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r - 包含缺失数据时如何在 "dplyr"中使用 rowSums()?

转载 作者:行者123 更新时间:2023-12-03 21:13:40 25 4
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我想使用功能rowSumsdplyr并遇到了一些数据缺失的困难。示例数据为 mtcars .目标是估计 mpg 的三个变量的总和, cyldisp按行。但是,当特定行中存在缺失值时,以下 R 代码的结果似乎不正确(请参阅输出中的变量 new1new2)。任何意见和建议表示赞赏!

data<-mtcars%>%
mutate(
mpg=case_when(mpg>25~NA_real_,TRUE~as.numeric(mpg)), # generate missing data in "mpg"
new1=rowSums(.[c("mpg","cyl","disp")],na.rm=FALSE), # method1: row sum, treat NA as NA?
new2=rowSums(.[c("mpg","cyl","disp")],na.rm=TRUE), # method2: row sum, treat NA as zero?
new3=mpg+cyl+disp # method3: row sum, by hand
)

data

下面列出了输出:
    mpg cyl  disp  hp drat    wt  qsec vs am gear carb  new1  new2  new3
1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 187.0 187.0 187.0
2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 187.0 187.0 187.0
3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 134.8 134.8 134.8
4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 285.4 285.4 285.4
5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 386.7 386.7 386.7
6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 249.1 249.1 249.1
7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 382.3 382.3 382.3
8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 175.1 175.1 175.1
9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 167.6 167.6 167.6
10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 192.8 192.8 192.8
11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 191.4 191.4 191.4
12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 300.2 300.2 300.2
13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 301.1 301.1 301.1
14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 299.0 299.0 299.0
15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 490.4 490.4 490.4
16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 478.4 478.4 478.4
17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 462.7 462.7 462.7
18 NA 4 78.7 66 4.08 2.200 19.47 1 1 4 1 115.1 115.1 NA
19 NA 4 75.7 52 4.93 1.615 18.52 1 1 4 2 110.1 110.1 NA
20 NA 4 71.1 65 4.22 1.835 19.90 1 1 4 1 109.0 109.0 NA
21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 145.6 145.6 145.6
22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 341.5 341.5 341.5
23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 327.2 327.2 327.2
24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 371.3 371.3 371.3
25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 427.2 427.2 427.2
26 NA 4 79.0 66 4.08 1.935 18.90 1 1 4 1 110.3 110.3 NA
27 NA 4 120.3 91 4.43 2.140 16.70 0 1 5 2 150.3 150.3 NA
28 NA 4 95.1 113 3.77 1.513 16.90 1 1 5 2 129.5 129.5 NA
29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 374.8 374.8 374.8
30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 170.7 170.7 170.7
31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 324.0 324.0 324.0
32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 146.4 146.4 146.4

最佳答案

结果不是您所期望的,因为在 mpg 中插入了 NA是在同一个 mutate语句作为变量 new* ;因此,R 仍然考虑变量 mpg 中的先前值。 .通过使用另一个 mutate前一步,我们设法获得所需的结果

mtcars %>%
mutate(
mpg = case_when(mpg > 25 ~ NA_real_, TRUE ~ as.numeric(mpg)) # generate missing data in "mpg"
) %>%
mutate(
new1 = rowSums(.[c("mpg","cyl","disp")], na.rm = FALSE), # method1: row sum, treat NA as NA?
new2 = rowSums(.[c("mpg","cyl","disp")], na.rm = TRUE), # method2: row sum, treat NA as zero?
new3 = mpg + cyl + disp # method3: row sum, by hand
)

输出
    mpg cyl  disp  hp drat    wt  qsec vs am gear carb  new1  new2  new3
1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 187.0 187.0 187.0
2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 187.0 187.0 187.0
3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 134.8 134.8 134.8
4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 285.4 285.4 285.4
5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 386.7 386.7 386.7
6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 249.1 249.1 249.1
7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 382.3 382.3 382.3
8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 175.1 175.1 175.1
9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 167.6 167.6 167.6
10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 192.8 192.8 192.8
11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 191.4 191.4 191.4
12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 300.2 300.2 300.2
13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 301.1 301.1 301.1
14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 299.0 299.0 299.0
15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 490.4 490.4 490.4
16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 478.4 478.4 478.4
17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 462.7 462.7 462.7
18 NA 4 78.7 66 4.08 2.200 19.47 1 1 4 1 NA 82.7 NA
19 NA 4 75.7 52 4.93 1.615 18.52 1 1 4 2 NA 79.7 NA
20 NA 4 71.1 65 4.22 1.835 19.90 1 1 4 1 NA 75.1 NA
21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 145.6 145.6 145.6
22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 341.5 341.5 341.5
23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 327.2 327.2 327.2
24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 371.3 371.3 371.3
25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 427.2 427.2 427.2
26 NA 4 79.0 66 4.08 1.935 18.90 1 1 4 1 NA 83.0 NA
27 NA 4 120.3 91 4.43 2.140 16.70 0 1 5 2 NA 124.3 NA
28 NA 4 95.1 113 3.77 1.513 16.90 1 1 5 2 NA 99.1 NA
29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 374.8 374.8 374.8
30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 170.7 170.7 170.7
31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 324.0 324.0 324.0
32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 146.4 146.4 146.4

关于r - 包含缺失数据时如何在 "dplyr"中使用 rowSums()?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61989590/

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