number of items to replace is not a multiple of replacement length + several plots in same graph error
0
I'm trying to draw several plots in the same graph in Rstudio, but to no avail. This is the code I'm using:
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(i in 2:(N+1)){
B[i] <- B[i-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[i] <- x+B-(t/T)*(B[N+1]-y+x)
df <- melt(data = BB, id.vars = "t")
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
}
Using ggplot2 as I seen recomended in several Stackoverflow post yieds "number of items to replace is not a multiple of replacement length".
I've seen several answers to that question but being quite a noob in R I don't see how it applies to my problem. Please and thank you in advance.
r ggplot2 plot
edited Nov 20 at 14:18
camille
6,92631427
asked Nov 20 at 14:12
Aaron G.
12
add a comment |
0
I'm trying to draw several plots in the same graph in Rstudio, but to no avail. This is the code I'm using:
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(i in 2:(N+1)){
B[i] <- B[i-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[i] <- x+B-(t/T)*(B[N+1]-y+x)
df <- melt(data = BB, id.vars = "t")
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
}
Using ggplot2 as I seen recomended in several Stackoverflow post yieds "number of items to replace is not a multiple of replacement length".
I've seen several answers to that question but being quite a noob in R I don't see how it applies to my problem. Please and thank you in advance.
r ggplot2 plot
edited Nov 20 at 14:18
camille
6,92631427
asked Nov 20 at 14:12
Aaron G.
12
2
You're using the same indexiin two different for loops. What kind of object isBB? If you defineBBas a list and assign values to it usingBB[[i]]maybe you'll get a better idea about what is going on. The error occurs in theBB[i]part of your code.
– Niek
Nov 20 at 14:38
add a comment |
0
0
0
I'm trying to draw several plots in the same graph in Rstudio, but to no avail. This is the code I'm using:
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(i in 2:(N+1)){
B[i] <- B[i-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[i] <- x+B-(t/T)*(B[N+1]-y+x)
df <- melt(data = BB, id.vars = "t")
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
}
Using ggplot2 as I seen recomended in several Stackoverflow post yieds "number of items to replace is not a multiple of replacement length".
I've seen several answers to that question but being quite a noob in R I don't see how it applies to my problem. Please and thank you in advance.
r ggplot2 plot
edited Nov 20 at 14:18
camille
6,92631427
asked Nov 20 at 14:12
Aaron G.
12
I'm trying to draw several plots in the same graph in Rstudio, but to no avail. This is the code I'm using:
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(i in 2:(N+1)){
B[i] <- B[i-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[i] <- x+B-(t/T)*(B[N+1]-y+x)
df <- melt(data = BB, id.vars = "t")
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
}
Using ggplot2 as I seen recomended in several Stackoverflow post yieds "number of items to replace is not a multiple of replacement length".
I've seen several answers to that question but being quite a noob in R I don't see how it applies to my problem. Please and thank you in advance.
r ggplot2 plot
r ggplot2 plot
edited Nov 20 at 14:18
camille
6,92631427
asked Nov 20 at 14:12
Aaron G.
12
edited Nov 20 at 14:18
camille
6,92631427
asked Nov 20 at 14:12
Aaron G.
12
edited Nov 20 at 14:18
camille
6,92631427
edited Nov 20 at 14:18
camille
6,92631427
edited Nov 20 at 14:18
camille
6,92631427
6,92631427
asked Nov 20 at 14:12
Aaron G.
12
asked Nov 20 at 14:12
Aaron G.
12
asked Nov 20 at 14:12
Aaron G.
12
12
2
You're using the same indexiin two different for loops. What kind of object isBB? If you defineBBas a list and assign values to it usingBB[[i]]maybe you'll get a better idea about what is going on. The error occurs in theBB[i]part of your code.
– Niek
Nov 20 at 14:38
add a comment |
2
You're using the same indexiin two different for loops. What kind of object isBB? If you defineBBas a list and assign values to it usingBB[[i]]maybe you'll get a better idea about what is going on. The error occurs in theBB[i]part of your code.
– Niek
Nov 20 at 14:38
2
2
You're using the same index
i in two different for loops. What kind of object is BB? If you define BB as a list and assign values to it using BB[[i]] maybe you'll get a better idea about what is going on. The error occurs in the BB[i] part of your code.– Niek
Nov 20 at 14:38
You're using the same index
i in two different for loops. What kind of object is BB? If you define BB as a list and assign values to it using BB[[i]] maybe you'll get a better idea about what is going on. The error occurs in the BB[i] part of your code.– Niek
Nov 20 at 14:38
add a comment |
1 Answer
1
active
oldest
votes
0
How about this:
BB <- list() # define BB as a list
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(q in 2:(N+1)){ # Change your index from i to q
B[q] <- B[q-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[[i]] <- x+B-(t/T)*(B[N+1]-y+x) # Assign each iteration to a list entry
}
# Exit the for loop
df <- as.data.frame(cbind(unlist(BB), # unlist the values in BB
rep(t,10), # define t variable by simply repeating it
rep(1:10,each= 101))) # define loop id in a similar manner
names(df) <- c('value','t','variable') # give names to the variables
df$variable <- as.factor(df$variable) # turn variable into a factor
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
The resulting plot:
answered Nov 20 at 15:19
Niek
734417
Thanks for your answer! Really helpful and now I realize where I did wrong. Thank you very much
– Aaron G.
