CODEWITHSUNDEEP
r
bic ton
bic ton

Reputation: 1408

Compute the max and min daily from hourly data?

If I have this data:

   dat=structure(list(year = c(1980L, 1980L, 1980L, 1980L, 1980L, 1980L
   ), DOY = c(144, 144.042, 144.083, 144.125, 144.167, 144.208), 
   time = c(0, 1, 2, 3, 4, 5), B = c(67, 7, 7, 
   0, 7, 6), gg = c(67, 7, 7, 
  0, 7, 6)),row.names = c(NA, 6L), class = "data.frame")

I would like to compute the max and min per day of all columns and return

      year DOY  max.B  min.B  mean.B max.gg min.gg mean.gg

I have several columns not only B and gg

Upvotes: 0

Views: 339

Answers (2)

Rui Barradas
Rui Barradas

Reputation: 76412

Maybe the followqing does what the question asks for.

library(dplyr)
library(tidyr)

dat %>%
  mutate(day = strptime(paste(year, DOY), format = "%Y %j"),
         day = as.Date(day)) %>%
  select(-year, -DOY) %>%
  pivot_longer(-day) %>%
  group_by(day, name) %>%
  summarise(max = max(value), min = min(value), .groups = "drop") %>%
  pivot_wider(
    id_cols = day, 
    names_from = name,
    names_glue = "{name}_{.value}",
    values_from = c(max, min)
  )

Edit

To compute the means of min and max, assign the previous result to a variable daily, keep only the year/month in day, reshape to long format, group and compute the value column means. Then reshape back to wide format.

daily %>% 
  mutate(month = format(day, "%Y-%m")) %>%
  select(-day) %>%
  pivot_longer(-month) %>%
  mutate(name = sub("_.*$", "", name)) %>%
  group_by(month, name) %>%
  summarise(mean = mean(value, na.rm = TRUE), .groups = "drop") %>%
  pivot_wider(
    id_cols = month,
    names_from = name,
    names_glue = "{name}_{.value}",
    values_from = mean
  )

Upvotes: 1

andschar
andschar

Reputation: 3973

With data.table you can do:

require(data.table)
setDT(dat)
dat[ ,
     .(min.B = min(B),
       max.B = max(B),
       mean.B = mean(B),
       min.gg = min(gg),
       max.gg = max(gg),
       mean.gg = mean(gg)),
     .(year, DOY) ]

Upvotes: 1

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