英文:
How to pivot table to wide based on two colums instead of single
问题
我有这样的数据
df <- data.frame(ptid = rep(1:2, each = 6),
num = rep(1:3, each = 4),
type = rep(c("type1", "type2"), each = 6),
time = rep(1:3, times = 4),
volume = rnorm(12))
现在，它是长格式，我应该如何将其改为宽格式，基于时间和类型，而不是只有一个: 如 time1type1, time2type1, time3type1, time1type2, time2type2, time3type3, time3type3...?
我写了这段代码
df_wide <- df %>%
pivot_wider(names_from = c("time", "type"),
values_from = "volume",
names_prefix = "time",
names_sep = "")
但它说体积不足以识别，并且无法工作。谢谢帮助~~！
英文:
I have data like this
df <- data.frame(ptid = rep(1:2, each = 6),
num = rep(1:3, each = 4),
type = rep(c("type1", "type2"), each = 6),
time = rep(1:3, times = 4),
volume = rnorm(12))
Now, it is long format, how should I change it to wide format based on both time and type instead one: like time1type1, time2type1, time3type1, time1type2, time2type2, time3type3, time3type3...?
I wrote this code
df_wide <- df %>%
pivot_wider(names_from = c("time", "type"),
values_from = "volume",
names_prefix = "time",
names_sep = "")
but it says the volume is not sufficient to identify, and could not work. Thanks for help~~!
答案1
得分: 2
library(dplyr)
library(tidyr)
df %>%
unite(time_type, c(time, type), sep = "") %>%
pivot_wider(names_from = time_type,
values_from = volume,
names_prefix = "time",
values_fn = list) %>%
unnest(cols = c(time1type1, time2type1, time3type1, time1type2, time2type2, time3type2))
ptid num time1type1 time2type1 time3type1 time1type2 time2type2 time3type2
<int> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 0.864 -1.02 0.577 NA NA NA
2 1 1 1.33 -1.02 0.577 NA NA NA
3 1 2 NA -0.909 -1.30 NA NA NA
4 2 2 NA NA NA 0.255 -0.549 NA
5 2 3 NA NA NA -0.213 0.139 -1.59
6 2 3 NA NA NA -0.213 0.139 -0.384
<details>
<summary>英文:</summary>
library(dplyr)
library(tidyr)
df %>%
unite(time_type, c(time, type), sep = "") %>%
pivot_wider(names_from = time_type,
values_from = volume,
names_prefix = "time",
values_fn = list) %>%
unnest(cols = c(time1type1, time2type1, time3type1, time1type2, time2type2, time3type2))
ptid num time1type1 time2type1 time3type1 time1type2 time2type2 time3type2
<int> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 0.864 -1.02 0.577 NA NA NA
2 1 1 1.33 -1.02 0.577 NA NA NA
3 1 2 NA -0.909 -1.30 NA NA NA
4 2 2 NA NA NA 0.255 -0.549 NA
5 2 3 NA NA NA -0.213 0.139 -1.59
6 2 3 NA NA NA -0.213 0.139 -0.384
</details>
# 答案2
**得分**: 0
你有观测数据，这些数据匹配了ptid/num/type/time。如果你想让它们成为单独的行而不是列表，请添加一个标识符来区分它们：
```R
df %>%
dplyr::mutate(iteration = dplyr::row_number(), .by = c(ptid:time)) %>%
pivot_wider(names_from = c("time", "type"),
values_from = "volume",
names_prefix = "time",
names_sep = "")
# 一个 tibble: 6 × 9
ptid num iteration time1type1 time2type1 time3type1 time1type2 time2type2 time3type2
<int> <int> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 0.0202 1.09 -1.72 NA NA NA
2 1 1 2 -0.473 NA NA NA NA NA
3 1 2 1 NA -1.06 -1.02 NA NA NA
4 2 2 1 NA NA NA -1.11 -0.494 NA
5 2 3 1 NA NA NA -1.34 -0.146 -1.08
6 2 3 2 NA NA NA NA NA -0.463
如果你需要更多帮助，请告诉我。
英文:
You have observations which match ptid/num/type/time. If you want them to be separate rows instead of a list in that spot, add an identifier to distinguish them:
df %>%
dplyr::mutate(iteration = dplyr::row_number(), .by = c(ptid:time)) %>%
pivot_wider(names_from = c("time", "type"),
values_from = "volume",
names_prefix = "time",
names_sep = "")
# A tibble: 6 × 9
ptid num iteration time1type1 time2type1 time3type1 time1type2 time2type2 time3type2
<int> <int> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 0.0202 1.09 -1.72 NA NA NA
2 1 1 2 -0.473 NA NA NA NA NA
3 1 2 1 NA -1.06 -1.02 NA NA NA
4 2 2 1 NA NA NA -1.11 -0.494 NA
5 2 3 1 NA NA NA -1.34 -0.146 -1.08
6 2 3 2 NA NA NA NA NA -0.463
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