我正在提取“宽”数据，打算使用tidyr::pivot_longer()进行整理。
library(tidyverse)
df1 <-
data.frame(
M = words[1:10],
N = rnorm(10, 3, 3),
O = rnorm(10, 3, 3),
P = rnorm(10, 3, 3)
)
df2 <-
data.frame(
M = words[1:10],
N = rnorm(10, 3, 3),
O = rnorm(10, 3, 3),
P = rnorm(10, 3, 3)
)
df3 <-
data.frame(
M = words[1:10],
N = rnorm(10, 3, 3),
O = rnorm(10, 3, 3),
P = rnorm(10, 3, 3)
)
df4 <-
data.frame(
M = words[1:10],
N = rnorm(10, 3, 3),
O = rnorm(10, 3, 3),
P = rnorm(10, 3, 3)
)
lst <- list(df1, df2, df3, df4)
colname <-
c("ticker", "2017", "2018", "2019")
header <- list("Leverage", "Gearing", "Capex.to.sales", "FCFex")
lst <- lst %>%
lapply(setNames, colname) %>%
lapply(pivot_longer, -ticker, names_to = "Period", values_to = header)
使用values_to = header会给我这个错误：
[[<-。data.frame（ tmp `，“ .value”，value = list（“ Leverage”，： 替换有4行，数据有3
相反，我必须使用默认的values_to = "value"，然后使用此代码重命名我的列：
lst <- lst %>%
lapply(setNames, colname) %>%
lapply(pivot_longer, -ticker, names_to = "Period", values_to = "value")
lst <- map(seq_along(lst), function(i){
x <- lst[[i]]
colnames(x)[3] <- header[[i]]
x
})
我的输出显示如下（重命名的列），但是我想知道是否有一种方法可以将向量送入values_to而不是使用map（因为这样做可以更好地进行管道传输）？还是有一种更有效的方法来解决这个问题？
> lst
[[1]]
# A tibble: 30 x 3
ticker Period Leverage
<fct> <chr> <dbl>
1 a 2017 6.01
2 a 2018 4.82
3 a 2019 1.58
4 able 2017 8.64
5 able 2018 6.70
6 able 2019 0.831
7 about 2017 -0.187
8 about 2018 0.549
9 about 2019 0.829
10 absolute 2017 1.26
# ... with 20 more rows
[[2]]
# A tibble: 30 x 3
ticker Period Gearing
<fct> <chr> <dbl>
1 a 2017 2.37
2 a 2018 3.58
3 a 2019 5.63
4 able 2017 0.311
5 able 2018 0.708
6 able 2019 -0.0651
7 about 2017 2.89
8 about 2018 6.25
9 about 2019 10.1
10 absolute 2017 6.48
# ... with 20 more rows
[[3]]
# A tibble: 30 x 3
ticker Period Capex.to.sales
<fct> <chr> <dbl>
1 a 2017 5.22
2 a 2018 1.88
3 a 2019 0.746
4 able 2017 -3.90
5 able 2018 3.06
6 able 2019 1.91
7 about 2017 1.35
8 about 2018 4.12
9 about 2019 11.1
10 absolute 2017 1.76
# ... with 20 more rows
[[4]]
# A tibble: 30 x 3
ticker Period FCFex
<fct> <chr> <dbl>
1 a 2017 1.76
2 a 2018 2.85
3 a 2019 1.86
4 able 2017 -3.38
5 able 2018 -3.02
6 able 2019 -1.52
7 about 2017 6.46
8 about 2018 5.39
9 about 2019 0.810
10 absolute 2017 8.08
# ... with 20 more rows
对于我的问题的第二部分，我打算使用bind_col()将所有四个数据帧组合为一个，但是两个公共列正在重复（如下所示）。
如何告诉R仅绑定重命名的最右边的列，即排除最后三个数据帧的前两列？谢谢。
Metrics <- bind_cols(lst)
> Metrics
# A tibble: 30 x 12
ticker Period Leverage ticker1 Period1 Gearing ticker2 Period2
<fct> <chr> <dbl> <fct> <chr> <dbl> <fct> <chr>
1 a 2017 6.01 a 2017 2.37 a 2017
2 a 2018 4.82 a 2018 3.58 a 2018
3 a 2019 1.58 a 2019 5.63 a 2019
4 able 2017 8.64 able 2017 0.311 able 2017
5 able 2018 6.70 able 2018 0.708 able 2018
6 able 2019 0.831 able 2019 -0.0651 able 2019
7 about 2017 -0.187 about 2017 2.89 about 2017
8 about 2018 0.549 about 2018 6.25 about 2018
9 about 2019 0.829 about 2019 10.1 about 2019
10 absol~ 2017 1.26 absolu~ 2017 6.48 absolu~ 2017
# ... with 20 more rows, and 4 more variables: Capex.to.sales <dbl>,
# ticker3 <fct>, Period3 <chr>, FCFex <dbl>
答案 0 :(得分：0)
您可以使用public static function boot()
{
parent::boot();
// create a event to happen on updating
static::updating(function ($table) {
$table->updated_by = Auth::user()->id;
});
// create a event to happen on saving
static::saving(function ($table) {
$table->created_by = Auth::user()->id;
});
}
来做到这一点：
purrr输出：
library(purrr)
lst <- map(lst, setNames, colname)
map2_dfc(lst, header, ~ pivot_longer(
.x, -ticker, names_to = "Period", values_to = .y)) %>%
select(c(1:3, 6, 9, 12))