Skip to content

Commit

Permalink
fix: add urls
Browse files Browse the repository at this point in the history
  • Loading branch information
@Layalchristine24
Layalchristine24 committed Sep 9, 2024
1 parent bc84bda commit 27ebcfb
Show file tree
Hide file tree
Showing 2 changed files with 8 additions and 7 deletions.
4 changes: 2 additions & 2 deletions _freeze/posts/2024-09-09_reduce/index/execute-results/html.json
View file
Original file line number Diff line number Diff line change
@@ -1,8 +1,8 @@
{
"hash": "a65f90837f1b74974384ed1e6fdea28d",
"hash": "72d7d5b27a5db46edf6a900308b08097",
"result": {
"engine": "knitr",
"markdown": "---\ntitle: How do you iterate a vector?\nauthor:\n - name:\n given: Layal Christine\n family: Lettry\n orcid: 0009-0008-6396-0523\n affiliations:\n - id: cynkra\n - name: cynkra GmbH\n city: Zurich\n state: CH\n - id: unifr\n - name: University of Fribourg, Dept. of Informatics, ASAM Group\n city: Fribourg\n state: CH\ndate: 2024-09-09\ncategories: [purrr, constructive, reduce, accumulate, loop, iteration]\nimage: image.jpg\ncitation: \n url: https://rdiscovery.netlify.app/posts/2024-09-09_reduce/\nformat:\n html:\n toc: true\n toc-depth: 6\n toc-title: Contents\n toc-location: right\n number-sections: false\neditor_options: \n chunk_output_type: console\n---\n\n\n\n*What functions can you use to replace a loop?*\n\n# Initial object\n\nLet's assume that we have a numeric vector.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec <- c(\"first\" = 1L, \"second\" = 2L, \"third\" = 3L, \"fourth\" = 4L)\nconstructive::construct(my_vec)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nc(first = 1L, second = 2L, third = 3L, fourth = 4L)\n```\n\n\n:::\n:::\n\n\n\n# Use `reduce()` from purrr\n\nLet's say we want to compute the sum and the product of all the elements of our vector.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`+`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 10\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`*`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 24\n```\n\n\n:::\n:::\n\n\n\nIt is the same as doing\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec |> sum()\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 10\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |> prod()\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 24\n```\n\n\n:::\n:::\n\n\n\nWe could also write a loop for the sum, which would perform exactly the same operations as `reduce()`, but in a more complicated way.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# Define the function f\nf <- function(x, y) {\n x + y\n}\n\n# Start with the first value\nmy_vec <- unname(my_vec)\nresult <- my_vec[1]\n\n# Loop through the rest of the values\nfor (i in 2:length(my_vec)) {\n result <- f(result, my_vec[i])\n}\n\nprint(result)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 10\n```\n\n\n:::\n:::\n\n\n\nThese results show that `reduce()` takes the vector values and applies the `+` or `*` function iteratively in the `forward` direction (default).\n\n## Difference between the forward and the backward direction\n\nLet the function `f` be defined by `f(a, b) = a + b`. Then, `reduce()` will apply this to `my_vec`, i.e.\n\n$$\nf(f(f(1, 2), 3), 4) = f(f(3, 3), 4) = f(6, 4) = 10.\n$$\n\nIf the direction had been set `backwards`, the following would have occurred:\n\n$$\nf(f(f(4, 3), 2), 1) = f(f(7, 2), 1) = f(9, 1) = 10.\n$$ \n\nTo see the difference between `forward` and `backward`, let's subtract the elements of our vector iteratively.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`-`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] -8\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`-`, .dir = \"backward\")\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] -2\n```\n\n\n:::\n:::\n\n\n\nLet the function `g` be defined by `g(a, b) = a - b`. Then, the `forward` subtraction corresponds to this operation:\n\n$$\ng(g(g(1, 2), 3), 4) = g(g(-1, 3), 4) = g(-4, 4) = -8. \n$$\n\nThe the `backward` subtraction would be:\n\n$$ \ng(g(g(4, 3), 2), 1) = g(g(1, 2), 1) = g(-1, 1) = -2.\n$$\n\n## Initial value of the accumulation\n\nWe could set the initial value of the accumulation.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec[[3]]\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 3\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`-`, .init = my_vec[[3]])\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] -7\n```\n\n\n:::\n:::\n\n\n\nHere, there is one additional argument and the iteration starts with the value of `3` that corresponds to `my_vec[[3]]`:\n\n$$\ng(g(g(g(3, 1), 2), 3), 4) = g(g(g(2, 2), 3), 4) = g(g(0, 3), 4) = g(-3, 4) = -7.