2012年12月20日 · Convert R vector to string vector of 1 element [duplicate] Ask Question Asked 12 years, 3 months ago.

2014年7月14日 · Following the above post you can even use and modify the elements of a vector satisfying the criteria. In my opinion if it's not more costly to compute faster one should always do it. x = seq(0.1,10,0.1) y <- rep(2,length(x)) y[x<5] <- x[x<5]*2

2018年5月9日 · For finding all occurances as logical vector: > 'a' == v [1] FALSE TRUE FALSE TRUE FALSE Edit: Removing grep() and grepl() from the list for reason mentioned in comments

To check that an object obj is a vector, but not a list or matrix, consider: is.vector(obj) && is.atomic(obj) Using only is.vector(obj) returns TRUE for lists but not for matrices; and is.atomic(obj) returns TRUE for matrices but not for lists. Here are some sample checks:

2018年4月25日 · I would like to randomly reorganize the order of the numbers in a vector, in a simple one-line command? My particular vector V has 150 entries for each value from 1 to 10: V &lt;- rep(1:10, each=...

In R, I have an element x and a vector v. I want to find the first index of an element in v that is equal to x . I know that one way to do this is: which(x == v)[[1]] , but that seems excessively inefficient.

2008年9月17日 · Suppose I have a vector that is nested in a dataframe with one or two levels. Is there a quick and dirty way to access the last value, without using the length() function? Something ala PERL's $#

Is it possible in R to say - I want all indices from position i to the end of vector/matrix? Say I want a submatrix from 3rd column onwards. I currently only know this way: A = matrix(rep(1:8, eac...

2019年11月8日 · Rodrigo, realize that in some languages, a vector is just a 1-dim array; this is not the case in R, which differentiates between the two. Therefore, apply(1:10,...) will not work. – r2evans

2021年10月19日 · R match part of string against vector of strings. 15. Filter rows that contain a string from a vector. 1. ...