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Removing character elements from a vector


Removing character elements from a vector

By : cytrax
Date : November 22 2020, 04:01 AM
it should still fix some issue The period is a meta character (see here), so try escaping it like so-
code :
str_remove(A3, “\\.”)
library(stringr)
a <- c("hello. this. is. text","this. is. also. text","here. is. even. more. text")
remove_these <- c("\\.","text")
str_remove_all(a, paste(remove_these, collapse = "|"))
"hello this is "     "this is also "      "here is even more "


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Removing an element from the vector without removing the elements after

Removing an element from the vector without removing the elements after


By : Sanzia
Date : March 29 2020, 07:55 AM
it should still fix some issue You're looking for n as an element, but are also using it as an index. I don't think this is what you want.
If you replace (n-1) by a, it should work:
code :
        list.erase(list.begin()+a);
#include <algorithm>
//...
void removeOne(int n){
    vector<int>::iterator found = std::find(list.begin(), list.end(), n) ;
    if (found!=list.end())
        list.erase(found);
}
How do I write a function to convert a character vector into a character vector of unique pairs of its elements?

How do I write a function to convert a character vector into a character vector of unique pairs of its elements?


By : mbztt
Date : March 29 2020, 07:55 AM
like below fixes the issue What the input would be: , We can use combn from base R with FUN argument as paste
code :
combn(x, 2, FUN = paste, collapse = ";")
#[1] "a;b" "a;c" "b;c"
x <- c("a", "b", "c")
Better solution to check elements of one character vector with another character vector using the tidyverse?

Better solution to check elements of one character vector with another character vector using the tidyverse?


By : user3341155
Date : March 29 2020, 07:55 AM
wish of those help It looks like you want the synonyms vector without values that have any overlap with mixNames. You can subset synonyms to remove the matches. Here str_c/paste collapse mixNames to create a pattern with all the mixNames. Then you just use partial string matching (i.e., str_detect and grepl from there).
Here using stringr - which is slightly tidy-er
code :
synonyms[str_detect(synonyms, str_c(mixNames, collapse = "|"), negate = T)]
synonyms[!grepl(paste(mixNames, collapse = "|"), synonyms)]
# OR
grep(paste(mixNames, collapse = "|"), synonyms, value = T, invert = T)
Iterating over and removing elements from a Vector. Error:Vector iterator not Incrementable

Iterating over and removing elements from a Vector. Error:Vector iterator not Incrementable


By : Amal
Date : March 29 2020, 07:55 AM
I wish did fix the issue. The auto in your code will return the type inside the vector container. It will not give you the iterator. Instead of a range-based for loop, you can use a classic for loop and erase your element with the iterator.
Removing elements in R vector to correspond with NAs in another R vector

Removing elements in R vector to correspond with NAs in another R vector


By : Thomas
Date : September 30 2020, 08:00 PM
will help you I have two vectors, say A and B , Use is.na
code :
A[!is.na(B)]
#[1] 1 3 4
B[!is.na(B)]
#[1] 6 8 9
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