\n$$\n\n## Apply a predefined function iteratively\n\n### With two arguments\n\nLet's say we want to compute the empiric variance for our vector.\n\nWe usually compute it with `var(my_vec) =` 1.6666667. This is an unbiased estimator of the variance because the denominator is `n-1 = length(my_vec) - 1 =` 3. The following operation:\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 / (length(my_vec) - 1) * sum((my_vec - mean(my_vec))^2)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1.666667\n```\n\n\n:::\n:::\n\n\n\ncorresponds to\n\n$$ \n1/(n-1) \\sum_{i = 1}^n (x_i - \\bar{x})^2.\n$$ However, it does not work when we try to apply this formula using `reduce()`, because we need to supply two arguments.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# does not work\nmy_vec |>\n purrr::reduce(\\(x) 1 / (length(x) - 1) * sum((x - mean(x))^2))\n```\n\n::: {.cell-output .cell-output-error}\n\n```\nError in fn(out, elt, ...): unused argument (elt)\n```\n\n\n:::\n:::\n\n\n\nLet's do it by entering two arguments into the `reduce()` function.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# works but is not the variance of my_vec\nmy_vec |>\n purrr::reduce(\\(x, y) var(c(x, y)))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.3828125\n```\n\n\n:::\n\n```{.r .cell-code}\n# or\nmy_var <- function(x, y) {\n 1 / (length(c(x, y)) - 1) * sum((c(x, y) - mean(c(x, y)))^2)\n}\nmy_vec |>\n purrr::reduce(\\(x, y) my_var(x, y))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.3828125\n```\n\n\n:::\n:::\n\n\n\nSo, what does `reduce()` compute?\n\nLet $x =$ `my_vec` and $h(x_1, x_2)$ be a function defined by\n\n$$\nh(x_1, x_2) = var(x_1, x_2) = 1/(2-1) \\sum_{i = 1}^2 (x_i - \\bar{x})^2 = \\left[x_1 - \\frac{(x_1 + x_2)}{2}\\right]^2 + \\left[x_2 - \\frac{(x_1 + x_2)}{2}\\right]^2.\n$$ Thus, we get 0.3828125 because `reduce()` computes $h(h(h(1, 2), 3), 4)$ which is equal to\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nvar(c(var(c(var(my_vec[1:2]), my_vec[3])), my_vec[4]))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.3828125\n```\n\n\n:::\n\n```{.r .cell-code}\n# or\n\nmy_var(my_var(my_var(my_vec[[1]], my_vec[[2]]), my_vec[[3]]), my_vec[[4]])\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.3828125\n```\n\n\n:::\n:::\n\n\n\n### With three arguments\n\nIf `.init` is not specified, the function `reduce2()` takes as arguments a first vector and a second one which is shorter than the former by one element. Otherwise, both arguments should have the same number of of elements.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_x <- my_vec\nmy_y <- my_vec\n\npurrr::reduce2(my_x, my_y, paste, .init = 5)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] \"5 1 1 2 2 3 3 4 4\"\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_x <- my_vec |> unname()\nmy_y <- my_x[-length(my_x)] + 6\n\npurrr::reduce2(my_x, my_y, paste)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] \"1 2 7 3 8 4 9\"\n```\n\n\n:::\n:::\n\n\n\nIf `.init` is not specified, the `reduce2()` function takes as the initial value the first element of the first argument. \n\nOtherwise, the argument `.init` gives the first element. This is the first argument. \n\nThen, the values going from the first non-used element of the two arguments (here `my_x` and `my_y`) will be the second and the third elements.\n\n# Use `accumulate()` from purrr\n\nThe function `accumulate()` will give the intermediate results of the `reduce()` function.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec |>\n purrr::accumulate(`+`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1 3 6 10\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::accumulate(`*`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1 2 6 24\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::accumulate(\\(x, y) var(c(x, y)))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1.0000000 0.5000000 3.1250000 0.3828125\n```\n\n\n:::\n:::\n\n\n\nYou can alsow use `accumulate2()` if you use 2 lists.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_x <- my_vec\nmy_y <- my_vec\n\npurrr::accumulate2(my_x, my_y, paste, .init = 5)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[[1]]\n[1] 5\n\n[[2]]\n[1] \"5 1 1\"\n\n[[3]]\n[1] \"5 1 1 2 2\"\n\n[[4]]\n[1] \"5 1 1 2 2 3 3\"\n\n[[5]]\n[1] \"5 1 1 2 2 3 3 4 4\"\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_x <- my_vec |> unname()\nmy_y <- my_x[-length(my_x)] + 6\n\npurrr::accumulate2(my_x, my_y, paste)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[[1]]\n[1] 1\n\n[[2]]\n[1] \"1 2 7\"\n\n[[3]]\n[1] \"1 2 7 3 8\"\n\n[[4]]\n[1] \"1 2 7 3 8 4 9\"\n```\n\n\n:::\n:::\n",
