英文:
Find the very last character of the previous row in R
问题
以下是翻译的内容：
让我们假设我有一个像这样的数据框：
d <- data.frame(order = c("101", "01", "10", "01", "101"),dur_1 = c(50, 20, 40, 25, 45),dur_2 = c(40, 30, 45, 34, 96),dur_3 = c(20, 0, 0, 0, 125))
我想要做的是创建新的列 pre_order 和 pre_dur。
pre_order 是前一个 order 的最后一个字符。所以第一行应该是 NA，因为我们不知道前一个 order 的值。
pre_dur 是前一个 dur_ 列的最后一个非负值。我想要表示为 dur_，因为在我的实际数据中，我必须处理一个通用版本。所以我必须将其表示为 dur_。我的预期输出应该如下所示：
d1<-data.frame(order=c("101","01","10","01","101"),dur_1=c(50,20,40,25,45),dur_2=c(40,30,45,34,96),dur_3=c(20,0,0,0,125),pre_order=c(NA,1,1,0,1),pre_dur=c(NA,20,30,45,34))
我不知道如何在R中实现这一点。
英文:
Let's say I have a dataframe like this:
d <- data.frame(order = c("101", "01", "10", "01", "101"),dur_1 = c(50, 20, 40, 25, 45),dur_2 = c(40, 30, 45, 34, 96),dur_3 = c(20, 0, 0, 0, 125))
What I want to do is make new columns pre_order and pre_dur.
pre_order is the very last character of the previous order. So the first row should be NA. Because we don't know the previous value of order
pre_dur is the very last nonnegative value of the previous dur_ column. I want to express as dur_ beacause, in my actual data, I have to deal with a generalized version. So I have to express it as dur_. My expected output should look like this:
d1<-data.frame(order=c("101","01","10","01","101"),dur_1=c(50,20,40,25,45),dur_2=c(40,30,45,34,96),dur_3=c(20,0,0,0,125),pre_order=c(NA,1,1,0,1),pre_dur=c(NA,20,30,45,34))
I don't know how to do that in R
答案1
得分: 3
使用 max.col 和 substring 的基本变体。
i <- which(startsWith(names(d), "dur"))j <- max.col(d[i] > 0, "last")cbind(d, rbind(NA, data.frame(pre_order = substring(d$order, nchar(d$order)),pre_dur = d[cbind(seq_len(nrow(d)), i[j])])[-nrow(d),]))# order dur_1 dur_2 dur_3 pre_order pre_dur#1 101 50 40 20 <NA> <NA>#2 01 20 30 0 1 20#3 10 40 45 0 1 30#4 01 25 34 0 0 45#5 101 45 96 125 1 34
英文:
A base variant using max.col and substring.
i <- which(startsWith(names(d), "dur"))j <- max.col(d[i] > 0, "last")cbind(d, rbind(NA,data.frame(pre_order = substring(d$order, nchar(d$order)),pre_dur = d[cbind(seq_len(nrow(d)), i[j])])[-nrow(d),]))# order dur_1 dur_2 dur_3 pre_order pre_dur#1 101 50 40 20 <NA> <NA>#2 01 20 30 0 1 20#3 10 40 45 0 1 30#4 01 25 34 0 0 45#5 101 45 96 125 1 34
答案2
得分: 1
以下是您要翻译的内容：
Edit:
This approach solves the first part of your question but doesn't account for the "what if I have an unknown number of 'dur_' columns?" part of the question. I tried a few different solutions using purrr::map() but I wasn't able to get the correct output with your criteria.
GKi's solution successfully answers both parts of your question.
Original Answer:
If I've understood correctly, this is one potential solution:
library(tidyverse)d <- data.frame(order = c("101", "01", "10", "01", "101"),dur_1 = c(50, 20, 40, 25, 45),dur_2 = c(40, 30, 45, 34, 96),dur_3 = c(20, 0, 0, 0, 125))d1<-data.frame(order=c("101","01","10","01","101"),dur_1=c(50,20,40,25,45),dur_2=c(40,30,45,34,96),dur_3=c(20,0,0,0,125),pre_order=c(NA,1,1,0,1),pre_dur=c(NA,20,30,45,34))d1#> order dur_1 dur_2 dur_3 pre_order pre_dur#> 1 101 50 40 20 NA NA#> 2 01 20 30 0 1 20#> 3 10 40 45 0 1 30#> 4 01 25 34 0 0 45#> 5 101 45 96 125 1 34d %>%mutate(pre_order = lag(str_sub(order, -1)),pre_dur = case_when(lag(dur_3) > 0 ~ lag(dur_3),lag(dur_2) > 0 ~ lag(dur_2),lag(dur_1) > 0 ~ lag(dur_1),TRUE ~ NA_real_))#> order dur_1 dur_2 dur_3 pre_order pre_dur#> 1 101 50 40 20 <NA> NA#> 2 01 20 30 0 1 20#> 3 10 40 45 0 1 30#> 4 01 25 34 0 0 45#> 5 101 45 96 125 1 34
Created on 2023-05-22 with reprex v2.0.2
英文:
Edit:
This approach solves the first part of your question but doesn't account for the "what if I have an unknown number of 'dur_' columns?" part of the question. I tried a few different solutions using purrr::map() but I wasn't able to get the correct output with your criteria.
GKi's solution successfully answers both parts of your question.
Original Answer:
If I've understood correctly, this is one potential solution:
library(tidyverse)d <- data.frame(order = c("101", "01", "10", "01", "101"),dur_1 = c(50, 20, 40, 25, 45),dur_2 = c(40, 30, 45, 34, 96),dur_3 = c(20, 0, 0, 0, 125))d1<-data.frame(order=c("101","01","10","01","101"),dur_1=c(50,20,40,25,45),dur_2=c(40,30,45,34,96),dur_3=c(20,0,0,0,125),pre_order=c(NA,1,1,0,1),pre_dur=c(NA,20,30,45,34))d1#> order dur_1 dur_2 dur_3 pre_order pre_dur#> 1 101 50 40 20 NA NA#> 2 01 20 30 0 1 20#> 3 10 40 45 0 1 30#> 4 01 25 34 0 0 45#> 5 101 45 96 125 1 34d %>%mutate(pre_order = lag(str_sub(order, -1)),pre_dur = case_when(lag(dur_3) > 0 ~ lag(dur_3),lag(dur_2) > 0 ~ lag(dur_2),lag(dur_1) > 0 ~ lag(dur_1),TRUE ~ NA_real_))#> order dur_1 dur_2 dur_3 pre_order pre_dur#> 1 101 50 40 20 <NA> NA#> 2 01 20 30 0 1 20#> 3 10 40 45 0 1 30#> 4 01 25 34 0 0 45#> 5 101 45 96 125 1 34
<sup>Created on 2023-05-22 with reprex v2.0.2</sup>
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