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问题描述
我正在尝试基于 Group 变量 item.map 编写代码，其中包含以下项信息：
I am trying to write code based on a Group variable, item.map that has item information that includes an q-matrix showing which item is associated with which group.
Group <- c(1,2,3,4)
item.map <- data.frame(
item.id = c(21,41,61,72),
group.1 = c(1,1,1,0),
group.2 = c(0,1,0,1),
group.3 = c(1,1,1,0),
group.4 = c(0,0,0,1))
> item.map
item.id group.1 group.2 group.3 group.4
1 21 1 0 1 0
2 41 1 1 1 0
3 61 1 0 1 0
4 72 0 1 0 1
在此 item.map中 group.1有3个项目，而group.2有2个项目， group.3 有3个项目，而 group.4 有1个项目。使用此item.map，我想在下面的代码块中分配这些项目，但无法插入 item.map
In this item.map group.1 had 3 items while group.2 has two items, group.3 has three and group.4 has 1 item.. Using this item.map I wanted to assign those items within the chunk of code below but I was not able to plug the item.map information.
OUTPUT <- as.data.frame(c())
for(i in 1:length(item.map$item.id)) {
for(k in 0:(length(Group))) { # here with the length(State) I gained the sequqnece of 0,1,2,3
output <- paste0("Equal = ",paste0(paste("(", "G1, ",item.map$item.id[i], ","," Slope[",k,"])",collapse=", ", sep=""),", ",
paste( "(", "G2, ",item.map$item.id[i], ","," Slope[",k,"])",collapse=", ", sep=""),
";"))
OUTPUT <- c(OUTPUT, output)
}
}
[1] "Equal = (G1, 21, Slope[0]), (G2, 21, Slope[0]), (G3, 21, Slope[0]), (G4, 21, Slope[0]);"
[1] "Equal = (G1, 21, Slope[1]), (G2, 21, Slope[1]), (G3, 21, Slope[1]), (G4, 21, Slope[1]);"
[1] "Equal = (G1, 21, Slope[2]), (G2, 21, Slope[2]), (G3, 21, Slope[2]), (G4, 21, Slope[2]);"
[1] "Equal = (G1, 21, Slope[3]), (G2, 21, Slope[3]), (G3, 21, Slope[3]), (G4, 21, Slope[3]);"
[1] "Equal = (G1, 21, Slope[4]), (G2, 21, Slope[4]), (G3, 21, Slope[4]), (G4, 21, Slope[4]);"
[1] "Equal = (G1, 41, Slope[0]), (G2, 41, Slope[0]), (G3, 41, Slope[0]), (G4, 41, Slope[0]);"
[1] "Equal = (G1, 41, Slope[1]), (G2, 41, Slope[1]), (G3, 41, Slope[1]), (G4, 41, Slope[1]);"
[1] "Equal = (G1, 41, Slope[2]), (G2, 41, Slope[2]), (G3, 41, Slope[2]), (G4, 41, Slope[2]);"
[1] "Equal = (G1, 41, Slope[3]), (G2, 41, Slope[3]), (G3, 41, Slope[3]), (G4, 41, Slope[3]);"
[1] "Equal = (G1, 41, Slope[4]), (G2, 41, Slope[4]), (G3, 41, Slope[4]), (G4, 41, Slope[4]);"
[1] "Equal = (G1, 61, Slope[0]), (G2, 61, Slope[0]), (G3, 61, Slope[0]), (G4, 61, Slope[0]);"
[1] "Equal = (G1, 61, Slope[1]), (G2, 61, Slope[1]), (G3, 61, Slope[1]), (G4, 61, Slope[1]);"
[1] "Equal = (G1, 61, Slope[2]), (G2, 61, Slope[2]), (G3, 61, Slope[2]), (G4, 61, Slope[2]);"
[1] "Equal = (G1, 61, Slope[3]), (G2, 61, Slope[3]), (G3, 61, Slope[3]), (G4, 61, Slope[3]);"
[1] "Equal = (G1, 61, Slope[4]), (G2, 61, Slope[4]), (G3, 61, Slope[4]), (G4, 61, Slope[4]);"
[1] "Equal = (G1, 72, Slope[0]), (G2, 72, Slope[0]), (G3, 72, Slope[0]), (G4, 72, Slope[0]);"
[1] "Equal = (G1, 72, Slope[1]), (G2, 72, Slope[1]), (G3, 72, Slope[1]), (G4, 72, Slope[1]);"
[1] "Equal = (G1, 72, Slope[2]), (G2, 72, Slope[2]), (G3, 72, Slope[2]), (G4, 72, Slope[2]);"
[1] "Equal = (G1, 72, Slope[3]), (G2, 72, Slope[3]), (G3, 72, Slope[3]), (G4, 72, Slope[3]);"
[1] "Equal = (G1, 72, Slope[4]), (G2, 72, Slope[4]), (G3, 72, Slope[4]), (G4, 72, Slope[4]);"
因此，在所需的输出中， G1 不应该包含项目 72 和G2不应包含项目 21 和 61 的信息在分组块中。
另外，我无法对 G1进行排序。和 G2；在我的代码中。考虑到 G1 ， G2 ， G3 和 G4 ？
So, in the desired output, G1 should not have item 72 and G2 should not have items 21 and 61 information in the grouping chunk.Also, I was not able to sequence "G1" and "G2" in my code. Is there a way to combine these two lines into one considering G1, G2, G3 and G4?
