How to make monthly data to daily data using random techniques in R

Question

I need to transform this monthly data to daily data by using some randomization technique, for example. Here is the dataframe:


library(dplyr)
library(lubridate)

month_year <- c( 
"08-2021",
"09-2021",
"10-2021",
"11-2021",
"12-2021"
)

monthly_values_var1 <- c(
598,
532,
736,
956,
780
)

monthly_values_var2 <- c(
18.3179,
62.6415,
11.1033,
30.7443,
74.2076
)

df <- data.frame(month_year, monthly_values_var1,  monthly_values_var2) 
df

That is the month dataset view:

And the expected result of this, is something like this:

How to do it using R ?

Is there some error you want on the random variable? or a distribution you're looking for? — AndS., Jan 07 '22 at 21:53
I think to use some rnorm with ~N(0,1) with some error moviments around the real values. — Rodrigo H. Ozon, Jan 07 '22 at 22:55

Jon Spring · Accepted Answer · 2022-01-07T23:44:05.427

Like this perhaps?

  df %>%
    mutate(mo_start = dmy(paste(1,month_year))) %>%
    tidyr::uncount(days_in_month(mo_start), .id = "day") %>%
    mutate(date = dmy(paste(day,month_year))) %>%
    mutate(across(contains("var"), ~rnorm(n(), mean = .x, sd = 1)))

# A tibble: 153 x 6
   month_year monthly_values_var1 monthly_values_var2 mo_start     day date      
   <chr>                    <dbl>               <dbl> <date>     <int> <date>    
 1 08-2021                   599.                18.8 2021-08-01     1 2021-08-01
 2 08-2021                   598.                17.4 2021-08-01     2 2021-08-02
 3 08-2021                   596.                18.0 2021-08-01     3 2021-08-03
 4 08-2021                   598.                19.2 2021-08-01     4 2021-08-04
 5 08-2021                   600.                18.3 2021-08-01     5 2021-08-05
 6 08-2021                   597.                19.8 2021-08-01     6 2021-08-06
 7 08-2021                   599.                18.9 2021-08-01     7 2021-08-07
 8 08-2021                   597.                17.9 2021-08-01     8 2021-08-08
 9 08-2021                   597.                16.0 2021-08-01     9 2021-08-09
10 08-2021                   596.                17.7 2021-08-01    10 2021-08-10
# … with 143 more rows

`rowwise` is not needed. You can vectorize the random number generation by `~rnorm(n(), .x, 1)` or `~.x + rnorm(n())` — ekoam, Jan 07 '22 at 23:42

score 0 · Answer 2 · answered Jan 08 '22 at 00:09

It is not a one function question.

There are more compact answers, but it is clearer step by step.

First the data:

    month_year <- c( 
      "08-2021",
      "09-2021",
      "10-2021",
      "11-2021",
      "12-2021"
    )

    monthly_values_var1 <- c(
      598,
      532,
      736,
      956,
      780
    )

    monthly_values_var2 <- c(
      18.3179,
      62.6415,
      11.1033,
      30.7443,
      74.2076
    )

    df <- data.frame(month_year, monthly_values_var1,  monthly_values_var2) 
    df

Some useful libraries:

    library(dplyr)
    library(lubridate)
    library(stringr)

It needs a similar data frame to save new data:

    df$month_year <- lubridate::dmy(paste0('01-',df$month_year))
    new.df <- df[0,]

Now the code

    counter <- 1
    for (i in 1:nrow(df)) {
      days_month <- lubridate::days_in_month(df[i, 'month_year'])
      mean1 <- df[i, 'monthly_values_var1']/days_month
      mean2 <- df[i, 'monthly_values_var2']/days_month
      for(j in 1:days_month){
        if (j < 10) {
          value <- str_pad(string = j, width = length(as.character(j))+1, pad = "0")
        } else {
          value <- as.character(j)
        }
        new.df[counter, 'month_year'] <- paste0(lubridate::year(df[i, 'month_year']),'-', lubridate::month(df[i, 'month_year']), '-', value)
        new.df[counter, 'monthly_values_var1'] <- rnorm(n = 1, mean = mean1, sd = mean1/3)
        new.df[counter, 'monthly_values_var2'] <- rnorm(n = 1, mean = mean2, sd = mean2/3)
        counter <- counter + 1
      }
    }
    View(new.df)

lubridate::days_in_month() function shows how many days are in a specific month.

rnorm assign a random number with normal distribution. I choose a mean around each data divided days in month, and a standard deviation mean/3.

How to make monthly data to daily data using random techniques in R

2 Answers2