我有这样一个数据框
set.seed(123)
对于一个向量,如果我想生成均值,以及上下 95% CI,我可以这样做:
x <- rnorm(20)
quantile(x, probs = 0.500) # mean
quantile(x, probs = 0.025) # lower
quantile(x, probs = 0.975) # upper bound
我有一个数据框
df <- data.frame(loc = rep(1:2, each = 4),
year = rep(1980:1983, times = 2),
x1 = rnorm(8), x2 = rnorm(8), x3 = rnorm(8), x4 = rnorm(8),
x5 = rnorm(8), x6 = rnorm(8), x7 = rnorm(8), x8 = rnorm(8))
对于每个位置和年份,我想使用 x1 到 x8 找到中值、下限和上限。
df %>% group_by(loc, year) %>%
dplyr::summarise(mean.x = quantile(x1, x2, x3, x4, x5, x6 , x7, x8, probs = 0.500),
lower.x = quantile(x1, x2, x3, x4, x5, x6 , x7, x8, probs = 0.025),
upper.x = quantile(x1, x2, x3, x4, x5, x6 , x7, x8, probs = 0.975))
但这给了我对所有人相同的答案。
# A tibble: 8 x 5
# Groups: loc [?]
loc year mean.x lower.x upper.x
<int> <int> <dbl> <dbl> <dbl>
1 1 1980 -1.07 -1.07 -1.07
2 1 1981 -0.218 -0.218 -0.218
3 1 1982 -1.03 -1.03 -1.03
4 1 1983 -0.729 -0.729 -0.729
5 2 1980 -0.625 -0.625 -0.625
6 2 1981 -1.69 -1.69 -1.69
7 2 1982 0.838 0.838 0.838
8 2 1983 0.153 0.153 0.153
此外,有什么方法可以不通过 x1、x2...x8 来引用列,而是通过索引来做类似的事情
3:ncol(df)
最佳答案
您可能希望先将宽数据转换为长数据:
require(dplyr)
require(tidyr)
df %>% gather(xvar, value, x1:x8) %>%
group_by(loc, year) %>%
summarise(mean.x = quantile(value, probs = 0.50),
lower.x = quantile(value, probs = 0.025),
upper.x = quantile(value, probs = 0.975))
你得到:
# A tibble: 8 x 5
# Groups: loc [?]
loc year mean.x lower.x upper.x
<int> <int> <dbl> <dbl> <dbl>
1 1 1980 0.152 -0.982 2.08
2 1 1981 -0.478 -1.33 0.825
3 1 1982 -0.0415 -1.95 1.02
4 1 1983 0.855 -0.180 1.43
5 2 1980 0.658 -1.24 2.23
6 2 1981 0.196 -0.782 0.827
7 2 1982 -0.629 -0.937 0.285
8 2 1983 -0.0737 -0.744 1.27
关于r - 使用 dplyr 从多列计算分位数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51287704/