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我有一个数据框，用于存储有关何时进行特定评估（“何时”）的信息。此评估发生在不同的时间（t1-t3），具体时间因参与者而异。
数据框还包含每个参与者曾经完成的所有评估（包括“时间”列中引用的评估）。我只想要“时间”列中表示的评估信息。因此，如果数字为1，我想保留与该评估有关的所有数据，并删除该评估未收集的所有数据。请注意，实际数据集中的变量比此缩短的数据集中的变量多，因此任何解决方案都不应依赖重复的变量名。
这是我能做的最好的事情。该解决方案的问题在于，必须为每个变量名称重复该解决方案。
df2 <- mutate(.data = df,
a1G_when = if_else(when == 1, a1G_t1, NA_real_))
# here is what we start with
df <- structure(list(id = 1:10, when = c(1, 3, 2, 1, 2, 1, 3, 2, 3,
1), a1G_t1 = c(0.78, 0.21, 0.04, 0.87, 0.08, 0.25, 0.9, 0.77,
0.51, 0.5), Stqo_t1 = c(0.68, 0.77, 0.09, 0.66, 0.94, 0.05, 0.97,
0.92, 1, 0.04), Twcdz_t1 = c(0.95, 0.41, 0.29, 0.54, 0.06, 0.45,
0.6, 0.24, 0.17, 0.55), Kgh_t1 = c(0.25, 0.86, 0.37, 0.34, 0.97,
0.75, 0.73, 0.68, 0.37, 0.66), `2xWX_t1` = c(0.47, 0.52, 0.23,
0.5, 0.88, 0.71, 0.21, 0.98, 0.76, 0.21), `2IYnS_t1` = c(0.32,
0.75, 0.03, 0.46, 0.89, 0.71, 0.51, 0.83, 0.34, 0.32), a1G_t2 = c(0.97,
0.01, 0.58, 0.33, 0.58, 0.37, 0.76, 0.33, 0.39, 0.56), Stqo_t2 = c(0.78,
0.42, 0.5, 0.69, 0.09, 0.72, 0.84, 0.94, 0.46, 0.83), Twcdz_t2 = c(0.62,
0.34, 0.72, 0.62, 0.8, 0.26, 0.3, 0.88, 0.42, 0.53), Kgh_t2 = c(0.99,
0.66, 0.02, 0.17, 0.51, 0.03, 0.03, 0.74, 0.1, 0.26), `2xWX_t2` = c(0.68,
0.97, 0.56, 0.27, 0.66, 0.71, 0.96, 0.24, 0.37, 0.76), `2IYnS_t2` = c(0.24,
0.88, 0.58, 0.31, 0.8, 0.92, 0.91, 0.9, 0.55, 0.52), a1G_t3 = c(0.73,
0.6, 0.66, 0.06, 0.33, 0.34, 0.09, 0.44, 0.73, 0.56), Stqo_t3 = c(0.28,
0.88, 0.56, 0.75, 0.85, 0.33, 0.88, 0.4, 0.63, 0.61), Twcdz_t3 = c(0.79,
0.95, 0.41, 0.07, 0.99, 0.06, 0.74, 0.17, 0.89, 0.4), Kgh_t3 = c(0.06,
0.52, 0.35, 0.91, 0.43, 0.74, 0.72, 0.96, 0.39, 0.4), `2xWX_t3` = c(0.25,
0.09, 0.64, 0.32, 0.15, 0.14, 0.18, 0.33, 0.97, 0.6), `2IYnS_t3` = c(0.92,
0.49, 0.09, 0.95, 0.3, 0.83, 0.82, 0.56, 0.29, 0.36)), row.names = c(NA,
-10L), class = "data.frame")
# here is an example of what I want with the first column. I would also want all other repeating columns to look like this (Stq0_when, Twcdz, etc.)
id when a1G_when
1 1 1 0.78
2 2 3 0.88
3 3 2 0.58
4 4 1 0.87
5 5 2 0.58
6 6 1 0.25
7 7 3 0.09
8 8 2 0.33
9 9 3 0.73
10 10 1 0.50
答案 0 :(得分：0)
这里有机会使用新的tidyr::pivot_longer。我们可以使用它来调整数据的形状，以使var和t在自己的列中，filter只是包含我们想要的数据的行（即t等于when），然后将数据旋转回宽范围。
library(tidyverse)
df1 <- structure(list(ID = c(101, 102, 103, 104, 105), when = c(1, 2, 3, 1, 2), var1_t1 = c(5, 6, 4, 5, 6), var2_t1 = c(2, 3, 4, 2, 3), var1_t2 = c(7, 8, 9, 7, 8), var2_t2 = c(5, 4, 5, 4, 5), var1_t3 = c(3, 4, 3, 4, 3), var2_t3 = c(6, 7, 6, 7, 6)), row.names = c(NA, 5L), class = "data.frame")
df1 %>%
pivot_longer(
cols = starts_with("var"),
names_to = c("var", "t"),
names_sep = "_t",
values_to = "val",
col_ptypes = list(var = character(), t = numeric())
) %>%
filter(when == t) %>%
select(-t) %>%
pivot_wider(names_from = "var", values_from = "val")
#> # A tibble: 5 x 4
#> ID when var1 var2
#> <dbl> <dbl> <dbl> <dbl>
#> 1 101 1 5 2
#> 2 102 2 8 4
#> 3 103 3 3 6
#> 4 104 1 5 2
#> 5 105 2 8 5
由reprex package（v0.3.0）于2019-07-16创建
答案 1 :(得分：0)
使用data.table，您可以执行以下操作：
library(data.table)
cols <- unique(paste0(gsub("_.*", "", setdiff(names(df), c("id", "when"))), "_when"))
setDT(df)[
, (cols) := lapply(cols, function(x) paste0(gsub("_.*", "", x), "_t", when))][
, (cols) := lapply(cols, function(x) as.character(.SD[[get(x)]])), by = cols][
, (cols) := lapply(.SD, as.numeric), .SDcols = cols
]
输出（仅前10行，仅相关的when列）：
a1G_when Stqo_when Twcdz_when Kgh_when 2xWX_when 2IYnS_when
1: 0.78 0.68 0.95 0.25 0.47 0.32
2: 0.60 0.88 0.95 0.52 0.09 0.49
3: 0.58 0.50 0.72 0.02 0.56 0.58
4: 0.87 0.66 0.54 0.34 0.50 0.46
5: 0.58 0.09 0.80 0.51 0.66 0.80
6: 0.25 0.05 0.45 0.75 0.71 0.71
7: 0.09 0.88 0.74 0.72 0.18 0.82
8: 0.33 0.94 0.88 0.74 0.24 0.90
9: 0.73 0.63 0.89 0.39 0.97 0.29
10: 0.50 0.04 0.55 0.66 0.21 0.32