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Hi, I have a school data frame and I need to calculate the evasion number. The program would be more or less that, only this way it takes a long time to finish executing. Does anyone know how I can do it otherwise using some library (preferably ddply)?
Evasao <- function(row1, df){
for(row2 in 1:nrow(df)){
if(df[row1,"Id"] == df[row2,"Id"] & (df[row2, "NU_ANO_CENSO"]+1) == df[row1, "NU_ANO_CENSO"])
FALSE
}
TRUE
}
#1 - evasão, 2 - não evasão
df <- dados_padronizados_ensino_medio_com_eja_2007_a_2018_com_id
df$EVASAO <- 0
for(row in 1:nrow(df)){
#print(df["TP_ETAPA_ENSINO"])
if(df[row,"TP_ETAPA_ENSINO"] == 25 | df[row,"TP_ETAPA_ENSINO"] == 26 ){
if(Evasao(row, df)){
df[row,"EVASAO"] <- 1
}
}
}
Try to do
Evasao <- cmpfun(Evasao)and after you have created the Evasion function. It creates a compiled version lower level than your Evasion, and compare computational time. And come back here, Bl?– Guilherme Parreira
@Guilhermeparreira vlw man! I’ll try here
– Renato Lopo
Can you explain some of the function tests? Or post one
dput(head(dados))? Which should improve with some vector operation instead of for– Jorge Mendes
The question code seems to be a clear case of easily vectorable code. It does not need cycles
forof certainty, nor (almost certainly) ofcmpfunto be much faster. But we need test data.– Rui Barradas
This code seems to have been written by a programmer not used to r, you can replace this Row check passed with some lag or lead function for example
– Bruno
Hello @Noisy, when it would be interesting to use the
cmpfun? Once my friend used this, and his code got much faster (his functions involved integration of Laplace and optimization). When I tried to apply a function I created, where I compare 7 different prediction methods viaforecastpackage, no computational gain– Guilherme Parreira