I have a function that outputs a 1 by 2 data frame like the below reproducible example:
g_1 <- 2
g_2 <- 3
tbl <- cbind(g_1, g_2)
tbl <- as.data.frame(tbl)
tbl
And I'm trying to run a simulation of this function 5000 times and map the output of the function to a matrix or in other words fill the matrix with the output of each iteration. I have this code which I know doesn't work because I get the error under it but also because I think it's trying to fill the output into one column maybe?
nreps <- 5000
#Creating workspace
df_sim <- matrix(-999, nreps, 2, dimnames = list( c(), c("g_1", "g_2")))
for (i in 1:nreps){
df_sim[i] <- sim_tab(x = 5, y = 6)
}
number of items to replace is not a multiple of replacement lengthnumber
Is there a way to fill the matrix with each 1,2 output from each iteration of the loop?
You can use replicate to repeat the function nreps times and combine the result using do.call.
result <- do.call(rbind, replicate(nreps,sim_tab(x = 5, y = 6),simplify = FALSE))
We could use rerun from purrr
library(purrr)
library(dplyr)
sim_tab(x = 5, y = 6) %>%
rerun(nreps) %>%
bind_rows
Related
I am trying to populate the output of a for loop into a data frame. The loop is repeating across the columns of a dataset called "data". The output is to be put into a new dataset called "data2". I specified an empty data frame with 4 columns (i.e. ncol=4). However, the output generates only the first two columns. I also get a warning message: "In matrix(value, n, p) : data length [2403] is not a sub-multiple or multiple of the number of columns [2]"
Why does the dataframe called "data2" have 2 columns, when I have specified 4 columns? This is my code:
a <- 0
b <- 0
GM <- 0
GSD <- 0
data2 <- data.frame(ncol=4, nrow=33)
for (i in 1:ncol(data))
{
if (i==34) {break}
a[i] <- colnames(data[i])
b <- data$cycle
GM[i] <- geoMean(data[,i], na.rm=TRUE)
GSD[i] <- geoSD(data[,i], na.rm=TRUE)
data2[i,] <- c(a[i], b, GM[i], GSD[i])
}
data2
If you look at the ?data.frame() help page, you'll see that it does not take arguments nrow and ncol--those are arguments for the matrix() function.
This is how you initialize data2, and you can see it starts with 2 columns, one column is named ncol, the second column is named nrow.
data2 <- data.frame(ncol=4, nrow=33)
data2
# ncol nrow
# 1 4 33
Instead you could try data2 <- as.data.frame(matrix(NA, ncol = 4, nrow = 33)), though if you share a small sample of data and your expected result there may be more efficient ways than explicit loops to get this job done.
Generally, if you do loop, you want to do as much outside of the loop as possible. This is just guesswork without having sample data, these changes seem like a start at improving your code.
a <- colnames(data)
b <- data$cycle ## this never changes, no need to redefine every iteration
GM <- numeric(ncol(data)) ## better to initialize vectors to the correct length
GSD <- numeric(ncol(data))
data2 <- as.data.frame(matrix(NA, ncol = 4, nrow = 33))
for (i in 1:ncol(data))
{
if (i==34) {break}
GM[i] <- geoMean(data[,i], na.rm=TRUE)
GSD[i] <- geoSD(data[,i], na.rm=TRUE)
## it's weird to assign a row of data.frame at once...
## maybe you should keep it as a matrix?
data2[i,] <- c(a[i], b, GM[i], GSD[i])
}
data2
Hello everyone I have two data frame trying to do bootstrapping with below script1 in my script1 i am taking number of rows from data frame one and two. Instead of taking rows number from entire data frame I wanted split individual columns as a data frame and remove the zero values and than take the row number than do the bootstrapping using below script. So trying with script2 where I am creating individual data frame from for loop as I am new to R bit confused how efficiently do add the script1 function to it
please suggest me below I am providing script which is running script1 and the script2 I am trying to subset each columns creating a individual data frame
Script1
set.seed(2)
m1 <- matrix(sample(c(0, 1:10), 100, replace = TRUE), 10)
m2 <- matrix(sample(c(0, 1:5), 50, replace = TRUE), 5)
m1 <- as.data.frame(m1)
m2 <- as.data.frame(m2)
nboot <- 1e3
n_m1 <- nrow(m1); n_m2 <- nrow(m2)
temp<- c()
for (j in seq_len(nboot)) {
boot <- sample(x = seq_len(n_m1), size = n_m2, replace = TRUE)
value <- colSums(m2)/colSums(m1[boot,])
temp <- rbind(temp, value)
}
boot_data<- apply(temp, 2, median)
script2
for (i in colnames(m1)){
m1_subset=(m1[m1[[i]] > 0, ])
m1_subset=m1_subset[i]
m2_subset=m2[m2[[i]] >0, ]
m2_subset=m2_subset[i]
num_m1 <- nrow(m1_subset); n_m2 <- nrow(m2_subset)# after this wanted add above script changing input
}
If I understand correctly, you want to do the sampling and calculation on each column individually, after removing the 0 values. I. modified your code to work on a single vector instead of a dataframe (i.e., using length() instead of nrow() and sum() instead of colSums(). I also suggest creating the empty matrix for your results ahead of time, and filling in -- it will be fasted.
temp <- matrix(nrow = nboot, ncol = ncol(m1))
for (i in seq_along(m1)){
m1_subset = m1[m1[,i] > 0, i]
m2_subset = m2[m2[,i] > 0, i]
n_m1 <- length(m1_subset); n_m2 <- length(m2_subset)
for (j in seq_len(nboot)) {
boot <- sample(x = seq_len(n_m1), size = n_m2, replace = TRUE)
temp[j, i] <- sum(m2_subset)/sum(m1_subset[boot])
}
}
boot_data <- apply(temp, 2, median)
boot_data <- setNames(data.frame(t(boot_data)), names(m1))
boot_data
I have a data set containing 526 rows nd 560 columns. In this data set, I want to run pca analysis for each 16 columns, respectively, in the loop and save the PCA scores for each row. I tried the below code but it did not work. I would be happy to get your advice.
