R 中有条件的 Cumsum

Question

set.seed(123)
df <- data.frame(loc.id = rep(1:3,each = 3*5), 
             year = rep(rep(1981:1983, each = 5), times = 3), 
             week = rep(rep(20:24, times = 3), times = 3),
             cumsum.val = runif(min  = -2, max = 4, 5*3*3))

数据包含 3 locations X 3 years X 5 weeks 和一个名为 cumsum.val 的值。对于每个位置和年份，我想找到 cumsum.val > 1 所在的星期。然后，如果连续 cumsum.val > 1 的两周出现 select 第一周。一个例子

  test <- df[df$loc.id == 1 & df$year == 1981,]
  test$cumsum.test <- test$cumsum.val > 1 # weeks where cumsum.val > 1
  head(test)
    loc.id year   week cumsum.val cumsum.test
 1      1 1981   20 -0.2745349       FALSE
 2      1 1981   21  2.7298308        TRUE
 3      1 1981   22  0.4538615       FALSE
 4      1 1981   23  3.2981044        TRUE
 5      1 1981   24  3.6428037        TRUE

现在 select 连续两次出现 TRUE 的第一周，在上面的例子中是周 23（因为周 23 和 24 都是 TRUE).

如何为 df 实现此功能。可能是 cumusm.val > 1 没有连续出现两周的情况。在这种情况下，简单 select 第一周 cumsum.val > 1

Answer 1

set.seed(123)
df <- data.frame(loc.id = rep(1:3,each = 3*5), 
                 year = rep(rep(1981:1983, each = 5), times = 3), 
                 week = rep(rep(20:24, times = 3), times = 3),
                 cumsum.val = runif(min  = -2, max = 4, 5*3*3))

View(df)
b <- unique(df$loc.id)
data <- data.frame()
for(i in seq_along(b)){
  check=0
  for(j in 1:length(df$loc.id)){
    if(df$cumsum.val[j]>1 && df$loc.id[j]==b[i]){
      check=check+1
    }
    else if(df$loc.id[j]==b[i]){
      check=0
    }
    if(check>=2){
      data1 <- data.frame(week1=df$week[j-1],idd=df$loc.id[j])
      data <- rbind(data,data1)
    }
  } 
}

Answer 2

一个基于dplyr的解决方案可以解决问题。请注意 cumsum.test 已被计算为 numeric，因此 lag 和 lead 的 default 值可以用于 0/1.[=18 以外的值=]

df %>% mutate(cumsum.test = as.numeric(cumsum.val>1)) %>%
  group_by(loc.id, year) %>%
  mutate(SelctCond = ifelse(cumsum.test == 1 & 
                       cumsum.test == lead(cumsum.test, default = -1L) &
                       cumsum.test != lag(cumsum.test, default = -1L), TRUE , FALSE )) %>%
  filter(SelctCond) %>%
  select(-SelctCond)
# # Groups: loc.id, year [6]
# loc.id  year  week cumsum.val cumsum.test
# <int> <int> <int>      <dbl>       <dbl>
# 1      1  1981    23       3.30        1.00
# 2      1  1982    21       1.17        1.00
# 3      1  1983    22       2.07        1.00
# 4      2  1982    20       3.34        1.00
# 5      2  1983    20       2.25        1.00
# 6      3  1981    20       3.78        1.00

Answer 3

一个data.table方法：

require(data.table) # load package
setDT(df) # Convert to data.table
df[, cumsum.test := cumsum.val > 1] # create new variable

# Find consecutive values, check they are indeed cumsum.val > 1, and return the first row of them:
df[c(diff(cumsum.test), NA) == 0 & cumsum.test == TRUE, .SD[1, ]]

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