Question

species <- c("frog1","frog1","frog1","frog1","frog1","frog1","frog1","frog1"
        ,"frog1","frog1","frog2","frog2","frog2","frog2","frog2",
        "frog2","frog2","frog2","frog2","frog2")
month <- c(1,12,5,8,10,3,5,7,9,4,2,4,6,7,6,3,8,9,11,1)
number <- c(3,4,5,1,2,3,4,7,6,7,3,5,6,7,8,9,9,5,3,1)
a<- data.frame(species,month,number)

我的数据框意味着我捕获了两种青蛙，frog1和frog2，不同月份的数字不同。

我想将月份转换为4个季节。第一季是第1个月，第12个月，第2个月，第二个是4,3,5，第三个是7,6,8，第四个是10,9,11。这四季有顺序，即第一季我想选择第1个月，第12个月，第2个月，第二个季节，同样地，在第二个季节我会先选择第4个月，第3个月第二个最终，第5个月，等等。例如，在青蛙1中，有2个月的1和12，我想在第一季中拿到第1个月而不是第12个月。

我想问一下如何创建一个可以在两种青蛙中依次为4个季节选择最重要月份的列。例如，在青蛙1中，有2个月1和12，我会喜欢在第一季中选择第1个月而不是第12个月。

我的预期输出是：

species <- c("frog1","frog1","frog1","frog1","frog1","frog1","frog1","frog1"
,"frog1","frog1","frog2","frog2","frog2","frog2","frog2",
"frog2","frog2","frog2","frog2","frog2")
month <- c(1,12,5,8,10,3,5,7,9,4,2,4,6,7,6,3,8,9,11,1)
number <- c(3,4,5,1,2,3,4,7,6,7,3,5,6,7,8,9,9,5,3,1)
choosemonth <- c("season1","","","","season4","","","season3","","season2",
         "","season2","","season3","","","","season4","","season1")
b<- data.frame(species,month,number,choosemonth)

Answer 1

我猜你最后想要的结果，但是这里是如何创造季节和重要性的，我正在解决每个物种最重要的月份

这是dplyr的一种方式：

library(dplyr)
a %>% 
  # Season is basically just one-off quarters
  mutate(season = trunc((month + 1)%%12 / 3),
  # for each month the value mod 3 goes in order 2,3,1
    importance = c(2,3,1)[month %% 3 + 1]) %>%
  group_by(season, species) %>%
  # keep only those with the max importance
  filter(importance == max(importance))

编辑：看起来你只想标记最重要的值，所以这里是如何做到的，

a %>% 
  # Season is basically just one-off quarters
  mutate(season = trunc((month + 1)%%12 / 3),
  # for each month the value mod 3 goes in order 2,3,1
         importance = c(2,3,1)[month %% 3 + 1]) %>%
  mutate(choosemonth = ifelse(importance == 3, paste0('season',season + 1),''))

编辑2：再编辑一次，分为3个季而不是4个季。

选择一个季节的最佳月份

1 个答案: