Vectorized If Statement in R

Vectorized IF statement in R?

x <- seq(0.1,10,0.1)

> x
  [1]  0.1  0.2  0.3  0.4  0.5  0.6  0.7  0.8  0.9  1.0  1.1  1.2  1.3  1.4  1.5
 [16]  1.6  1.7  1.8  1.9  2.0  2.1  2.2  2.3  2.4  2.5  2.6  2.7  2.8  2.9  3.0
 [31]  3.1  3.2  3.3  3.4  3.5  3.6  3.7  3.8  3.9  4.0  4.1  4.2  4.3  4.4  4.5
 [46]  4.6  4.7  4.8  4.9  5.0  5.1  5.2  5.3  5.4  5.5  5.6  5.7  5.8  5.9  6.0
 [61]  6.1  6.2  6.3  6.4  6.5  6.6  6.7  6.8  6.9  7.0  7.1  7.2  7.3  7.4  7.5
 [76]  7.6  7.7  7.8  7.9  8.0  8.1  8.2  8.3  8.4  8.5  8.6  8.7  8.8  8.9  9.0
 [91]  9.1  9.2  9.3  9.4  9.5  9.6  9.7  9.8  9.9 10.0

> ifelse(x < 5, 1, 2)
  [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
 [38] 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
 [75] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

R - vectorized if for multiple conditions

How about this?

It's the same function as above, but without the explicit if and else?

> Teff=seq(30,40,1)
> hsa<- 1*(Teff<Tcr) + (1 - (Teff - Tcr)/(Tlim - Tcr))*(Teff >= Tcr & Teff < Tlim)
> hsa
 [1] 1.0000000 1.0000000 1.0000000 1.0000000 0.8571429 0.7142857 0.5714286 0.4285714 0.2857143 0.1428571 0.0000000

** Note that you can add + 0*(Teff>=Tlim) in the end, but it wouldn't change anything, because the assigned value would be 0 anyways.

If you really want to use ifelse, then you have to nest them, so it should be something like this:

> hsa<- ifelse(Teff<Tcr, 1,
               ifelse(Teff >= Tcr & Teff < Tlim, 
               (1 - (Teff - Tcr)/(Tlim - Tcr)), 0))
> hsa
     [1] 1.0000000 1.0000000 1.0000000 1.0000000 0.8571429 0.7142857 0.5714286 0.4285714 0.2857143 0.1428571 0.0000000

Vectorized 'if' statement without 'ifelse'

To summarize the info I got in this thread, the answer to my first problem was 'don't try to set values inside of an ifelse(), use ifelse() to return a value and set it that way.

The second problem I was having with the else portion of my statement overwriting previous statements, the answer was maddeningly simple: just return the current value. So the following

df$flag <- ifelse((df$year==2011 | df$year==2012) & df$code==df2$yes_code &
df$year==df2$year, 'Y', 'N')

becomes this

df$flag <- ifelse((df$year==2011 | df$year==2012) & df$code==df2$yes_code &
df$year==df2$year, 'Y', df$flag)

Thanks to all who helped, this was a very difficult question to articulate.

R how to vectorize a function with multiple if else conditions

Here is a vectorized way. It creates logical vectors i1, i2, i3 and i4 corresponding to the 4 conditions. Then it assigns the new values to the positions indexed by them.

Trial_func2 <- function(df1){
  i1 <- df1[["Obs_Type"]] == 1
  i2 <- df1[["Obs_Type"]] == 2
  i3 <- df1[["Obs_Type"]] == 3
  i4 <- df1[["Obs_Type"]] == 4

  #If Type == 1; then a=-Inf, b = Upper_Bound
  df1[i1, "draw_value"] <- rtruncnorm(sum(i1), a =-Inf, 
                                      b = df1[i1, "Upper_bound"], 
                                      mean = df1[i1, "mean"], sd = 1)
  #If Type == 2; then a=-10, b = Upper_Bound
  df1[i2, "draw_value"] <- rtruncnorm(sum(i2), a = -10,
                                      b = df1[i2 , "Upper_bound"],
                                      mean = df1[i2, "mean"], sd = 1)
  #If Type == 3; then a=Lower_bound, b = Inf
  df1[i3,"draw_value"] <- rtruncnorm(sum(i3), 
                                     a = df1[i3, "Lower_bound"],
                                     b = Inf, mean = df1[i3, "mean"], 
                                     sd = 1)
  #If Type == 3; then a=Lower_bound, b = 10
  df1[i4, "draw_value"] <- rtruncnorm(sum(i4), 
                                      a = df1[i4, "Lower_bound"],
                                      b = 10,
                                      mean = df1[i4,"mean"],
                                      sd = 1)
  df1
}

In the speed test I have named @Dave2e's answer Trial_func3.

mbm <- microbenchmark(
  loop = Trial_func(df1 = df1),
  vect = Trial_func2(df1 = df1),
  cwhen = Trial_func3(df1 = df1),
  times = 10)

print(mbm, order = "median")
#Unit: milliseconds
#  expr         min          lq       mean      median          uq         max neval cld
#  vect    4.349444    4.371169    4.40920    4.401384    4.450024    4.487453    10  a 
# cwhen   13.458946   13.484247   14.16045   13.528792   13.787951   19.363104    10  a 
#  loop 2125.665690 2138.792497 2211.20887 2157.185408 2201.391083 2453.658767    10   b

Using if/ifelse statement with vector conditions in R

You can also do a normal if statement, since it returns a value:

my_result = if(all(Vector1 == c(T,T,T))) {"Combo1"} else {"Combo2"}

The ifelse function is made for vectorized conditional statements.

By using the standard if statement, you remove potential ambiguity/misinterpretation because the standard if only evaluates one condition.

Vectorized IF statement combined with logic AND in R

b <- ifelse(a==1 & b<1, 1, b)

Create a vector using if...else if...else statements in R

Here's a couple of ways to do it:

the most analogous to your existing method is:

X <- ifelse(MATRIX$Col1==1,
            ifelse(MATRIX$Col2==1,"both","col1"),
            ifelse(MATRIX$Col2==1,"col2","none"))

It can be slightly quicker to do:

x <- rep(NA,nrow(MATRIX))
x[MATRIX$Col1[i]==1 && MATRIX$Col2[i]==1] <- "both"
x[MATRIX$Col1[i]==1 && !MATRIX$Col2[i]==1] <- "col1"
x[!MATRIX$Col1[i]==1 && MATRIX$Col2[i]==1] <- "col2"
x[!MATRIX$Col1[i]==1 && !MATRIX$Col2[i]==1] <- "none"

but it's harder to see whether all cases have been covered by the code

Note:

• It looks like MATRIX really is a data.frame; learning to be
  precise about you data types can really help when debugging code.
• If MATRIX$Col1 really is Boolean, you can drop the ==1 comparison,
  that's wasting time by converting the matrix to numeric and then
  testing for equality.
• To me, the most transparant method is to create
  a small data.frame with the possible values of Col1, Col2 and
  expected output, and merge this with the existing data.frame, but
  this may not be as efficient.

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