I cannot figure out how to do this without throwing errors. I have a set of column names for my data frame I want to create and add to that looks like this:
x <- c("A", "B", "C")
So, I go down through the loop and I calculate some numerical values in a vector, say:
z <- c(1, 5, 7, 8, 34, 5)
z is the same dimension each time through the loop.
The first time through (or even outside the loop) I want to initialize a data frame by doing something like:
df$x[1] <- z
so I have a data frame that looks like:
A
1 1
2 5
3 7
4 8
5 34
6 5
The next time through the loop I want to add another column to df with a column heading being the second element of x, and a set of new z values. If the data frame has to be completely dimensioned ahead of time, I could calculate variables outside the loop to do this, say, M and N, but these may change from one run to the next.
I cannot seem to figure out how to do this. Suggestions much appreciated.
Try this:
set.seed(1)
#set the column names
x <- c("A", "B", "C")
#create the list that later we will convert to a data.frame
df<-setNames(vector("list",length(x)),x)
#loop to produce the various z
for (i in 1:length(x)) {
#do some stuff to evaluate z
z<-sample(5)
#assign to an element of df
df[[i]]<-z
}
#coerce to a data.frame
df<-as.data.frame(df)
# A B C
#1 2 5 2
#2 5 4 1
#3 4 2 3
#4 3 3 4
#5 1 1 5
Related
I'd like to make a data frame using only the last computed values from a Repeat loop.
For the repeat and sample functions, I'm using this data. The numbers in Prob column are the probabilities of each number to occur.
enter image description here
b <- 1
repeat {
c <- sample(a$Plus, size=1, prob=(a$Prob))
cat(b, '\t', c, '\n')
b <- b + 1
if (c >= 10) {
{
break
}
}
}
#I'm interested in the result greater than 10 only
If I run the code above, then it will compute something like
1 4
2 8
3 13
If I run this again, it will compute different results like..
1 9
2 3
3 7
4 3
5 11
What I'd like to do is to make a data frame using only the last outputs of each loop.
For example, using the computed data above, I'd like to make a frame that looks like
Trial Result
3 13
5 11
Is there any way to repeat this loop the number of times I want to and make a data frame using only the last outputs of each repeated function?
You can use a user defined function to do this. Since you haven't given your dataframe a, I've defined it as follows:
library(tidyverse)
a <- tibble(
Plus = 1:15,
Prob = seq(from = 15, to = 1, by = -1)
)
The following function does the same thing as your repeat loop, but stores the relevant results in a tibble. I've left your variable b out of this because as far as I can see, it doesn't contribute to your desired output.
samplefun <- function(a) {
c <- sample(a$Plus, size=length(a$Plus), prob=a$Prob)
res <- tibble(
Trial = which(c >= 10)[1],
Result = c[which(c >= 10)[1]]
)
return(res)
}
Then use map_dfr to return as many samples as you like:
nsamples <- 5
map_dfr(1:nsamples, ~ samplefun(a))
Output:
# A tibble: 5 x 2
Trial Result
<int> <int>
1 4 11
2 6 14
3 5 11
4 2 10
5 4 15
I have written a function and am trying to iterate so that each value from a list becomes the input to the function once.
repeat100 <- function(B) {
se100 <- replicate(100, bootandse(B))
bandse <- data.frame("B" = B, "SE" = se100)
}
The output of this function is a dataframe with one column for the value of B that was inputted and another column for the SE. Like this, but with 100 rows.
B SE
1 2
1 4
1 3
I have an object B with several values. I am trying to iterate repeat100() over each value of B.
B <- c(1, 4, 6, 20, 30)
How can I get the resulting output to be a dataframe that has the output of each repeat100() for each value of B stacked like this?
