这个问题把我难住了。
我有以下数据框:
library(dplyr)
# approximation of data frame
x <- data.frame(doy = sample(c(seq(200, 300)), 20, replace = T),
year = sample(c("2000", "2005"), 20, replace = T),
phase = sample(c("pre", "post"), 20, replace = T))
和一个简单的“汇总”函数,它将列名作为变量接收,并且运行良好:
getStats <- function(df, col) {
col <- as.name(col)
df %>%
group_by(year, phase) %>%
summarize(n = sum(!is.na(col)),
mean = mean(col, na.rm = T),
sd = sd(col, na.rm = T),
se = sd/sqrt(n))
}
> getStats(x, "doy")
Source: local data frame [4 x 6]
Groups: year [?]
year phase n mean sd se
<fctr> <fctr> <int> <dbl> <dbl> <dbl>
1 2000 post 8 248.625 30.42526 10.75695
2 2000 pre 2 290.000 14.14214 10.00000
3 2005 post 5 231.400 32.86031 14.69558
4 2005 pre 5 274.200 29.79429 13.32441
但是,如果我修改函数以获取中位数,则会返回错误:
getStats <- function(df, col) {
col <- as.name(col)
df %>%
group_by(year, phase) %>%
summarize(n = sum(!is.na(col)),
mean = mean(col, na.rm = T),
med = median(col, na.rm = T), # new line
sd = sd(col, na.rm = T),
se = sd/sqrt(n))
}
> getStats(x, "doy")
Error in median (doy, na.rm = TRUE): object "doy" not found
我尝试了许多名称和位置更改,但都产生相同的结果:'median' 不接受列名作为传递的变量。我想我错过了一些非常基本的东西,当有人向我指出时我会做一个手掌,但在此期间我觉得我正在失去理智。我很感激任何见解!