Cut numeric vector into intervals, but only returning the lower boundary for each element as numeric-CodePudding

Cut numeric vector into intervals, but only returning the lower boundary for each element as numeric vector

Below is my attempt. It works, but I am looking for a less hacky and more general solution. I prefer a solution relying more on math than on functions.

library(tidyverse)
x = 1943:2023
y = cut(x, seq(1943, 2023, 5), include.lowest = TRUE, right = FALSE) |> as.character() |> str_sub(2, 5) |> as.numeric()
tibble(x, y) |> print(n=15)
#> # A tibble: 81 x 2
#>        x     y
#>    <int> <dbl>
#>  1  1943  1943
#>  2  1944  1943
#>  3  1945  1943
#>  4  1946  1943
#>  5  1947  1943
#>  6  1948  1948
#>  7  1949  1948
#>  8  1950  1948
#>  9  1951  1948
#> 10  1952  1948
#> 11  1953  1953
#> 12  1954  1953
#> 13  1955  1953
#> 14  1956  1953
#> 15  1957  1953
#> # ... with 66 more rows

Any help appreciated!

CodePudding user response：

Does this method work for you?

x %>%
  enframe() %>%
  group_by(x1 = ceiling(name/5)) %>%
  mutate(y = min(value)) %>%
  ungroup() %>%
  select(x = value, y)

       x     y
   <int> <int>
 1  1943  1943
 2  1944  1943
 3  1945  1943
 4  1946  1943
 5  1947  1943
 6  1948  1948
 7  1949  1948
 8  1950  1948
 9  1951  1948
10  1952  1948

CodePudding user response：

You could do:

breaks <- seq(1943, 2023, 5)
breaks[findInterval(x, breaks, rightmost.closed = TRUE)]

[1] 1943 1943 1943 1943 1943 1948 1948 1948 1948 1948 1953 1953 1953 1953 1953 1958 1958 1958 1958 1958 1963 1963 1963 1963 1963 1968 1968 1968 1968 1968 1973 1973 1973 1973 1973 1978 1978 1978 1978 1978 1983 1983 1983 1983 1983
[46] 1988 1988 1988 1988 1988 1993 1993 1993 1993 1993 1998 1998 1998 1998 1998 2003 2003 2003 2003 2003 2008 2008 2008 2008 2008 2013 2013 2013 2013 2013 2018 2018 2018 2018 2018 2018

For a math approach when the intervals are evenly spaced, you could do something like:

min(x)   (x - min(x)) %/% 5 * 5

But would need additional logic depending on the boundaries desired.