output <- paste0("Equal = ",paste0(paste("(", "G1, ",item.map$item.id[i], ","," Slope[",k,"])",collapse=", ", sep=""),", ",
paste("(", "G2, ",item.map$item.id[i], ","," Slope[",k,"])",collapse=", ", sep=""),", ",
paste("(", "G3, ",item.map$item.id[i], ","," Slope[",k,"])",collapse=", ", sep=""),", ",
paste( "(", "G4, ",item.map$item.id[i], ","," Slope[",k,"])",collapse=", ", sep=""),
";"))
所需的输出为：
[1] "Equal = (G1, 21, Slope[0]), (G3, 21, Slope[0]);"
[1] "Equal = (G1, 21, Slope[1]), (G3, 21, Slope[1]);"
[1] "Equal = (G1, 21, Slope[2]), (G3, 21, Slope[2]);"
[1] "Equal = (G1, 21, Slope[3]), (G3, 21, Slope[3]);"
[1] "Equal = (G1, 21, Slope[4]), (G3, 21, Slope[4]);"
[1] "Equal = (G1, 41, Slope[0]), (G2, 41, Slope[0]), (G3, 41, Slope[0]);"
[1] "Equal = (G1, 41, Slope[1]), (G2, 41, Slope[1]), (G3, 41, Slope[1]);"
[1] "Equal = (G1, 41, Slope[2]), (G2, 41, Slope[2]), (G3, 41, Slope[2]);"
[1] "Equal = (G1, 41, Slope[3]), (G2, 41, Slope[3]), (G3, 41, Slope[3]);"
[1] "Equal = (G1, 41, Slope[4]), (G2, 41, Slope[4]), (G3, 41, Slope[4]);"
[1] "Equal = (G1, 61, Slope[0]), (G3, 61, Slope[0]);"
[1] "Equal = (G1, 61, Slope[1]), (G3, 61, Slope[1]);"
[1] "Equal = (G1, 61, Slope[2]), (G3, 61, Slope[2]);"
[1] "Equal = (G1, 61, Slope[3]), (G3, 61, Slope[3]);"
[1] "Equal = (G1, 61, Slope[4]), (G3, 61, Slope[4]);"
[1] "Equal = (G2, 72, Slope[0]), (G4, 72, Slope[0]);"
[1] "Equal = (G2, 72, Slope[1]), (G4, 72, Slope[1]);"
[1] "Equal = (G2, 72, Slope[2]), (G4, 72, Slope[2]);"
[1] "Equal = (G2, 72, Slope[3]), (G4, 72, Slope[3]);"
[1] "Equal = (G2, 72, Slope[4]), (G4, 72, Slope[4]);"
有人有什么想法吗？
谢谢
Does anyone have any ideas?Thanks
推荐答案
这里是 tidyverse 的一种选择我们遍历组列名称，选择来自列表，重命名到'G1'，'G2'，然后执行交叉扩展数据集， filter 基于逻辑组列，使用 glue_data （来自 grlue ）和展平 列表到向量
Here is one option with tidyverse where we loop over the 'group' column names, select those from 'item.map in a list, rename it to 'G1', 'G2', then do crossing to expand the dataset, filter based on the logical group column, create the expression with glue_data (from grlue) and flatten the list to a vector
library(dplyr)
library(purrr)
library(stringr)
out <- map(c('group.1', 'group.2'),
~ item.map %>%
select(item.id, .x) %>%
rename_at(.x, ~ str_c('G', str_remove(., "\\D+"))) %>%
crossing(k = 0:2) %>%
filter(across(starts_with('G'), as.logical)) %>%
glue::glue_data("Equal = ({names(.)[2]}, {item.id}, Slope[{k}]);")%>%
as.character) %>%
flatten_chr
-输出
out
#[1] "Equal = (G1, 21, Slope[0]);" "Equal = (G1, 21, Slope[1]);" "Equal = (G1, 21, Slope[2]);" "Equal = (G1, 41, Slope[0]);"
#[5] "Equal = (G1, 41, Slope[1]);" "Equal = (G1, 41, Slope[2]);" "Equal = (G1, 61, Slope[0]);" "Equal = (G1, 61, Slope[1]);"
#[9] "Equal = (G1, 61, Slope[2]);" "Equal = (G2, 41, Slope[0]);" "Equal = (G2, 41, Slope[1]);" "Equal = (G2, 41, Slope[2]);"
#[13] "Equal = (G2, 72, Slope[0]);" "Equal = (G2, 72, Slope[1]);" "Equal = (G2, 72, Slope[2]);"
如果我们希望将两个都为1的分组，
If we want to group those that are 1 in both groups,
i1 <- ave(seq_along(out), sub("G\\d+", "", out), FUN = length)
out[i1 > 1] <- ave(out[i1 > 1], sub("Equal = \\(G\\d+", "", out[i1 > 1]),