"markdown": "---\ntitle: How do you iterate a vector?\nauthor:\n - name:\n given: Layal Christine\n family: Lettry\n orcid: 0009-0008-6396-0523\n affiliations:\n - id: cynkra\n - name: cynkra GmbH\n city: Zurich\n state: CH\n - id: unifr\n - name: University of Fribourg, Dept. of Informatics, ASAM Group\n city: Fribourg\n state: CH\ndate: 2024-09-09\ncategories: [purrr, constructive, reduce, accumulate, loop, iteration]\nimage: image.jpg\ncitation: \n url: https://rdiscovery.netlify.app/posts/2024-09-09_reduce/\nformat:\n html:\n toc: true\n toc-depth: 6\n toc-title: Contents\n toc-location: right\n number-sections: false\neditor_options: \n chunk_output_type: console\n---\n\n\n\n*What functions can you use to replace a loop?*\n\n# Initial object\n\nLet's assume that we have a numeric vector.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec <- c(\"first\" = 1L, \"second\" = 2L, \"third\" = 3L, \"fourth\" = 4L)\nconstructive::construct(my_vec)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nc(first = 1L, second = 2L, third = 3L, fourth = 4L)\n```\n\n\n:::\n:::\n\n\n\n# Use `reduce()` from purrr\n\nLet's say we want to compute the sum and the product of all the elements of our vector. \nWe can use the [`reduce()`](https://purrr.tidyverse.org/reference/reduce.html) function.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`+`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 10\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`*`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 24\n```\n\n\n:::\n:::\n\n\n\nIt is the same as doing\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec |> sum()\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 10\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |> prod()\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 24\n```\n\n\n:::\n:::\n\n\n\nWe could also write a loop for the sum, which would perform exactly the same operations as `reduce()`, but in a more complicated way.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# Define the function f\nf <- function(x, y) {\n x + y\n}\n\n# Start with the first value\nmy_vec <- unname(my_vec)\nresult <- my_vec[1]\n\n# Loop through the rest of the values\nfor (i in 2:length(my_vec)) {\n result <- f(result, my_vec[i])\n}\n\nprint(result)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 10\n```\n\n\n:::\n:::\n\n\n\nThese results show that `reduce()` takes the vector values and applies the `+` or `*` function iteratively in the `forward` direction (default).\n\n## Difference between the forward and the backward direction\n\nLet the function `f` be defined by `f(a, b) = a + b`. Then, `reduce()` will apply this to `my_vec`, i.e.\n\n$$\nf(f(f(1, 2), 3), 4) = f(f(3, 3), 4) = f(6, 4) = 10.\n$$\n\nIf the direction had been set `backwards`, the following would have occurred:\n\n$$\nf(f(f(4, 3), 2), 1) = f(f(7, 2), 1) = f(9, 1) = 10.\n$$\n\nTo see the difference between `forward` and `backward`, let's subtract the elements of our vector iteratively.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`-`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] -8\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`-`, .dir = \"backward\")\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] -2\n```\n\n\n:::\n:::\n\n\n\nLet the function `g` be defined by `g(a, b) = a - b`. Then, the `forward` subtraction corresponds to this operation:\n\n$$\ng(g(g(1, 2), 3), 4) = g(g(-1, 3), 4) = g(-4, 4) = -8. \n$$\n\nThe the `backward` subtraction would be:\n\n$$ \ng(g(g(4, 3), 2), 1) = g(g(1, 2), 1) = g(-1, 1) = -2.\n$$\n\n## Initial value of the accumulation\n\nWe could set the initial value of the accumulation.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec[[3]]\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 3\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::reduce(`-`, .init = my_vec[[3]])\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] -7\n```\n\n\n:::\n:::\n\n\n\nHere, there is one additional argument and the iteration starts with the value of `3` that corresponds to `my_vec[[3]]`:\n\n$$\ng(g(g(g(3, 1), 2), 3), 4) = g(g(g(2, 2), 3), 4) = g(g(0, 3), 4) = g(-3, 4) = -7.\n$$\n\n## Apply a predefined function iteratively\n\n### With two arguments\n\nLet's say we want to compute the empiric variance for our vector.\n\nWe usually compute it with `var(my_vec) =` 1.6666667. This is an unbiased estimator of the variance because the denominator is `n-1 = length(my_vec) - 1 =` 3. The following operation:\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 / (length(my_vec) - 1) * sum((my_vec - mean(my_vec))^2)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1.666667\n```\n\n\n:::\n:::\n\n\n\ncorresponds to\n\n$$ \n1/(n-1) \\sum_{i = 1}^n (x_i - \\bar{x})^2.