Displaying 6 results from an estimated 6 matches for "0for".
Did you mean:
20for
2015 Feb 24
3
Problema con bucle for
...el bucle y de los índices...
Saludos,
Carlos Ortega
www.qualityexcellence.es
El 24 de febrero de 2015, 10:36, Francisco Rodríguez <fjroar en hotmail.com>
escribió:
> Si he entendido bien el problema, lo que quieres hacer realmente es esto:
> x<- c(24,12,45,68,45)n<-length(x)res=0for(i in 2:n-1){ for(j in
> (i+1):n){ res<- res + (x[i]*x[j]) print(res) }}
> Cuya salida es:
> [1] 288[1] 1368[1] 3000[1] 4080[1] 4620[1] 5436[1] 5976[1] 9036[1]
> 11061[1] 14121
> Varias observaciones:
> 1:n-1 define un vector que empieza en 0, cuando la posición 1 en R e...
2015 Feb 24
5
Problema con bucle for
Hola, quiero obtener la suma del producto de los elementos de un vector
y cuando construyo el código me aparecen una serie de NA que me impiden
calcular la suma. ¿Alguna sugerencia?
El código es el siguiente:
x<- c(24,12,45,68,45)
n<-length(x)
res<-numeric()
for(i in 1:n-1){
for(j in i+1:n){
res<- sum(x[i]*x[j])
print(res)
}
}
res
[[alternative HTML version deleted]]
2016 Jan 25
4
Corregir mismo ID para individuos diferentes en una serie temporal
...grid, args = list(args, stringsAsFactors = F)) x <- x[, rev(names(x)), drop = F] x <- do.call(paste0, x) if (n <= length(x)) return(x[1:n]) return(c(x, letterwrap(n - length(x), depth = depth + 1)))}
ltrs <- letterwrap(nrow(database)) # Create as many letters as unique IDs
ltrn<-0for(i in 1:nrow(new_df)){ if(new_df[i,3]==0) {ltrn<-ltrn+1;new_df[i,4]<-ltrs[ltrn]} else {ind<-which(new_df[,1]==new_df[i,1]) ind<-ind[ind<i] new_df[i,4]<-tail(new_df[ind,4],1)}}
[[alternative HTML version deleted]]
2008 Aug 16
0
use of all row elements
...Generating random time or x coordinate (through random time of displacement - random recall)
## First generating random time within recall period
randomrecall<-round(runif(n=1,min=recallmin,max=recallmax))
## Second generating random time of displacement
todrange<-todmax-todminrandomtod<-0for(i in 1:todrange){if (proprand>data[[9+i]] & proprand<=data[[10+i]]) randomtod=i}
Following messages are given:
Warning messages:1: In 1:todrange : numerical expression has 50 elements: only the first used2: In if (proprand > data[[9 + i]] & proprand <= data[[10 + i]]) rando...
2016 Jan 26
2
Corregir mismo ID para individuos diferentes en una serie temporal
...d, args = list(args, stringsAsFactors = F)) x <- x[, rev(names(x)), drop = F] x <- do.call(paste0, x) if (n <= length(x)) return(x[1:n]) return(c(x, letterwrap(n - length(x), depth = depth + 1)))}
ltrs <- letterwrap(nrow(database)) # Create as many letters as unique IDs
ltrn<-0for(i in 1:nrow(new_df)){ if(new_df[i,3]==0) {ltrn<-ltrn+1;new_df[i,4]<-ltrs[ltrn]} else {ind<-which(new_df[,1]==new_df[i,1]) ind<-ind[ind<i] new_df[i,4]<-tail(new_df[ind,4],1)}}
[[alternative HTML version deleted]]
______________________________________...
2016 Jan 26
2
Corregir mismo ID para individuos diferentes en una serie temporal
...)) x <- x[, rev(names(x)), drop = F] x <-
>> do.call(paste0, x) if (n <= length(x)) return(x[1:n]) return(c(x,
>> letterwrap(n - length(x), depth = depth + 1)))}
>> ltrs <- letterwrap(nrow(database)) # Create as many letters as unique IDs
>>
>> ltrn<-0for(i in 1:nrow(new_df)){ if(new_df[i,3]==0)
>> {ltrn<-ltrn+1;new_df[i,4]<-ltrs[ltrn]} else
>> {ind<-which(new_df[,1]==new_df[i,1]) ind<-ind[ind<i]
>> new_df[i,4]<-tail(new_df[ind,4],1)}}
>>
>> [[alternative HTML version deleted]]
>>...