r - dplyr:使用行而不是列进行变异

我的数据集看起来像这样

year Spp  CO2 plantN_mean  N15_mean plantN_sd    N15_sd plantN_se    N15_se
1 2004   A  amb   17.136667 10.723333 1.2615202 0.7507552 0.7283391 0.4334487
2 2004   A elev   23.310000 13.043333 2.7160081 2.6595175 1.5680880 1.5354731
3 2004  AB  amb   14.410000 10.156667 1.1363538 1.7773670 0.6560742 1.0261633
4 2004  AB elev   19.470000 14.786667 2.9173790 3.7358979 1.6843495 2.1569217
5 2004  AM  amb    9.603333 13.510000 0.5515735 1.7176437 0.3184511 0.9916821
6 2004  AM elev   16.333333  9.743333 2.3622306 1.8825869 1.3638345 1.0869120

对于 plantN_mean 和 N15_mean ，我需要计算治疗效果比，即 elev/amb 。我可以使用 plyr 对其中一个变量进行处理，如下所示:

effect <- ddply(data, .(year,Spp), function (x){
  plantN_ratio <- x$plantN_mean[x$CO2 == "elev"]/x$plantN_mean[x$CO2 == "amb"]
  data.frame(plantN_ratio)
})

什么是 dplyr 版本，以及 plantN_mean 和 N15_mean ？

我想答案可以从这样的开始:

effect <- summary %>% group_by(year,Spp) %>% mutate(
plantN_ratio=plantN_mean[CO2 == "elev"]/plantN_mean[CO2 == "amb"],
N15_ratio= N15_mean[CO2 == "elev"]/N15_mean[CO2 == "amb"])

最佳答案

tidyr 在这里很有帮助。

install_packages("tidyr")
library(tidyr)
library(dplyr) # for %>% from [magrittr][http://cran.r-project.org/web/packages/magrittr/vignettes/magrittr.html]

可重复数据:

df <- structure(list(year = c(2004L, 2004L, 2004L, 2004L, 2004L, 2004L
), Spp = c("A", "A", "AB", "AB", "AM", "AM"), CO2 = c("amb",
"elev", "amb", "elev", "amb", "elev"), plantN_mean = c(17.136667,
23.31, 14.41, 19.47, 9.603333, 16.333333), N15_mean = c(10.723333,
13.043333, 10.156667, 14.786667, 13.51, 9.743333)), .Names = c("year",
"Spp", "CO2", "plantN_mean", "N15_mean"), row.names = c(NA, -6L
), class = c("tbl_df", "tbl", "data.frame"), .internal.selfref = <pointer: 0x0>)

df 看起来像:

  year Spp  CO2 plantN_mean  N15_mean
1 2004   A  amb   17.136667 10.723333
2 2004   A elev   23.310000 13.043333
3 2004  AB  amb   14.410000 10.156667
4 2004  AB elev   19.470000 14.786667
5 2004  AM  amb    9.603333 13.510000
6 2004  AM elev   16.333333  9.743333

1. 让我们收集所有的平均值变量:

gdf <- df %>% group_by(year,Spp) %>% gather(mean_id,mean_val,plantN_mean:N15_mean)

gdf 看起来像:

    year Spp  CO2     mean_id  mean_val
1  2004   A  amb plantN_mean 17.136667
2  2004   A elev plantN_mean 23.310000
3  2004  AB  amb plantN_mean 14.410000
4  2004  AB elev plantN_mean 19.470000
5  2004  AM  amb plantN_mean  9.603333
6  2004  AM elev plantN_mean 16.333333
7  2004   A  amb    N15_mean 10.723333
8  2004   A elev    N15_mean 13.043333
9  2004  AB  amb    N15_mean 10.156667
10 2004  AB elev    N15_mean 14.786667
11 2004  AM  amb    N15_mean 13.510000
12 2004  AM elev    N15_mean  9.743333

2. 让我们根据 CO2 变量展开平均值:

sdf <- gdf %>% spread(CO2,mean_val)

sdf 看起来像:

  year Spp     mean_id       amb      elev
1 2004   A plantN_mean 17.136667 23.310000
2 2004   A    N15_mean 10.723333 13.043333
3 2004  AB plantN_mean 14.410000 19.470000
4 2004  AB    N15_mean 10.156667 14.786667
5 2004  AM plantN_mean  9.603333 16.333333
6 2004  AM    N15_mean 13.510000  9.743333

3. 现在计算 elev/amb 的比率:

sdf %>% mutate(elev_o_amb = elev / amb)

要得到:

  year Spp     mean_id       amb      elev elev_o_amb
1 2004   A plantN_mean 17.136667 23.310000  1.3602412
2 2004   A    N15_mean 10.723333 13.043333  1.2163506
3 2004  AB plantN_mean 14.410000 19.470000  1.3511450
4 2004  AB    N15_mean 10.156667 14.786667  1.4558582
5 2004  AM plantN_mean  9.603333 16.333333  1.7007984
6 2004  AM    N15_mean 13.510000  9.743333  0.7211942

关于r - dplyr:使用行而不是列进行变异，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/29828124/