英文:
Parse a function input name as an output name
问题
我试图在我的函数"result"中将数据输入变量名作为输出值添加到一个单独的列($V5)中。
输入是不同的数据值，我想将这些名称保存在输出中，以便我可以追踪数据的来源。
我以为这会很简单，我可以使用print(deparse(substitute(input))来实现，但这不起作用。
有人能推荐一个解决方案吗？
英文:
I am trying to add the data input variable name as an output value in a separate column ($V5 within my function "result).
The inputs are different data values, and would like to save these names in the outputs so I can track where the data is coming from.
I thought this would be fairly straightforward, and that I could use print(deparse(substitute(input))) but this doesn't work.
Can anyone recommend a solution?
library(dplyr)library(tidyr)## Inputs ##input_1 = structure(list(V1 = c("Team_2022", "Team_2022", "Team_2022"), V2 = c("Frank", "Mary", "John"), V3 = c("Sydney", "Sydney", "Sydney"), V4 = c(55, 76, 14)), row.names = c(NA, -3L), class = c("data.table", "data.frame"))input_2 = structure(list(V1 = c("Team_2023", "Team_2023", "Team_2023"), V2 = c("Bill", "Mary", "John"), V3 = c("Sydney", "Sydney", "Sydney"), V4 = c(113, 23, 10)), row.names = c(NA, -3L), class = c("data.table", "data.frame"))input_3 = structure(list(V1 = c("Team_2024", "Team_2024", "Team_2024"), V2 = c("Frank", "Mary", "Bill"), V3 = c("Sydney", "Sydney", "Sydney"), V4 = c(7, 19, 52)), row.names = c(NA, -3L), class = c("data.table", "data.frame"))input_4 = structure(list(V1 = c("Team_2025", "Team_2025", "Team_2025"), V2 = c("Frank", "Mary", "John"), V3 = c("Sydney", "Sydney", "Sydney"), V4 = c(46, 44, 88)), row.names = c(NA, -3L), class = c("data.table", "data.frame"))## Teams ##teams = structure(list(V1 = c("team1", "team2", "team3"), V2 = c("Mary + Frank","Mary + John", "Mary + Bill")), class = "data.frame", row.names = c(NA, -3L))## Group the inputs into one ##all_objects = ls()input_objects = grep("^input", all_objects, value = T)input_test = as.data.frame(input_obj)## Function ##result = function(input, teams) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>% fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4),V5 = print(deparse(substitute(input))))return(data)}all_objects <- ls()input_objects <- grep("^input_\\d", all_objects, value = T)input_test <- lapply(input_objects, get)output = input_test %>%lapply(result, teams) %>%bind_rows()### Current output ###structure(list(V1.x = c("team1", "team2", "team3", "team1", "team2","team3", "team1", "team2", "team3", "team1", "team2", "team3"), V1.y = c("Team_2022", "Team_2022", "Team_2022", "Team_2023","Team_2023", "Team_2023", "Team_2024", "Team_2024", "Team_2024","Team_2025", "Team_2025", "Team_2025"), V2 = c("Mary + Frank","Mary + John", "Mary + Bill", "Mary + Frank", "Mary + John","Mary + Bill", "Mary + Frank", "Mary + John", "Mary + Bill","Mary + Frank", "Mary + John", "Mary + Bill"), V3 = c("Sydney","Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney","Sydney", "Sydney", "Sydney", "Sydney"), V4 = c(131, 90, 76,23, 33, 136, 26, 19, 71, 90, 132, 44), V5 = c("input", "input","input", "input", "input", "input", "input", "input", "input","input", "input", "input")), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -12L))### Desired Output ###structure(list(V1.x = c("team1", "team2", "team3", "team1", "team2","team3", "team1", "team2", "team3", "team1", "team2", "team3"), V1.y = c("Team_2022", "Team_2022", "Team_2022", "Team_2023","Team_2023", "Team_2023", "Team_2024", "Team_2024", "Team_2024","Team_2025", "Team_2025", "Team_2025"), V2 = c("Mary + Frank","Mary + John", "Mary + Bill", "Mary + Frank", "Mary + John","Mary + Bill", "Mary + Frank", "Mary + John", "Mary + Bill","Mary + Frank", "Mary + John", "Mary + Bill"), V3 = c("Sydney","Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney","Sydney", "Sydney", "Sydney", "Sydney"), V4 = c(131, 90, 76,23, 33, 136, 26, 19, 71, 90, 132, 44), V5 = c("input_1", "input_1","input_1", "input_2", "input_2", "input_2", "input_3", "input_3", "input_3","input_4", "input_4", "input_4")), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -12L))
答案1
得分: 1
如果我理解你的意思正确：purrr::map_df() 函数具有一个很好的特性，可以识别最终 data.frames 中的输入列表（最好是有命名的）：
