-08 15:36:06
IIUC，您可以使用布尔掩码：
df2 = df.iloc[:, [2,4,6,8,10,12,14,16,18]].astype(float)
m = df2.ge(0.05)
df['countif'] = m.sum(axis=1)
df['sumif'] = df2.where(m).sum(axis=1)输出(仅最后3列)：
sum countif sumif
0 0.299866 3 0.202346
1 0.241317 2 0.133848
2 0.196547 1 0.050390
3 0.092159 0 0.000000
4 0.075174 0 0.000000
5 0.206376 1 0.120739
6 0.026360 0 0.000000
7 0.147246 1 0.058161
8 0.105403 0 0.000000
9 0.090998 0 0.000000页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/73997934
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