\n$$ However, it does not work when we try to apply this formula using `reduce()`, because we need to supply two arguments.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# does not work\nmy_vec |>\n purrr::reduce(\\(x) 1 / (length(x) - 1) * sum((x - mean(x))^2))\n```\n\n::: {.cell-output .cell-output-error}\n\n```\nError in fn(out, elt, ...): unused argument (elt)\n```\n\n\n:::\n:::\n\n\n\nLet's do it by entering two arguments into the `reduce()` function.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# works but is not the variance of my_vec\nmy_vec |>\n purrr::reduce(\\(x, y) var(c(x, y)))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.3828125\n```\n\n\n:::\n\n```{.r .cell-code}\n# or\nmy_var <- function(x, y) {\n 1 / (length(c(x, y)) - 1) * sum((c(x, y) - mean(c(x, y)))^2)\n}\nmy_vec |>\n purrr::reduce(\\(x, y) my_var(x, y))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.3828125\n```\n\n\n:::\n:::\n\n\n\nSo, what does `reduce()` compute?\n\nLet $x =$ `my_vec` and $h(x_1, x_2)$ be a function defined by\n\n$$\nh(x_1, x_2) = var(x_1, x_2) = 1/(2-1) \\sum_{i = 1}^2 (x_i - \\bar{x})^2 = \\left[x_1 - \\frac{(x_1 + x_2)}{2}\\right]^2 + \\left[x_2 - \\frac{(x_1 + x_2)}{2}\\right]^2.\n$$ Thus, we get 0.3828125 because `reduce()` computes $h(h(h(1, 2), 3), 4)$ which is equal to\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nvar(c(var(c(var(my_vec[1:2]), my_vec[3])), my_vec[4]))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.3828125\n```\n\n\n:::\n\n```{.r .cell-code}\n# or\n\nmy_var(my_var(my_var(my_vec[[1]], my_vec[[2]]), my_vec[[3]]), my_vec[[4]])\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.3828125\n```\n\n\n:::\n:::\n\n\n\n### With three arguments\n\nIf `.init` is not specified, the function `reduce2()` takes as arguments a first vector and a second one which is shorter than the former by one element. Otherwise, both arguments should have the same number of of elements.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_x <- my_vec\nmy_y <- my_vec\n\npurrr::reduce2(my_x, my_y, paste, .init = 5)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] \"5 1 1 2 2 3 3 4 4\"\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_x <- my_vec |> unname()\nmy_y <- my_x[-length(my_x)] + 6\n\npurrr::reduce2(my_x, my_y, paste)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] \"1 2 7 3 8 4 9\"\n```\n\n\n:::\n:::\n\n\n\nIf `.init` is not specified, the `reduce2()` function takes as the initial value the first element of the first argument.\n\nOtherwise, the argument `.init` gives the first element. This is the first argument.\n\nThen, the values going from the first non-used element of the two arguments (here `my_x` and `my_y`) will be the second and the third elements.\n\n# Use `accumulate()` from purrr\n\nThe function [`accumulate()`](https://purrr.tidyverse.org/reference/accumulate.html) will give the intermediate results of the [`reduce()`](https://purrr.tidyverse.org/reference/reduce.html) function.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_vec |>\n purrr::accumulate(`+`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1 3 6 10\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::accumulate(`*`)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1 2 6 24\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_vec |>\n purrr::accumulate(\\(x, y) var(c(x, y)))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1.0000000 0.5000000 3.1250000 0.3828125\n```\n\n\n:::\n:::\n\n\n\nYou can alsow use `accumulate2()` if you use 2 lists.\n\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmy_x <- my_vec\nmy_y <- my_vec\n\npurrr::accumulate2(my_x, my_y, paste, .init = 5)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[[1]]\n[1] 5\n\n[[2]]\n[1] \"5 1 1\"\n\n[[3]]\n[1] \"5 1 1 2 2\"\n\n[[4]]\n[1] \"5 1 1 2 2 3 3\"\n\n[[5]]\n[1] \"5 1 1 2 2 3 3 4 4\"\n```\n\n\n:::\n\n```{.r .cell-code}\nmy_x <- my_vec |> unname()\nmy_y <- my_x[-length(my_x)] + 6\n\npurrr::accumulate2(my_x, my_y, paste)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[[1]]\n[1] 1\n\n[[2]]\n[1] \"1 2 7\"\n\n[[3]]\n[1] \"1 2 7 3 8\"\n\n[[4]]\n[1] \"1 2 7 3 8 4 9\"\n```\n\n\n:::\n:::\n",
"supporting": [],
"filters": [
"rmarkdown/pagebreak.lua"
Expand Down
11 changes: 6 additions & 5 deletions posts/2024-09-09_reduce/index.qmd
View file
Original file line number Diff line number Diff line change
Expand Up @@ -43,7 +43,8 @@ constructive::construct(my_vec)

