Question

我有数据集

mydat=structure(list(time = structure(c(6L, 7L, 8L, 9L, 1L, 2L, 3L, 
4L, 5L), .Label = c("01.01.2008", "01.02.2008", "01.03.2008", 
"01.04.2008", "01.05.2008", "01.09.2007", "01.10.2007", "01.11.2007", 
"01.12.2007"), class = "factor"), account_a = structure(c(6L, 
4L, 3L, 2L, 9L, 8L, 5L, 7L, 1L), .Label = c("7725.00848360078", 
"7904.51066973023", "8000.05688342733", "8020.91725643046", "8032.80824397166", 
"8107.79491750336", "8111.57284600134", "8250.23617172539", "8341.89192978947"
), class = "factor"), account_b = structure(c(7L, 7L, 8L, 6L, 
5L, 4L, 3L, 2L, 1L), .Label = c("4878.34404162271", "4883.90444211266", 
"4889.84119615347", "4892.14279920565", "4893.31732735194", "4894.12141627531", 
"4897.0059129273", "4897.01754483248"), class = "factor"), account_c = structure(c(4L, 
7L, 1L, 6L, 8L, 9L, 3L, 5L, 2L), .Label = c("1026.6141549422", 
"238.489052868377", "362.833115212652", "426.728323306974", "510.785643175662", 
"695.680008726439", "871.207211560508", "895.998302762546", "978.620137201732"
), class = "factor")), .Names = c("time", "account_a", "account_b", 
"account_c"), class = "data.frame", row.names = c(NA, -9L))

变量： account_a account_b account_c

我需要按年份汇总

我这样做

library(data.table)
DT <- data.table(date = as.Date(mydat$time), time[-1])
DT[, list(mean = mean(account_a,account_b,account_c),

   by = year(date)]

我得到了错误

> DT <- data.table(date = as.Date(mydat$time), time[-1])
Error in charToDate(x) : 
  character string is not in a standard unambiguous format
> DT[, list(mean = mean(account_a,account_b,account_c),
+ 
+    by = year(date)]
Error: unexpected ']' in:
"
   by = year(date)]"

如何解决？另外当使用as.Date时，我有同样的错误如何按年份汇总这三个变量？

Answer 1

可以做到：

library(data.table)

setDT(mydat)

mydat[, year := year(as.Date(as.character(time), "%d.%m.%Y"))][
  , lapply(.SD, function(x) as.numeric(as.character(x))), by = year, .SDcols = -1][
    , lapply(.SD, mean), by = year
  ]

在第一行中，我们首先将time转换为Date（需要指定格式，因为它不是通常的格式），然后提取year；在第二行中，我们将所有account列转换为numeric（它们是factors）；在最后一行，我们通过mean得到所需的year。

输出：

   year account_a account_b account_c
1: 2007  8008.320  4896.288  755.0574
2: 2008  8092.304  4887.510  597.3453

这将是一种data.table的方法，您也可以在dplyr中执行以下操作：

library(dplyr)

mydat %>%
  mutate(year = format(as.Date(as.character(time), "%d.%m.%Y"), "%Y")) %>%
  mutate_at(vars(starts_with("account")), list(~ as.numeric(as.character(.)))) %>%
  group_by(year) %>%
  summarise_at(vars(starts_with("account")), list(~ mean))

charToDate（x）中的错误：R中按年份进行性能汇总时

1 个答案: