[英]How do I access a member of a list within a data.frame based on a column in the data.frame?
[英]How can I reference a list based on a variable within a data.frame?
我有一個帶有emp_id
和job_code
的簡單表。 我想根據job_code
返回正確的payout
我用嵌套的ifelse管理了這個,但是如果我有更多的job_code
呢?
library(dplyr)
set.seed(1)
emp_id <- round(rnorm(100, 500000, 10000))
job_code <- sample(c('a', 'b', 'c'), 100, replace = TRUE)
result <- sample(c(1,2,3,4), 100, replace = TRUE)
df <- data.frame(emp_id = emp_id, job_code = job_code, result = result)
job_a <- c(0, 500, 1000, 5000)
job_b <- c(0, 200, 500, 750)
job_c <- c(0, 250, 750, 1000)
# Works but sucky
df %>% mutate(payout = ifelse(job_code == 'a', job_a[result],
ifelse(job_code == 'b', job_b[result],
job_c[result])))
和dput
如果你喜歡:
structure(list(emp_id = c(493735, 501836, 491644, 515953, 503295,
491795, 504874, 507383, 505758, 496946, 515118, 503898, 493788,
477853, 511249, 499551, 499838, 509438, 508212, 505939, 509190,
507821, 500746, 480106, 506198, 499439, 498442, 485292, 495218,
504179, 513587, 498972, 503877, 499462, 486229, 495850, 496057,
499407, 511000, 507632, 498355, 497466, 506970, 505567, 493112,
492925, 503646, 507685, 498877, 508811, 503981, 493880, 503411,
488706, 514330, 519804, 496328, 489559, 505697, 498649, 524016,
499608, 506897, 500280, 492567, 501888, 481950, 514656, 501533,
521726, 504755, 492901, 506107, 490659, 487464, 502914, 495567,
500011, 500743, 494105, 494313, 498648, 511781, 484764, 505939,
503330, 510631, 496958, 503700, 502671, 494575, 512079, 511604,
507002, 515868, 505585, 487234, 494267, 487754, 495266), job_code = structure(c(1L,
1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 3L, 3L, 1L, 3L, 3L, 3L, 1L, 2L,
3L, 3L, 2L, 1L, 1L, 1L, 2L, 3L, 2L, 1L, 1L, 2L, 3L, 2L, 1L, 2L,
2L, 2L, 3L, 3L, 2L, 2L, 2L, 1L, 2L, 3L, 1L, 2L, 1L, 2L, 1L, 2L,
3L, 3L, 3L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 3L, 2L, 1L, 1L, 3L, 3L,
1L, 1L, 3L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 3L, 1L,
2L, 3L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 2L, 3L, 1L,
1L, 1L, 3L), .Label = c("a", "b", "c"), class = "factor"), result = c(3,
1, 2, 2, 2, 4, 1, 4, 1, 2, 1, 1, 4, 3, 2, 2, 1, 2, 4, 3, 3, 2,
2, 4, 4, 4, 4, 4, 2, 4, 4, 2, 2, 4, 1, 2, 2, 1, 3, 4, 4, 1, 3,
2, 3, 2, 2, 1, 2, 3, 2, 1, 4, 2, 4, 2, 4, 1, 4, 2, 1, 2, 4, 2,
3, 4, 1, 3, 3, 2, 2, 3, 4, 1, 1, 2, 2, 4, 1, 2, 2, 3, 3, 4, 1,
1, 4, 4, 1, 4, 1, 1, 4, 3, 1, 2, 3, 2, 2, 1)), .Names = c("emp_id",
"job_code", "result"), row.names = c(NA, -100L), class = "data.frame")
我理想的做法是在data.frame中獲得支出,但不確定如何正確引用它:
job_payouts <- data.frame(a = job_a, b = job_b, c = job_c)
# Won't work...
df %>% mutate(payout = job_payouts$job_code[result])
這可以通過基礎R中的矩陣索引的超酷方法來實現,這非常快速和有效。
# build jobs payout lookup matrix, by hand (see edit below for an extension)
jobs <- rbind(job_a, job_b, job_c)
# add row names to the matrix for convenient reference
rownames(jobs) <- levels(df$job_code)
# get payout using matrix indexing
df$payout <- jobs[cbind(df$job_code, df$result)]
這回來了
# print out first 6 observations
head(df)
emp_id job_code result payout
1 493735 a 3 1000
2 501836 a 1 0
3 491644 b 2 200
4 515953 a 2 500
5 503295 a 2 500
6 491795 b 4 750
# print out jobs matrix for comparison
jobs
[,1] [,2] [,3] [,4]
a 0 500 1000 5000
b 0 200 500 750
c 0 250 750 1000
有一些值得一提的細節。
data.frame
函數轉換job_code字符向量,因此df$job_code
是一個因子變量,其中標簽與自然數1,2,3,...相關聯。默認情況下,因子的級別按標簽按字母順序排序,因此在此示例中,標簽“a”對應於1,“b”對應於2,“c”對應於3.您可以使用levels
函數查找因子變量的順序,並在該模板之后構造作業矩陣。 cbind(df$job_code, df$result)
形成一個2乘nrow(df)
(100)矩陣,用於使用矩陣索引從作業矩陣中查找每個雇員的nrow(df)
支付值。 R intro手冊有一個關於矩陣索引的很好的介紹部分,其他細節可以在help("[")
。 編輯: 自動構建查找矩陣
在對這個答案的評論中,OP表示擔心手工構建查找矩陣(我稱之為“作業”)將是乏味的並且容易出錯。 為了解決這些有效的問題,我們可以對mget
函數使用一個有點模糊的參數,“ifnotfound”。 這個參數允許我們控制mget
返回的列表元素的輸出,當它們出現在名稱向量中時,但不存在於環境中。
在評論中,我建議使用NA
填寫下面評論中的缺失級別。 我們可以通過使用NA
作為“ifnotfound”的輸入來擴展它。
假設df$job_code
是df$job_code
具有級別“a”,“aa”,“b”和“c”的因子。 然后我們構建查找矩陣如下:
# build vector for example, the actual code, using levels(), follows as a comment
job_codes <- c("a", "aa", "b", "c") # job_codes <- levels(df$jobcodes)
# get ordered list of payouts, with NA for missing payouts
payoutList <- mget(paste0("job_", job_codes), ifnotfound=NA)
它返回一個命名列表。
payoutList
$job_a
[1] 0 500 1000 5000
$job_aa
[1] NA
$job_b
[1] 0 200 500 750
$job_c
[1] 0 250 750 1000
請注意, payoutList$job_aa
是一個NA。 現在,從此列表構建矩陣。
# build lookup matrix using do.call() and rbind()
jobs.lookupMat <- do.call(rbind, payoutList)
jobs.lookupMat
[,1] [,2] [,3] [,4]
job_a 0 500 1000 5000
job_aa NA NA NA NA
job_b 0 200 500 750
job_c 0 250 750 1000
矩陣的行根據因子df$job_code
的級別正確排序,方便地命名,並且NA
在任何沒有支付的地方填充行。
在不更改數據結構的情況下,可以通過定義函數來實現:
job_search <- function(code){
var_name <- paste0("job_",code)
if (exists(var_name)){
return(get(var_name))
}else{
return(NA)
}
}
library(data.table)
setDT(df)
df[, payout := job_search(job_code)[result], by = .(emp_id)]
df
emp_id job_code result payout
1: 493735 a 3 1000
2: 501836 a 1 0
3: 491644 b 2 200
4: 515953 a 2 500
5: 503295 a 2 500
6: 491795 b 4 750
7: 504874 b 1 0
8: 507383 a 4 5000
9: 505758 a 1 0
10: 496946 c 2 250
11: 515118 c 1 0
12: 503898 a 1 0
...
但是,這是保存數據的一種相當不穩定的方法,而且粘貼+獲取語法是錯綜復雜的。
存儲數據的更好方法是在查找表中:
library(data.table)
job_a <- data.frame(payout = c(0, 500, 1000, 5000))
job_b <- data.frame(payout = c(0, 200, 500, 750))
job_c <- data.frame(payout = c(0, 250, 750, 1000))
job_lookup <- rbindlist( #this is a data.table
l = list(a = job_a,b = job_b,c = job_c),
idcol = TRUE
)
# create your result index
job_lookup[, result := 1:.N, by = .id]
job_lookup
.id payout result
1: a 0 1
2: a 500 2
3: a 1000 3
4: a 5000 4
5: b 0 1
6: b 200 2
7: b 500 3
8: b 750 4
9: c 0 1
10: c 250 2
11: c 750 3
12: c 1000 4
# merge to your initial data.frame
merge(df, job_lookup, by.x = c("job_code","result"), by.y = c(".id","result"), all.x = TRUE)
job_code result emp_id payout
1 a 1 505758 0
2 a 1 501836 0
3 a 1 503898 0
4 a 1 494575 0
5 a 1 487464 0
6 a 1 503700 0
7 a 1 505939 0
8 a 1 503330 0
9 a 1 512079 0
10 a 1 481950 0
11 a 1 507685 0
12 a 1 490659 0
...
使用tidyverse的工具:
library(dplyr)
library(stringr)
library(tidyr)
# your data
set.seed(1)
emp_id <- round(rnorm(100, 500000, 10000))
job_code <- sample(c('a', 'b', 'c'), 100, replace = TRUE)
result <- sample(c(1,2,3,4), 100, replace = TRUE)
# construct a data frame
df <-
data.frame(emp_id = emp_id,
job_code = job_code,
result = result,
stringsAsFactors = FALSE)
# your jobs
job_a <- c(0, 500, 1000, 5000)
job_b <- c(0, 200, 500, 750)
job_c <- c(0, 250, 750, 1000)
# construct a data frame
my_job <-
data.frame(job_a, job_b, job_c) %>%
gather(job, value) %>%
group_by(job) %>%
mutate(result = 1:n(),
job_code = str_replace(job, "job_", "")) %>%
ungroup %>%
select(-job)
# join df and my_job into my_results table
my_results <-
left_join(df, my_job)
結果 :
my_results %>% tbl_df
Source: local data frame [100 x 4]
emp_id job_code result value
(dbl) (chr) (dbl) (dbl)
1 493735 a 3 1000
2 501836 a 1 0
3 491644 b 2 200
4 515953 a 2 500
5 503295 a 2 500
6 491795 b 4 750
7 504874 b 1 0
8 507383 a 4 5000
9 505758 a 1 0
10 496946 c 2 250
.. ... ... ... ...
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