# Use `reduce()` from purrr

Let's say we want to compute the sum and the product of all the elements of our vector.
Let's say we want to compute the sum and the product of all the elements of our vector.
We can use the [`reduce()`](https://purrr.tidyverse.org/reference/reduce.html) function.

```{r}
my_vec |>
Expand Down Expand Up @@ -95,7 +96,7 @@ If the direction had been set `backwards`, the following would have occurred:

$$
f(f(f(4, 3), 2), 1) = f(f(7, 2), 1) = f(9, 1) = 10.
$$
$$

To see the difference between `forward` and `backward`, let's subtract the elements of our vector iteratively.

Expand Down Expand Up @@ -209,15 +210,15 @@ my_y <- my_x[-length(my_x)] + 6
purrr::reduce2(my_x, my_y, paste)
```

If `.init` is not specified, the `reduce2()` function takes as the initial value the first element of the first argument.
If `.init` is not specified, the `reduce2()` function takes as the initial value the first element of the first argument.

Otherwise, the argument `.init` gives the first element. This is the first argument.
Otherwise, the argument `.init` gives the first element. This is the first argument.

Then, the values going from the first non-used element of the two arguments (here `my_x` and `my_y`) will be the second and the third elements.

# Use `accumulate()` from purrr

The function `accumulate()` will give the intermediate results of the `reduce()` function.
The function [`accumulate()`](https://purrr.tidyverse.org/reference/accumulate.html) will give the intermediate results of the [`reduce()`](https://purrr.tidyverse.org/reference/reduce.html) function.

```{r}
my_vec |>
Expand Down

0 comments on commit 27ebcfb

Please sign in to comment.