FUN = function(x) {
x[1] <- sub(";", "", x[1])
paste(x[1], sub("Equal = ", "", x[2]), sep =", ")
})
out1 <- unique(out)
out1
#[1] "Equal = (G1, 21, Slope[0]);" "Equal = (G1, 21, Slope[1]);"
#[3] "Equal = (G1, 21, Slope[2]);" "Equal = (G1, 41, Slope[0]), (G2, 41, Slope[0]);"
#[5] "Equal = (G1, 41, Slope[1]), (G2, 41, Slope[1]);" "Equal = (G1, 41, Slope[2]), (G2, 41, Slope[2]);"
#[7] "Equal = (G1, 61, Slope[0]);" "Equal = (G1, 61, Slope[1]);"
#[9] "Equal = (G1, 61, Slope[2]);" "Equal = (G2, 72, Slope[0]);"
#[11] "Equal = (G2, 72, Slope[1]);" "Equal = (G2, 72, Slope[2]);"
更新
使用已更新的数据集
Update
With the updated dataset
out <- map(c('group.1', 'group.2', 'group.3', 'group.4'),
~ item.map %>%
select(item.id, .x) %>%
rename_at(.x, ~ str_c('G', str_remove(., "\\D+"))) %>%
crossing(k = 0:4) %>%
filter(across(starts_with('G'), as.logical)) %>%
glue::glue_data("Equal = ({names(.)[2]}, {item.id}, Slope[{k}]);")%>%
as.character) %>%
flatten_chr
out[i1 > 1] <- ave(out[i1 > 1], sub("Equal = \\(G\\d+", "", out[i1 > 1]),
FUN = function(x) {
x[-length(x)] <- sub(";", "", x[-length(x)])
paste(x[1], paste(sub("Equal = ", "", x[-1]), collapse = ", "), sep=", ")
})
unique(out)
[1] "Equal = (G1, 21, Slope[0]), (G3, 21, Slope[0]);"
[2] "Equal = (G1, 21, Slope[1]), (G3, 21, Slope[1]);"
[3] "Equal = (G1, 21, Slope[2]), (G3, 21, Slope[2]);"
[4] "Equal = (G1, 21, Slope[3]), (G3, 21, Slope[3]);"
[5] "Equal = (G1, 21, Slope[4]), (G3, 21, Slope[4]);"
[6] "Equal = (G1, 41, Slope[0]), (G2, 41, Slope[0]), (G3, 41, Slope[0]);"
[7] "Equal = (G1, 41, Slope[1]), (G2, 41, Slope[1]), (G3, 41, Slope[1]);"
[8] "Equal = (G1, 41, Slope[2]), (G2, 41, Slope[2]), (G3, 41, Slope[2]);"
[9] "Equal = (G1, 41, Slope[3]), (G2, 41, Slope[3]), (G3, 41, Slope[3]);"
[10] "Equal = (G1, 41, Slope[4]), (G2, 41, Slope[4]), (G3, 41, Slope[4]);"
[11] "Equal = (G1, 61, Slope[0]), (G3, 61, Slope[0]);"
[12] "Equal = (G1, 61, Slope[1]), (G3, 61, Slope[1]);"
[13] "Equal = (G1, 61, Slope[2]), (G3, 61, Slope[2]);"
[14] "Equal = (G1, 61, Slope[3]), (G3, 61, Slope[3]);"
[15] "Equal = (G1, 61, Slope[4]), (G3, 61, Slope[4]);"
[16] "Equal = (G2, 72, Slope[0]), (G4, 72, Slope[0]);"
[17] "Equal = (G2, 72, Slope[1]), (G4, 72, Slope[1]);"
[18] "Equal = (G2, 72, Slope[2]), (G4, 72, Slope[2]);"
[19] "Equal = (G2, 72, Slope[3]), (G4, 72, Slope[3]);"
[20] "Equal = (G2, 72, Slope[4]), (G4, 72, Slope[4]);"
或嵌套的循环
OUTPUT <- c()
# // loop over the sequence of rows
for(i in seq_len(nrow(item.map))) {
# // nested loop to expand on a sequence
for(k in 0:2) {
# // do a second nest based on the 'Group'
for(j in seq_along(Group)) {
# // create a logical expression based on the 'group' column
i1 <- as.logical(item.map[[paste0("group.", j)]][i])
# // if it is TRUE, then only do the below
if(i1) {
# // create the expression with paste
output <- paste0("Equal = ", paste("(", "G", j,
", ", item.map$item.id[i], ", Slope[", k, "])",
collapse=", ", sep=""))
# // concatenate the NULL vector with the temporary output
OUTPUT <- c(OUTPUT, output)
}
}
}
}
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