The Condition Has Length > 1 and Only the First Element Will Be Used in If Else Statement

Interpreting condition has length 1 warning from `if` function

maybe you want ifelse:

a <- c(1,1,1,1,0,0,0,0,2,2)
ifelse(a>0,a/sum(a),1)

 [1] 0.125 0.125 0.125 0.125 1.000 1.000 1.000 1.000
 [9] 0.250 0.250

Getting error the condition has length 1 and only the first element will be used when using if else loop

if and else should be used on logical conditions that are length 1. You are asking if a vector (OldTimeColumn) of some length n is equal to a vector of a single value TRUE -- so that's why the warning message says what it says, and it only evaluates the first element of OldTimeColumn by comparing it to TRUE.

A better way for your particular example might be to use the case_when structure.

library(dplyr)
library(stringr)
df <- data.frame(
  OldTimeColumn = c("12", "1", "2", "3", "4"),
  stringsAsFactors = F
)
df <- df %>%
  mutate(NewTimeColumn = case_when(
    str_detect(OldTimeColumn, "^12") ~ "12",
    str_detect(OldTimeColumn, "^1") ~ "1",
    str_detect(OldTimeColumn, "^2") ~ "2",
    str_detect(OldTimeColumn, "^3") ~ "3",
    TRUE ~ "11"
  ))

df

  OldTimeColumn NewTimeColumn
1            12            12
2             1             1
3             2             2
4             3             3
5             4            11

Trying to do this with ifelse (which does work on evaluating vectors) would be more cumbersome but can be done:

df <- df %>%
  mutate(NewTimeColumn = ifelse(str_detect(OldTimeColumn, "^12") == TRUE,
         "12", ifelse(str_detect(OldTimeColumn, "^1") == TRUE, 
         "1", ifelse(str_detect(OldTimeColumn, "^2") == TRUE,
         "2", ifelse(str_detect(OldTimeColumn, "^3") == TRUE, "3", "11")))))

What does the error the condition has length 1 and only the first element will be used mean?

the problem is that you are using an if-statement with a vector. This is not allowed and doesn't work as you would expect. You can use the case_when function from dplyr.

library(dplyr)

split_firstname <- function(full_name){
  x <- stringr::str_count(full_name, " ")
  case_when(
    x == 1 ~ stringr::word(full_name, 1),
    x == 2 ~ paste(stringr::word(full_name,1), stringr::word(full_name,2), sep = " "),
    x == 4 ~ paste(stringr::word(full_name,1), stringr::word(full_name,2), stringr::word(full_name,3), stringr::word(full_name,4), sep = " ")
  )
}

“the condition has length 1 and only the first element will be used” warning from nested `if elses' over a dataframe

1) Personally I try to do as much R as I can with only a small subset of its many commands. Maybe a simple apply is an easier way to manage this. apply with MARGIN = 1 will give each row pf your data.frame to a function. So I made this slight change to your function (just the first 3 lines are of interest here, the rest is copy&paste):

incidence_headaches<-function(row){
  x <- row[1]
  y <- row[2]
  if (is.na(x)|is.na(y)){
    output<-NA
  }
  else if (x==2){
    if (y==2){
      output<-'previous_headache_maintained'
    }else if(y==0){
      output<-'previous_headache_ceased'
    }
  }else if(x %in% c(0,774,775,776)){
    if (y==2){
      output<-'new_onset_headache'
    }else if (y %in% c(0, 774, 775, 776)){
      output<-'no_headache'
    }
  }
}

You can then use simple apply like this:

apply(df_headache_tibble, MARGIN = 1, incidence_headaches)

To get something like this:

> apply(df_headache_tibble, MARGIN = 1, incidence_headaches)
  [1] "no_headache"                  "previous_headache_ceased"     "previous_headache_maintained"
  [4] "previous_headache_maintained" "new_onset_headache"           "no_headache"                 
  [7] "no_headache"                  "no_headache"                  "previous_headache_ceased"    
 [10] "new_onset_headache"           "previous_headache_ceased"     "previous_headache_maintained"
 [13] "no_headache"                  "previous_headache_ceased"     "no_headache" 
...

2) mapply is obviously a perfectly working function and there is no reason not to use it. Your problem was: tibbles are data.frames but they do not behave like data.frames. This works well:

mapply(incidence_headaches, 
       as.data.frame(df_headache_tibble)[,1],
       as.data.frame(df_headache_tibble)[,2])

When you subset only one row from a data.frame, it will give you a vector, when you subset only one row from a tibble, it will give you a tibble. Hadley has a different opinion on how things should work then the people who invented the R data.frame. There are ways around this as in

mapply(incidence_headaches, 
       df_headache_tibble[,1, drop = TRUE],
       df_headache_tibble[,2, drop = TRUE])

Read details here but mostly be always aware, that although tibbles are data.frames they do not behave exactly like data.frames: https://tibble.tidyverse.org/reference/subsetting.html

If statement error - the condition has length 1 and only the first element will be used

You can try use ifelse the vectorized version of if

df$Year1 <- ifelse(df$pos1 == 3 ,  str_sub(df$Date,-4,-1) ,
 str_sub(df$Date, 1, 4))

The condition has length 1 and only the first element will be used

You get the error because if can only evaluate a logical vector of length 1.

Maybe you miss the difference between & (|) and && (||). The shorter version works element-wise and the longer version uses only the first element of each vector, e.g.:

c(TRUE, TRUE) & c(TRUE, FALSE)
# [1] TRUE FALSE

# c(TRUE, TRUE) && c(TRUE, FALSE)
[1] TRUE

You don't need the if statement at all:

mut1 <- trip$Ref.y=='G' & trip$Variant.y=='T'|trip$Ref.y=='C' & trip$Variant.y=='A'
trip[mut1, "mutType"] <- "G:C to T:A"

What does warning mean: the condition has length 1 and only the first element will be used

This,

paste(df.short$AD, df.short$ED, sep="")

is a vector, not of length 1, so when you pass that to your function, you are testing a vector against a scalar:

nchar(x) == 5

I suggest looping over your function using an apply function, something like

mapply(fixED, x = paste(df.short$AD, df.short$ED, sep=""))

Error in the if function: the condition has length 1 and only the first element will be used

You got that error because you can only evaluate a single T/F in if (...). However, data[1, ] < ... is a vectorised evaluation that returns a T/F vector of a length n (i.e. 1000 for your case above). Nevertheless, I think your second function (est_L) does not match the equation shown in that image. Consider the following implementation instead:

L <- function(n) {
  del <- 0.9 * sqrt(2 * log(n))
  data <- rnorm(n)
  mean(exp(del * data - 0.5 * del * del))
}

L_tilde <- function(n) {
  del <- 0.9 * sqrt(2 * log(n))
  data <- rnorm(n)
  mean(exp(del * data * ifelse(data < sqrt(2 * log(n)), 1, 0) - 0.5 * del * del))
}

Then you can just

hist(replicate(1000, L_tilde(1000)))

Output

---

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