library(dplyr)library(tidyr)# dropping V5 as it will be "automatically" computedresult = function(input, teams) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>%fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4)# we do not need V5 here anymore)return(data)}l_objects <- ls()input_objects <- grep("^input_\\d", l_objects, value = TRUE)input_test <- lapply(input_objects, get)# name the object list to use the name for identificationnames(input_test) <- input_objects# use purrr map to data.frame with the .id featurepurrr::map_df(input_test, ~result(.x, teams), .id = "V5")
请注意，这也可以在不命名列表的情况下工作，尽管你只会得到列表项的编号，这可能不够。
此外，plyr::ldply 函数也可以用于绑定命名列表，生成一个新列，其中包含结果数据帧中的列表名称。
英文:
If I understand you correctly: the purrr::map_df() function has a nice feature to identify inputs lists (ideally named) in final data.frames:
library(dplyr)library(tidyr)# dropping V5 as it will be "automatically" computedresult = function(input, teams) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>%fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4)# we do not need V5 here anymore)return(data)}l_objects <- ls()input_objects <- grep("^input_\\d", all_objects, value = T)input_test <- lapply(input_objects, get)# name the object list to use the name for identificationnames(input_test) <- input_objects# use purrr map to data.frame with the .id featurepurrr::map_df(input_test, ~result(.x, teams), .id = "V5")V5 V1.x V1.y V2 V3 V4<chr> <chr> <chr> <chr> <chr> <dbl>1 input_1 team1 Team_2022 Mary + Frank Sydney 1312 input_1 team2 Team_2022 Mary + John Sydney 903 input_1 team3 Team_2022 Mary + Bill Sydney 764 input_2 team1 Team_2023 Mary + Frank Sydney 235 input_2 team2 Team_2023 Mary + John Sydney 336 input_2 team3 Team_2023 Mary + Bill Sydney 1367 input_3 team1 Team_2024 Mary + Frank Sydney 268 input_3 team2 Team_2024 Mary + John Sydney 199 input_3 team3 Team_2024 Mary + Bill Sydney 7110 input_4 team1 Team_2025 Mary + Frank Sydney 9011 input_4 team2 Team_2025 Mary + John Sydney 13212 input_4 team3 Team_2025 Mary + Bill Sydney 44
Note that this works without naming the list aswell, though you will only get the list item number, which might be insufficient.
Also the plyr::ldply function can be used to bind named lists, generating a new column with the list names in the result data.frame.
答案2
得分: 1
不使用deparse/substitute，而是在result函数中为名称创建一个参数，然后使用它。
library(dplyr)library(purrr)library(tidyr)result <- function(input, teams, inputnm) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>%fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4),V5 = inputnm)return(data)}
测试
input_test %>%pull(input_objects) %>%mget(inherits = TRUE) %>%imap_dfr(~ result(.x, teams, .y))
输出
# A tibble: 12 × 6V1.x V1.y V2 V3 V4 V5<chr> <chr> <chr> <chr> <dbl> <chr>1 team1 Team_2022 Mary + Frank Sydney 131 input_12 team2 Team_2022 Mary + John Sydney 90 input_13 team3 Team_2022 Mary + Bill Sydney 76 input_14 team1 Team_2023 Mary + Frank Sydney 23 input_25 team2 Team_2023 Mary + John Sydney 33 input_26 team3 Team_2023 Mary + Bill Sydney 136 input_27 team1 Team_2024 Mary + Frank Sydney 26 input_38 team2 Team_2024 Mary + John Sydney 19 input_39 team3 Team_2024 Mary + Bill Sydney 71 input_310 team1 Team_2025 Mary + Frank Sydney 90 input_411 team2 Team_2025 Mary + John Sydney 132 input_412 team3 Team_2025 Mary + Bill Sydney 44 input_4
这是您提供的代码的翻译。
英文:
Instead of doing the deparse/substitute, create an argument in result for the names as well, and then use that
library(dplyr)library(purrr)library(tidyr)result <- function(input, teams, inputnm) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>% fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4),V5 = inputnm)return(data)}
-testing
input_test %>%pull(input_objects) %>%mget(inherits = TRUE) %>%imap_dfr(~ result(.x, teams, .y))
-output
# A tibble: 12 × 6V1.x V1.y V2 V3 V4 V5<chr> <chr> <chr> <chr> <dbl> <chr>1 team1 Team_2022 Mary + Frank Sydney 131 input_12 team2 Team_2022 Mary + John Sydney 90 input_13 team3 Team_2022 Mary + Bill Sydney 76 input_14 team1 Team_2023 Mary + Frank Sydney 23 input_25 team2 Team_2023 Mary + John Sydney 33 input_26 team3 Team_2023 Mary + Bill Sydney 136 input_27 team1 Team_2024 Mary + Frank Sydney 26 input_38 team2 Team_2024 Mary + John Sydney 19 input_39 team3 Team_2024 Mary + Bill Sydney 71 input_310 team1 Team_2025 Mary + Frank Sydney 90 input_411 team2 Team_2025 Mary + John Sydney 132 input_412 team3 Team_2025 Mary + Bill Sydney 44 input_4
答案3
