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Accepted
Juan Luis
Asked: 2020-04-27 11:30:55 +0800 CST 2020-04-27 11:30:55 +0800 CST

Create a dataframe from a column that I want to modify, add another column and a third one that compares them in R

  • 772

Be:

set.seed(2020)
y_fitted<-runif(10,min=0,max = 1)
y_fitted
0.197 0.078 0.818 0.942 0.884 0.166 0.355 0.748 0.451 0.556

y<-sample(0:1,10,replace=T)
y
1 1 1 0 0 1 1 1 1 0

I want to create a dataframe such that:

  • Column 1: if y_fitted>= 0.5it is 1 and if y_fitted<0.5it is 0
  • Column 2: yunchanged
  • Column 3: TRUEyes y_fitted==y; FALSEYesy_fitted!==y

Finally, get the percentage of values TRUE​​that are in the third column.

The problem I am having when creating column 1 : I don't understand why when executing the following lines of code, the values ​​are modified by 1 as I want, but the values ​​between 0:0.4999 are not changed to 0 .

y_fitted[y_fitted==0:0.499999999]<-0
y_fitted[y_fitted==0.5:1]<-1

I want to do all this to analyze the goodness of fit of my binomial logit model, where y_fittedare the values ​​estimated by my model and the values ​​of yare the values ​​of my variable to explain (endogenous).

I put 0.499999999in to make sure no value is left out of the change I want to make to column 1, but I'm sure there's a cleaner way to do it.

The percentage that I request would indicate, then, the percentage of correct answers that my model has within the sample with which I am working.

What is the most efficient way to do this?

r

1 Answers

  1. Best Answer

    This clause: y_fitted[y_fitted==0:0.499999999]<-0(and the following one as well) do not have the meaning you seek to give it, some reasons I can tell you:

    • 0:0.499999999I understand that what you are looking for is to generate a sequence of values ​​from 0 to 0.49..., however, it :works as a sequence of step 1, it does not work for a sequence with a smaller step, so the result will be a vector with a only value 0, you can check it by evaluating 0:0.49999.

    • To generate NON-integer sequences you have to use seq(), however, even if you generate this sequence like this seq(from=0, to=0.49, by=.1), it will always be a set of discrete values, you can add more values ​​by increasing the precision/step with bybut it will always be a vector of discrete values.

    • Also when you compare the a == bwhat happens is that you compare each value of awith each value of bthat is to say something like a[1] == b[1], a[2] == b[2]that, including recycling of values ​​if one vector is greater than the other, which certainly does not solve the problem.

    What you are looking for, you should actually write it like this:

    y_fitted[y_fitted < 0.5] <- 0
y_fitted[y_fitted >= 0.5] <-1

    Or much more concise:

    y_fitted <- ifelse(y_fitted >= 0.5, 1, 0)

    Or why not, taking advantage of the conversion from logical to integers, this, thanks to the fact that you want the values ​​1 and 0 that are consistent with the TRUEor FALSE:

    y_fitted <- as.integer(y_fitted >= 0.5)
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