Nov 21 at 12:37
You're welcome :). If you like the answer, could you please click ''accept'' (green check under the vote counter) and/or upvote?
– Niek
Nov 21 at 12:53
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
0
How about this:
BB <- list() # define BB as a list
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(q in 2:(N+1)){ # Change your index from i to q
B[q] <- B[q-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[[i]] <- x+B-(t/T)*(B[N+1]-y+x) # Assign each iteration to a list entry
}
# Exit the for loop
df <- as.data.frame(cbind(unlist(BB), # unlist the values in BB
rep(t,10), # define t variable by simply repeating it
rep(1:10,each= 101))) # define loop id in a similar manner
names(df) <- c('value','t','variable') # give names to the variables
df$variable <- as.factor(df$variable) # turn variable into a factor
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
The resulting plot:
answered Nov 20 at 15:19
Niek
734417
Thanks for your answer! Really helpful and now I realize where I did wrong. Thank you very much
– Aaron G.
Nov 21 at 12:37
You're welcome :). If you like the answer, could you please click ''accept'' (green check under the vote counter) and/or upvote?
– Niek
Nov 21 at 12:53
add a comment |
0
How about this:
BB <- list() # define BB as a list
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(q in 2:(N+1)){ # Change your index from i to q
B[q] <- B[q-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[[i]] <- x+B-(t/T)*(B[N+1]-y+x) # Assign each iteration to a list entry
}
# Exit the for loop
df <- as.data.frame(cbind(unlist(BB), # unlist the values in BB
rep(t,10), # define t variable by simply repeating it
rep(1:10,each= 101))) # define loop id in a similar manner
names(df) <- c('value','t','variable') # give names to the variables
df$variable <- as.factor(df$variable) # turn variable into a factor
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
The resulting plot:
answered Nov 20 at 15:19
Niek
734417
Thanks for your answer! Really helpful and now I realize where I did wrong. Thank you very much
– Aaron G.
Nov 21 at 12:37
You're welcome :). If you like the answer, could you please click ''accept'' (green check under the vote counter) and/or upvote?
– Niek
Nov 21 at 12:53
add a comment |
0
0
0
How about this:
BB <- list() # define BB as a list
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(q in 2:(N+1)){ # Change your index from i to q
B[q] <- B[q-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[[i]] <- x+B-(t/T)*(B[N+1]-y+x) # Assign each iteration to a list entry
}
# Exit the for loop
df <- as.data.frame(cbind(unlist(BB), # unlist the values in BB
rep(t,10), # define t variable by simply repeating it
rep(1:10,each= 101))) # define loop id in a similar manner
names(df) <- c('value','t','variable') # give names to the variables
df$variable <- as.factor(df$variable) # turn variable into a factor
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
The resulting plot:
answered Nov 20 at 15:19
Niek
734417
How about this:
BB <- list() # define BB as a list
for (i in 1:10){
require(ggplot2)
N <- 100
T <- 3
Delta <- T/N
B <- numeric(N+1)
t <- seq(0,T,length=N+1)
for(q in 2:(N+1)){ # Change your index from i to q
B[q] <- B[q-1]+rnorm(1) * sqrt(Delta)
}
x <- 0
y <- 3
BB[[i]] <- x+B-(t/T)*(B[N+1]-y+x) # Assign each iteration to a list entry
}
# Exit the for loop
df <- as.data.frame(cbind(unlist(BB), # unlist the values in BB
rep(t,10), # define t variable by simply repeating it
rep(1:10,each= 101))) # define loop id in a similar manner
names(df) <- c('value','t','variable') # give names to the variables
df$variable <- as.factor(df$variable) # turn variable into a factor
ggplot(data = df, aes(x = t, y = value, colour = variable)) + geom_line()
The resulting plot:
answered Nov 20 at 15:19
Niek
734417
answered Nov 20 at 15:19
Niek
734417
answered Nov 20 at 15:19
Niek
734417
answered Nov 20 at 15:19
Niek
734417
734417
Thanks for your answer! Really helpful and now I realize where I did wrong. Thank you very much
– Aaron G.
Nov 21 at 12:37
You're welcome :). If you like the answer, could you please click ''accept'' (green check under the vote counter) and/or upvote?
– Niek
Nov 21 at 12:53
add a comment |
Thanks for your answer! Really helpful and now I realize where I did wrong. Thank you very much
– Aaron G.
Nov 21 at 12:37
You're welcome :). If you like the answer, could you please click ''accept'' (green check under the vote counter) and/or upvote?
– Niek
Nov 21 at 12:53
Thanks for your answer! Really helpful and now I realize where I did wrong. Thank you very much
– Aaron G.
Nov 21 at 12:37
Thanks for your answer! Really helpful and now I realize where I did wrong. Thank you very much
– Aaron G.
Nov 21 at 12:37
You're welcome :). If you like the answer, could you please click ''accept'' (green check under the vote counter) and/or upvote?
– Niek
Nov 21 at 12:53
You're welcome :). If you like the answer, could you please click ''accept'' (green check under the vote counter) and/or upvote?
– Niek
Nov 21 at 12:53
add a comment |
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2
You're using the same index
iin two different for loops. What kind of object isBB? If you defineBBas a list and assign values to it usingBB[[i]]maybe you'll get a better idea about what is going on. The error occurs in theBB[i]part of your code.– Niek
Nov 20 at 14:38