使用 dplyr 从多列计算分位数

Question

我有一个这样的数据框

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，而是可以通过索引来做类似

的事情

3:ncol(df)

Answer 1

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

Answer 2

函数quantile只需要一个输入向量。当你做

quantile(x1, x2, x3, x4, x5, x6 , x7, x8, probs = 0.5)

你给它提供了 8 个输入向量，它只会使用 x1 而忽略 x2 到 x8。

示例：

x <- rnorm(20)
y = rnorm(20) + 100

quantile(x, probs = 0.025) # lower 
#   2.5% 
# -1.633378 
quantile(x, y, probs = 0.025) # y will be ignored. This yields same result as quantile(x, probs = 0.025). A warning explains this
#    2.5% 
# -1.633378 
# Warning message:
#     In if (na.rm) x <- x[!is.na(x)] else if (anyNA(x)) stop("missing values and NaN's not allowed if 'na.rm' is FALSE") :
#     the condition has length > 1 and only the first element will be used

要解决您的具体问题，请将 x1 到 x8 放在 c() 中以形成向量：

df %>% group_by(loc, year) %>% 
dplyr::summarise(lower.x = quantile(c(x1, x2, x3, x4, x5, x6 , x7, x8), probs = 0.025),
                 mean.x = quantile(c(x1, x2, x3, x4, x5, x6 , x7, x8), probs = 0.5),
                 upper.x = quantile(c(x1, x2, x3, x4, x5, x6 , x7, x8), probs = 0.975))

产量：

# A tibble: 8 x 5
# Groups:   loc [?]
    loc  year     lower.x     mean.x   upper.x
  <int> <int>       <dbl>      <dbl>     <dbl>
1     1  1980 -1.12583212  0.1683845 1.1579655
2     1  1981 -1.20363611 -0.1399433 1.9308253
3     1  1982 -0.93238412 -0.3195850 0.3835611
4     1  1983 -2.08331501 -0.4235632 1.2267823
5     2  1980 -1.46528453 -0.3096375 0.9863813
6     2  1981 -1.51563211  0.1100798 0.8267675
7     2  1982 -1.16435350  0.1885864 0.8349510
8     2  1983 -0.01427533  0.4301591 1.9688637

顺便说一句，上限应该是 0.975，你打错了 0.0975

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