R语言迭代过滤器列表

fdx2calv 于 12个月前发布在其他

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我想在一个过滤器的组合列表中添加一个过滤器组合，这些过滤器组合应用于一个类似于以下的嵌套框架的多个列：

df <- data.frame(    
    ID1 = c("Asset1", "Asset2", "Asset1", "Asset2", "Asset1", "Asset2", "Asset1", "Asset2"),
    ID2 = c("Asset1", "Asset1", "Asset2", "Asset2", "Asset1", "Asset1", "Asset2", "Asset2"),
    var1= c(0.011,      0.012,   0.012,    0.022,   .0011,      0.0012,   0.0012,    0.0022)
)

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举个例子：
我想计算以下组合的var 1的值：

filter1: ID1=='Asset1' and ID2=='Asset1'

型
和

filter2: ID1=='Asset1' and ID2=='Asset2'

型
对于一个组合，我显然可以这样做，用一个dspur过滤器：

library(dplyr)
df %>% filter(ID1=='Asset1', ID2=='Asset1')

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但是我想用一个for循环来迭代所有的过滤器：

filter_list <- c(filter1, filter2)

for (filter_i in filter_list){
    df %>% filter(filter_i)
}

型

来源：https://stackoverflow.com/questions/77657514/iterate-over-a-list-of-filters

2条答案

按热度按时间

iaqfqrcu1#

使用quos和bang-bang-bang aka !!!，您可以执行以下操作：

library(dplyr, warn = FALSE)

filters <- list(
  filter1 = quos(ID1 == "Asset1", ID2 == "Asset1"),
  filter2 = quos(ID1 == "Asset1", ID2 == "Asset2")
)

lapply(filters, \(x) {
  df %>%
    filter(!!!x)
})
#> $filter1
#>      ID1    ID2   var1
#> 1 Asset1 Asset1 0.0110
#> 2 Asset1 Asset1 0.0011
#> 
#> $filter2
#>      ID1    ID2   var1
#> 1 Asset1 Asset2 0.0120
#> 2 Asset1 Asset2 0.0012

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赞(0）回复(0）举报 12个月前

hk8txs482#

您可以从1:2列中的unique资产创建过滤器列表。接下来，使用expand.grid使用所有2-排列生成filter_list。

> assets <- unique(unlist(df[1:2]))
> filter_list <- expand.grid(assets, assets) |> asplit(1)

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最后，你可以用subset覆盖列表。

> lapply(filter_list, \(x) subset(x=df, ID1 == x[1] & ID2 == x[2]))
[[1]]
     ID1    ID2   var1
1 Asset1 Asset1 0.0110
5 Asset1 Asset1 0.0011

[[2]]
     ID1    ID2   var1
2 Asset2 Asset1 0.0120
6 Asset2 Asset1 0.0012

[[3]]
     ID1    ID2   var1
3 Asset1 Asset2 0.0120
7 Asset1 Asset2 0.0012

[[4]]
     ID1    ID2   var1
4 Asset2 Asset2 0.0220
8 Asset2 Asset2 0.0022

型
或者，如果您坚持使用for循环，

> res <- vector('list', length(filter_list))  ## initialize
> for (i in seq_along(filter_list)) {
+   res[[i]] <- subset(x=df, ID1 == filter_list[[i]][1] & ID2 == filter_list[[i]][2])
+ }
> res
[[1]]
     ID1    ID2   var1
1 Asset1 Asset1 0.0110
5 Asset1 Asset1 0.0011

[[2]]
     ID1    ID2   var1
2 Asset2 Asset1 0.0120
6 Asset2 Asset1 0.0012

[[3]]
     ID1    ID2   var1
3 Asset1 Asset2 0.0120
7 Asset1 Asset2 0.0012

[[4]]
     ID1    ID2   var1
4 Asset2 Asset2 0.0220
8 Asset2 Asset2 0.0022

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数据类型：*

> dput(df)
structure(list(ID1 = c("Asset1", "Asset2", "Asset1", "Asset2", 
"Asset1", "Asset2", "Asset1", "Asset2"), ID2 = c("Asset1", "Asset1", 
"Asset2", "Asset2", "Asset1", "Asset1", "Asset2", "Asset2"), 
    var1 = c(0.011, 0.012, 0.012, 0.022, 0.0011, 0.0012, 0.0012, 
    0.0022)), class = "data.frame", row.names = c(NA, -8L))

型

赞(0）回复(0）举报 12个月前

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R语言迭代过滤器列表

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