B SE
1 2
1 3
1 2
4 5
4 3
4 2
Use lapply with rbind to combine the data in one dataframe.
result <- do.call(rbind, lapply(B, repeat100))
Or purrr::map_df which is shorter.
purrr::map_df(B, repeat100)
We can use rbindlist from data.table
library(data.table)
rbindlist(lapply(B, repeat100))
Let's say i have the following list of df's (in reality i have many more dfs).
seq <- c("12345","67890")
li <- list()
for (i in 1:length(seq)){
li[[i]] <- list()
names(li)[i] <- seq[i]
li[[i]] <- data.frame(A = c(1,2,3),
B = c(2,4,6))
}
What i would like to do is calculate the mean within the same cell position between the lists, keeping the same amount of rows and columns as the original lists. How could i do this? I believe I can use the apply() function, but i am unsure how to do this.
The expected output (not surprising):
A B
1 1 2
2 2 4
3 3 6
In reality, the values within each list are not necessarily the same.
If there are no NAs, then we can Reduce to get the sum of observations for each element and divide by the length of the list
Reduce(`+`, li)/length(li)
# A B
#1 1 2
#2 2 4
#3 3 6
If there are NA values, then it may be better to use mean (which has na.rm argument). For this, we can convert it to array and then use apply
apply(array(unlist(li), dim = c(dim(li[[1]]), length(li))), c(1, 2), mean)
An equivalent option in tidyverse would be
library(tidyverse)
reduce(li, `+`)/length(li)
I have a table like:
a
n_msi2010 n_msi2011
1 -0.122876 1.818750
2 1.328930 0.931426
3 -0.111653 4.400060
4 1.222900 4.500450
5 3.604160 6.110930
I would like to merge these two columns into one column to obtain (I don't want to keep column names):
a
n_msi2010
1 -0.122876
2 1.328930
3 -0.111653
4 1.222900
5 3.604160
6 1.818750
7 0.931426
8 4.400060
9 4.500450
10 6.110930
When I am using prefabricated data like
x <- cbind(c(1, 2, 3), c(4, 5, 6))
colnames(x)<-c("a","b")
c(t(x))
# 1 4 2 5 3 6
c((x))
# 1 2 3 4 5 6
the column merging works fine. Only in "a" exemple id doesn't work and it creates 2 separate vectors. I don't really understand why. Any help? Thanks
It seems like your question is about column versus row order vector creation from a data.frame.
Using t() on a data.frame converts the data.frame to a matrix, and using c() on the matrix removes its dimensions.
With that knowledge, you can try:
# create a vector of values, column by column
c(as.matrix(a)) # you are missing the `as.matrix` in your current approach
# create a vector of values, row by row
c(t(a)) # you already know this works
Other approaches to get the "column by column" result would be:
unlist(a, use.names = FALSE)
stack(a)[, "values"] # add `drop = FALSE` if you want to retain a data.frame
Not a elegant way but it seems it can combine two or several columns to one.
n_msi2010 <- 1:5
n_msi2011 <- 6:10
a <- data.frame(n_msi2010, n_msi2011)
vector <- vector()
for (i in 1:dim(a)[2]){
vector <- append(vector, as.vector(a[,i]))
vector
}
You may do
as.matrix(vector) or data.frame(vector)
I've got a seemingly simple question that I can't answer: I've got three vectors:
x <- c(1,2,3,4)
weight <- c(5,6,7,8)
y <- c(1,1,1,2,2,2)
I want to create a new vector that replicates the values of weight for each time an element in x matches y such that it produces the following new weight vector associated with y:
y_weight <- c(5,5,5,6,6,6)
Any thoughts on how to do this (either loop or vectorized)? Thanks
You want the match function.
match(y, x)
to return the indicies of the matches, the use that to build your new weight vector
weight[match(y, x)]
#Using plyr
library(plyr)
df<-as.data.frame(cbind(x,weight)) # converting to dataframe
df<-rename(df,c(x="y")) # rename x as y for joining dataframes
y<-as.data.frame(y) # converting to dataframe
mydata <- join(df, y, by = "y",type="right")
> mydata
y weight
1 1 5
2 1 5
3 1 5
4 2 6
5 2 6
6 2 6