得分: 0
如果您一次只有一个输入，简单地将分配V5的行移到您的summarize和dplyr管道之外就可以了：
result = function(input, teams) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>% fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4),#V5 = print(deparse(substitute(input))))data$V5 <- deparse(substitute(input))return(data)}result(input_1, teams)# V1.x V1.y V2 V3 V4 V5# <chr> <chr> <chr> <chr> <dbl> <chr># 1 team1 Team_2022 Mary + Frank Sydney 131 input_1# 2 team2 Team_2022 Mary + John Sydney 90 input_1# 3 team3 Team_2022 Mary + Bill Sydney 76 input_1
但您使用列表的方式有些挑战（对于该代码会输出X[[i]]）。
为了解决这个问题，使用lapply，我建议简单地添加一个额外的输入来获取名称，并对lapply函数进行额外的调整以适应它：
result = function(input, teams, nme) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>% fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4),#V5 = print(deparse(substitute(input))))data$V5 <- nmereturn(data)}all_objects <- ls()input_objects <- grep("^input_\\d", all_objects, value = T)input_test <- lapply(input_objects, get)# 添加分配名称给列表names(input_test) <- input_objectsoutput = lapply(input_objects, function(x) result(input_test[[x]], teams, nme = x)) %>%bind_rows()# V1.x V1.y V2 V3 V4 V5# <chr> <chr> <chr> <chr> <dbl> <chr># 1 team1 Team_2022 Mary + Frank Sydney 131 input_1# 2 team2 Team_2022 Mary + John Sydney 90 input_1# 3 team3 Team_2022 Mary + Bill Sydney 76 input_1# 4 team1 Team_2023 Mary + Frank Sydney 23 input_2# 5 team2 Team_2023 Mary + John Sydney 33 input_2# 6 team3 Team_2023 Mary + Bill Sydney 136 input_2# 7 team1 Team_2024 Mary + Frank Sydney 26 input_3# 8 team2 Team_2024 Mary + John Sydney 19 input_3# 9 team3 Team_2024 Mary + Bill Sydney 71 input_3# 10 team1 Team_2025 Mary + Frank Sydney 90 input_4# 11 team2 Team_2025 Mary + John Sydney 132 input_4# 12 team3 Team_2025 Mary + Bill Sydney 44 input_4
英文:
If you just had one input at a time, simply moving the line for assigning V5 outside of your summarize and dplyr pipes would do it:
result = function(input, teams) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>% fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4),#V5 = print(deparse(substitute(input))))data$V5 <- deparse(substitute(input))return(data)}result(input_1, teams)# V1.x V1.y V2 V3 V4 V5# <chr> <chr> <chr> <chr> <dbl> <chr># 1 team1 Team_2022 Mary + Frank Sydney 131 input_1# 2 team2 Team_2022 Mary + John Sydney 90 input_1# 3 team3 Team_2022 Mary + Bill Sydney 76 input_1
But your use of lists makes that a bit challenging (it will output X[[i]] for that code).
To address this for using lapply, I would suggest simply adding an additional input that takes the name and simply assigning it that name, with additional tweaks to the lapply function to accommodate it:
result = function(input, teams, nme) {data = teams %>%separate_rows(V2) %>%left_join(input, by = c("V2" = "V2")) %>%replace_na(list(V4 = 0)) %>%group_by(V1.x) %>% fill(V1.y, V3) %>%summarize(V1.y = first(V1.y),V2 = paste(V2, collapse = " + "),V3 = first(V3),V4 = sum(V4),#V5 = print(deparse(substitute(input))))data$V5 <- nmereturn(data)}all_objects <- ls()input_objects <- grep("^input_\\d", all_objects, value = T)input_test <- lapply(input_objects, get)# add in assigning names to the listnames(input_test) <- input_objectsoutput = lapply(input_objects, function(x) result(input_test[[x]], teams, nme = x)) %>%bind_rows()# V1.x V1.y V2 V3 V4 V5# <chr> <chr> <chr> <chr> <dbl> <chr># 1 team1 Team_2022 Mary + Frank Sydney 131 input_1# 2 team2 Team_2022 Mary + John Sydney 90 input_1# 3 team3 Team_2022 Mary + Bill Sydney 76 input_1# 4 team1 Team_2023 Mary + Frank Sydney 23 input_2# 5 team2 Team_2023 Mary + John Sydney 33 input_2# 6 team3 Team_2023 Mary + Bill Sydney 136 input_2# 7 team1 Team_2024 Mary + Frank Sydney 26 input_3# 8 team2 Team_2024 Mary + John Sydney 19 input_3# 9 team3 Team_2024 Mary + Bill Sydney 71 input_3# 10 team1 Team_2025 Mary + Frank Sydney 90 input_4# 11 team2 Team_2025 Mary + John Sydney 132 input_4# 12 team3 Team_2025 Mary + Bill Sydney 44 input_4
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