Thanks in advance for your help.
for(i in 1:ncol(df)) {
df[ , i:(i+15)] <- prcomp(df[, i:(i+15)], scale. = TRUE, center = T)
}
Here is a way with a lapply loop. Create a vector f of consecutive integers, each repeated 16 times. Then split the data.frame names by this vector and lapply function prcomp to each subset. Finally, extract the scores.
f <- c(1, rep(0, 15))
f <- rep(f, length(names(df1))/16)
f <- cumsum(f)
nms <- split(names(df1), f)
pca_list <- lapply(nms, function(x){
prcomp(df1[x], center = TRUE, scale. = TRUE)
})
scores_list <- lapply(pca_list, '[[', 'x')
Test data creation code
set.seed(2021)
df1 <- replicate(560, rnorm(526))
df1 <- as.data.frame(df1)
reprod:
df1 <- data.frame(X = c(0:9), Y = c(10:19))
df2 <- data.frame(X = c(0:9), Y = c(10:19))
df3 <- data.frame(X = c(0:9), Y = c(10:19))
list_of_df <- list(A = df1, B = df2, C = df3)
list_of_df
I'm trying to apply the rollmean function from zoo to every 'Y' column in this list of dataframes.
I've tried lapply with no success, It seems no matter which way i spin it, there is no way to get around specifying the dataframe you want to apply to at some point.
This does one of the dataframes
roll_mean <- rollmean(list_of_df$A, 2)
roll_mean
obviously this doesn't work:
roll_mean1 <- rollmean(list_of_df, 2)
roll_mean1
I also tried this:
subset(may not be necessary)
Sub1 <- lapply(list_of_df, "[", 2)
roll_mean1 <- rollmean(Sub1, 2)
roll_mean1
there doesn't seem to be a way to do it without having to
specify the particular dataframe in the rollmean function
lapply(list_of_df), function(x) rollmean(list_of_df, 2))
for loop? also no success
For (i in list_of_df) {roll_mean1 <- rollmean(Sub1, 2)
Exp
}
Stating the obvious but I'm very new to coding in general and would appreciate some pointers.
It has occurred to me that even if it did work, the column that has been averaged would be one value longer than the rest of the dataframe; how would I get around that?
The question at one point says that it wants to perform the rollmean only on Y and at another point says that this works roll_mean <- rollmean(list_of_df$A, 2) but that does all columns.
1) Assuming that you want to apply rollmean to all columns:
Use lapply like this:
lapply(list_of_df, rollmean, 2)
This also works:
for(i in seq_along(list_of_df)) list_of_df[[i]] <- rollmean(list_of_df[[i]], 2)
2) If you only want to apply it to the Y column:
lapply(list_of_df, transform, Y = rollmean(Y, 2, fill = NA))
or
for(i in seq_along(list_of_df)) {
list_of_df[[i]]$Y <- rollmean(list_of_df[[i]]$Y, 2, fill = NA)
}
I tried to create a matrix from a list which consists of N unequal matrices...
The reason to do this is to make R individual bootstrap samples.
In the example below you can find e.g. 2 companies, where we have 1 with 10 & 1 with just 5 observations.
Data:
set.seed(7)
Time <- c(10,5)
xv <- matrix(c(rnorm(10,5,2), rnorm(5,20,1), rnorm(10,5,2), rnorm(5,20,1)), ncol=2);
y <- matrix( c(rnorm(10,5,2), rnorm(5,20,1)));
z <- matrix(c(rnorm(10,5,2), rnorm(5,20,1), rnorm(10,5,2), rnorm(5,20,1)), ncol=2)
# create data frame of input variables which helps
# to conduct the rowise bootstrapping
data <- data.frame (y = y, xv = xv, z = z);
rows <- dim(data)[1];
cols <- dim(data)[2];
# create the index to sample from the different panels
cumTime <- c(0, cumsum (Time));
index <- findInterval (seq (1:rows), cumTime, left.open = TRUE);
# draw R individual bootstrap samples
bootList <- replicate(R = 5, list(), simplify=F);
bootList <- lapply (bootList, function(x) by (data, INDICES = index, FUN = function(x) dplyr::sample_n (tbl = x, size = dim(x)[1], replace = T)));
---------- UNLISTING ---------
Currently, I try do it incorrectly like this:
Example for just 1 entry of the list:
matrix(unlist(bootList[[1]], recursive = T), ncol = cols)
The desired output is just
bootList[[1]]
as a matrix.
Do you have an idea how to do this & if possible reasonably efficient?
The matrices are then processed in unfortunately slow MLE estimations...
i found a solution for you. From what i gather, you have a Dataframe containing all observations of all companies, which may have different panel lengths. And as a result you would like to have a Bootstap sample for each company of same size as the original panel length.
You mearly have to add a company indicator
data$company = c(rep(1, 10), rep(2, 5)) # this could even be a factor.
L1 = split(data, data$company)
L2 = lapply(L1, FUN = function(s) s[sample(x = 1:nrow(s), size = nrow(s), replace = TRUE),] )
stop here if you would like to have saperate bootstap samples e.g. in case you want to estimate seperately
bootdata = do.call(rbind, L2)
Best wishes,
Tim