CODEWITHSUNDEEP
rloops
Sean Wheeler
Sean Wheeler

Reputation: 15

Is there a loop I can use to tidy up this code?

replace.value(zdata, c(1), DL[1], 0)  
replace.value(zdata, c(2), DL[2], 0)  
replace.value(zdata, c(3), DL[3], 0)  
replace.value(zdata, c(4), DL[4], 0)  
replace.value(zdata, c(5), DL[5], 0)  
replace.value(zdata, c(6), DL[6], 0)  
replace.value(zdata, c(7), DL[7], 0)  
replace.value(zdata, c(8), DL[8], 0)

I need to change a specific value (which are given in the DL vector) with 0 for each column of the data frame. For example the first value of the DL vector is 0.5 so the first line of the code above changes every 0.5 in the first column of the data frame to 0.

Upvotes: 0

Views: 52

Answers (2)

Gregor Thomas
Gregor Thomas

Reputation: 145870

Yes. Let's make some improvements. First of all, c() stands for concatenate, which means to "stick things together". If you only have one thing, you don't need c(). 5 is the same as c(5), "hello" is the same as c("hello"), etc.

A forloop lets you loop over specific values. The values you change in each line of code go from 1 to 8, so we'll loop over that:

for (val in 1:8) {
  replace.value(zdata, val, DL[val], 0)
}

I don't know the replace.value function (and you don't mention what package it's in), but it seem strange that you aren't assigning the results. In base R, you'd do something like this:

zdata[1][zdata[1] == DL[1]] <- 0
zdata[2][zdata[2] == DL[2]] <- 0
...

Using <- to assign the result so it is saved. That would turn into this for loop:

for (val in 1:8) {
  zdata[val][zdata[val] == DL[val]] <- 0
}

Upvotes: 2

zack
zack

Reputation: 5405

If you want to use the tidyverse more explicitly, you can use something like walk

library(tidyverse)
walk(1:8, ~replace.value(zdata, .x, DL[.x], 0))

Upvotes: 0

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