问题
I have a vector of factors given by a sequence of numbers. These factors are also found in separate data seta, called test_set
and train_set
. What the following code does is find where the factor in the data sets matches in the vector of factors and puts a 1 in the place of the matrix. Multiplying this matrix compound_test
by test_set$Compound
should give you compare_comp
.
compare_comp <- rbind(dcm,cmp1)[,1]
compound_test <- matrix(0,nrow(test_set),length(compare_comp)) # test indicator matrix
compound_train <-matrix(0,nrow(train_set),length(compare_comp))
for (i in 1:length(compare_comp)){
compound_test[which(compare_comp[i]==test_set$Compound),i]=1
compound_train[which(compare_comp[i]==train_set$Compound),i]=1}
It does this for a train and test set, and compare_comp is the vector of factors.
Is there a function in R that lets me create the same thing without the need for a for loop? I have tried model.matrix(~Compound,data=test_set)
without much luck.
回答1:
While you may not be able to completely avoid iteration since you are comparing each element of compare_comp
vector to the full vector of Compound
in each test_set
and train_set
, you can however use more compact assignment with apply family functions.
Specifically, sapply
returns a logical matrix of booleans (TRUE
, FALSE
) that we assign in corresponding position to initialized matrices where TRUE
converts to 1 and FALSE
to 0.
# SAPPLY AFTER MATRIX INITIALIZATION
compound_test2 <- matrix(0, nrow(test_set), length(compare_comp))
compound_train2 <- matrix(0, nrow(train_set), length(compare_comp))
compound_test2[] <- sapply(compare_comp, function(x) x == test_set$Compound)
compound_train2[] <- sapply(compare_comp, function(x) x == train_set$Compound)
Alternatively, the rarely used and known vapply
(similar to sapply
but must define the output type), returns an equivalent matrix but as numeric type.
# VAPPLY WITHOUT MATRIX INITIALIZATION
compound_test3 <- vapply(compare_comp, function(x) x == test_set$Compound,
numeric(length(compare_comp)))
compound_train3 <- vapply(compare_comp, function(x) x == train_set$Compound,
numeric(length(compare_comp)))
Testing confirms with random data (see demo below), both versions are identical to your looped version
identical(compound_test1, compound_test2)
identical(compound_train1, compound_train2)
# [1] TRUE
# [1] TRUE
identical(compound_test1, compound_test3)
identical(compound_train1, compound_train3)
# [1] TRUE
# [1] TRUE
Online Demo
来源:https://stackoverflow.com/questions/59408283/how-can-i-code-this-indicator-matrix-without-using-a-for-loop-in-r