Question

我有以下代码，在其中我无法计算代码的最后一行（第17列中前5个值的平均值），从而导致出现警告消息Warning message:In mean.default(df[1:5, 17], na.rm = TRUE) : argument is not numeric or logical: returning NA

有人建议我在做什么错吗？谢谢！

library(eurostat)
library(dplyr)
library(tidyr)


earn_mw_avgr2 <- get_eurostat("earn_mw_avgr2", stringsAsFactors = FALSE)

df <- earn_mw_avgr2 %>% 
    filter(geo %in% c("BE","BG","CZ","DK","DE","EE","IE","EL","ES","FR","HR",
                   "IT","CY","LV","LU","HU","MT","NL","AT","PL","PT","RO",
                   "SI","SK","FI","SE","UK"),
        indic_se=="MW_MEAGE",
        nace_r2=="B-S") %>% 
    spread(time, values)%>%
    select(-c(17,18))%>%
    mutate(avg=rowMeans(.[14:16], na.rm = TRUE))%>%
    arrange(desc(avg))


mean(df[1:5,17], na.rm = TRUE)

Answer 1

问题是它是newUser.contactDetails.countryName = "UK";，并且不会删除尺寸为tibble（其中data.frame为?Extract的尺寸）

drop = TRUE

mean(df[1:5,][[17]]) #[1] 47.29667适用于mean s。如果我们检查vector

x-一个R对象。当前，存在用于数字/逻辑矢量以及日期，日期时间和时间间隔对象的方法。仅允许trim = 0的复数向量。

Answer 2

将sed添加到$avg可以解决此问题。完整代码如下。

df[1:5,17]

无法使用dplyr（）计算mean（）

2 个答案: