R 计算行中多个值的总和

嗨，我有一个i/p数据框，如下所示

df = data.frame('value' = c("(1_00),(0_04),(1_08),(0_12),(1_14)","(1_15),(0_22),(1_29)",
                            "(0_30),(1_38),(0_40),(1_44)","(0_45),(1_57),(0_59)",
                            "(0_15),(1_21),(0_26),(0_29)"),stringsAsFactors = F)

for row:1-  i.e `(1_00),(0_04),(1_08),(0_12),(1_14)` calculated as second set  
 of character in second block after "_" (04) minus second set (00) in the first 
 block multply by first set in   first block  (i.e: (4-0)*1 )    
                  similarly from 3rd block to 2nd block .finally we need need to   add all blocks.`(4-0)*1 + (8-4)*0 +(12-8)*1 + (14-12)*0`

从上面的数据框中，我无法找到“样本计算”列中的“输出”列 “样本计算”的计算如下所示

df = data.frame('value' = c("(1_00),(0_04),(1_08),(0_12),(1_14)","(1_15),(0_22),(1_29)",
                            "(0_30),(1_38),(0_40),(1_44)","(0_45),(1_57),(0_59)",
                            "(0_15),(1_21),(0_26),(0_29)"),stringsAsFactors = F)

for row:1-  i.e `(1_00),(0_04),(1_08),(0_12),(1_14)` calculated as second set  
 of character in second block after "_" (04) minus second set (00) in the first 
 block multply by first set in   first block  (i.e: (4-0)*1 )    
                  similarly from 3rd block to 2nd block .finally we need need to   add all blocks.`(4-0)*1 + (8-4)*0 +(12-8)*1 + (14-12)*0`

o/p df1

df1 = data.frame('value' = c("(1_00),(0_04),(1_08),(0_12),(1_14)",
                             "(1_15),(0_22),(1_29)",
                             "(0_30),(1_38),(0_40),(1_44)",
                             "(0_45),(1_57),(0_59)","(1_00),(0_07),(1_14)",
                             "(0_15),(1_21),(0_26),(0_29)"),
                              "sample_calc"=c("(4-0)*1 + (8-4)*0 +(12-8)*1 + (14-12)*0",
                                              "(22-15)*1 + (29-22)*0",
                                              "(38-30)*0 + (40-38)*1 + (44-40)*0",
                                              "(57-45)*0 + (59-57)*1",
                                              "(7-0)*1 + (14-7)*0",
                                              "(21-15)*0  + (26-21)*1 + (29-26)*0"),
                            "output"=c(8,7,2,2,7,5),stringsAsFactors = F)

---

首先，我将使用以下代码将字符串转换为数字字符串：

foo <- lapply(strsplit(gsub("\\(|\\)", "", df$value), ","),
              function(x) as.numeric(unlist(strsplit(x, "_"))))

接下来，我们只需要对

foo

的子列表应用您的计算（我们得到“第二个集合”中元素的“id”（

y

）并应用所需的逻辑：

sum（（x[y]-x[y-2]）*x[y-3]）

）

---

sapply（foo，函数（x）{y
process.df.column下面是一个基本的R解决方案：

df$output <-
  sapply(strsplit(gsub('[()]','',df$value),'[_,]'),
       function(x) {
         x<-as.numeric(x)
         sum(-x[!seq_along(x)%%2]*diff(c(0,head(x[!!seq_along(x)%%2],-1),0)))})
# df
#                                value output
# 1 (1_00),(0_04),(1_08),(0_12),(1_14)      8
# 2               (1_15),(0_22),(1_29)      7
# 3        (0_30),(1_38),(0_40),(1_44)      2
# 4               (0_45),(1_57),(0_59)      2
# 5        (0_15),(1_21),(0_26),(0_29)      5

问题很清楚（或多或少），OP提供了可复制的示例和想要的输出。关闭将过于苛刻。比我的解决方案更好。；）
df$output <-
  sapply(strsplit(gsub('[()]','',df$value),'[_,]'),
       function(x) {
         x<-as.numeric(x)
         sum(-x[!seq_along(x)%%2]*diff(c(0,head(x[!!seq_along(x)%%2],-1),0)))})
# df
#                                value output
# 1 (1_00),(0_04),(1_08),(0_12),(1_14)      8
# 2               (1_15),(0_22),(1_29)      7
# 3        (0_30),(1_38),(0_40),(1_44)      2
# 4               (0_45),(1_57),(0_59)      2
# 5        (0_15),(1_21),(0_26),(0_29)      5

library(tidyverse)

df %>%
  rowid_to_column() %>%
  separate_rows(value,sep=",") %>%
  mutate_at("value",~gsub('[()]','',.x)) %>%
  separate(value,c("a","b"),convert = T) %>%
  group_by(rowid) %>%
  mutate_at("a",~-diff(c(0,.x[-length(.x)],0))) %>%
  summarize(output = sum(a*b)) %>%
  select(-rowid) %>%
  cbind(df,.)

#                                value output
# 1 (1_00),(0_04),(1_08),(0_12),(1_14)      8
# 2               (1_15),(0_22),(1_29)      7
# 3        (0_30),(1_38),(0_40),(1_44)      2
# 4               (0_45),(1_57),(0_59)      2
# 5        (0_15),(1_21),(0_26),(0_29)      5