I have data that has multiple sequences that I'd like to replace by sampling from another data frame. In my head, it would work something like
x = seq(1,100, 0.5)
sample_set = rnorm(20,1,1)
# here I want to replace certain values in x and replace them with values sampled from the normal distribution
x[c(2:5,30:32,50:56),1] = sample(sample_set, length([c(2:5,30:32,50:56)]), replace = TRUE)
In my data, this replacement only works for the first sequence specified in
x[c(2:5,30:32,50:56),1] # i.e. items 2:5
I've explored recode() and several other options, but nothing has completed the replacement at all locations. Thanks in advance! I'm probably overthinking this...
You have some inconsistencies in the way you refer to x. First you declare it as a one-dimensional object and then you refer to it as a matrix. I believe if you fix that and remove the dat from inside the sample function, everything works as you described:
x[c(2:5,30:32,50:56)] = sample(sample_set, length(c(2:5,30:32,50:56)), replace = TRUE)
Related
I have been using Stata and the loops are easily executed there. However, in R I have faced some errors in looping over variables. I tried some of the codes over here and it does not work. Basically, I am trying to clean the data by logging the values. I had to convert negative values to positive first before logging them.
I intend to loop over multiple firm statistics on the dataframe but I faced errors in doing so.
varlist <- c("revenue", "profit", "cost")`
for (v in varlist) {
data$log_v <- log(abs(ifelse(data$v>1, data$v, NA)))
data$log_v <- ifelse(data$v<0, data$log_v*-1,data$log_v)
}
Error in $<-.data.frame(tmp,"log_v", value = numeric(0)) : replacement has 0 rows, data has 9
It looks like you might be assuming that data$log_v is getting read as data$log_profit, but R's going to take it own it's own and read it as "log_v" all 3 times. This example might not be quite everything you're trying to do but it might help you. It's taking a list of variables and referencing them via their string names.
df <- data.frame(x = rnorm(15), y = rnorm(15))
vars <- c("x", "y")
for (v in vars) {
df[paste0("log_", v)] <- log(abs(df[v]))
}
Here's roughly the same thing in data.table.
library(data.table)
dt <- data.table(x = rnorm(15), y = rnorm(15))
dt[, `:=`(log_x = log(abs(x)), log_y = log(abs(y)))]
Here is an explanation to the source of your confusion:
A data.frame is a special type of list, it's elements are vectors of the same length – columns. Normally, you access an element of a list using the [[ function, for example df[["revenue"]]. Instead of "revenue", you can also use a variable, such as df[[varlist[1]]]. So far, so good.
However, lists have a convenience operator, $, which allows you to access the elements with less typing: df$revenue. Unfortunately, you cannot use variables this way: this by design. Since you don't have to use quotes with $, the operator cannot know whether you mean revenue as the literal name of the element or revenue as the variable that holds the literal name of the element.
Therefore, if you want to use variables, you need to use the [[ function, and not the $. Since programmers hate typing and want to make code as terse as possible, various ways around it have been invented, such as data.tables and tidyverse (I am exaggerating a bit here).
Also, here is a tidyverse solution.
library(tidyverse)
varlist <- c("revenue", "profit", "cost")
df <- data.frame(revenue=rnorm(100), profit=rnorm(100), cost=rnorm(100))
df <- df %>% mutate_at(varlist, list(log10 = ~ log10(abs(.))))
Explanation:
mutate_all applies log10(abs(.)) to every column. The dot . is a temporary variable that hold the column values for each of the columns.
by default, mutate_all will replace the existing variables. However, if instead of providing a function (~ log10(abs(.))) you provide a named list (list(log10 = ~ log10(abs(.)))), it will add new columns using log10 as a suffix in column name.
this method makes it easy to apply several functions to your columns, not only the one.
See? No (obvious) loops at all!
I am converting my for-loops in R for a model that has multiple input datasets. In the for-loop I use the current loop value to retrieve values from other datasets. I am looking to replicate this using an apply function (over columns in a dataset) however I'm struggling to establish index of the apply function in order to retrieve the appropriate variables from other data
The apply function references the column by the variable in the function which is fine and I've tried to use both colname (after having named my various columns by number) but have not had any joy. Below is an example dataset and for loop with what I'd like to achieve (simplified somewhat). The length of the vectors and the number of columns in the tabular dataset will always be equal.
iteration<-1:3
df <- data.frame("column1" = 6:10, "column2" = 12:16, "column3" = 31:35)
variable1<-rnorm(3,mean = 25)
variable2<-rnorm(3, mean = 0.21)
outcome<-numeric()
for (i in iteration) {
intermediate<-(mean(df[,i])*variable1[i])^variable2[i]
outcome<-c(outcome,intermediate)
}
outcome
The expected results are outcome above...trying this in apply
What I imagine it to be is this:
apply(df, 2, function(x) (mean(x)*variable1[colnumber(x)])^variable2[colnumber(x)]
or perhaps
apply(df, 2, function(x) (mean(x)*variable1[x])^variable2[x])
but these two obviously do not work.
first time user so apologies for any etiquette issues but found the answer to my own problem using the purrr package, but maybe this helps someone else
pmap(list(df, variable1, variable2), function(df, variable1, variable2) (mean(df)*variable1)^variable2)
I am quite new to R and have quite a challenging Question. I have a large dataframe consisting of 110,000 rows representing high-Resolution data from a Sediment core. I would like to select multiple rows based on Depth (which is recorded in mm to 3 decimal points). Of Course, I have not the time to go through the entire dataframe and pick the rows that I Need. I would like to be able to select the rows I would like based on the decimal Point part of the number and not the first Digit. I.e. I would like to be able to subset to a dataframe where all the .035 values would be returned. I have so far tried using the which() function but had no luck
newdata <- Linescan_EN18218[which(Linescan_EN18218$Position.mm.== .035),]
Can anyone offer any hints/suggestions how I can solve this Problem. Link to the first part of the dataframe csv
Welcome to stack overflow
Can you please further describe what you mean with had no luck. Did you get an error message or an empty data.frame?
In principle, your method should work. I have replicated it with simulated data.
n = 100
test <- data.frame(
a = 1:n,
b = rnorm(n = n),
c = sample(c(0.1,0.035, 0.0001), size = n, replace =T)
)
newdata <- test[which(test$c == 0.035),]
I am trying to compare multiple columns in two different dataframes in R. This has been addressed previously on the forum (Compare group of two columns and return index matches R) but this is a different scenario: I am trying to compare if a column in dataframe 1 is between the range of 2 columns in dataframe 2. Functions like match, merge, join, intersect won't work here. I have been trying to use purr::pluck but didn't get far. The dataframes are of different sizes.
Below is an example:
temp1.df <- mtcars
temp2.df <- data.frame(
Cyl = sample (4:8, 100, replace = TRUE),
Start = sample (1:22, 100, replace = TRUE),
End = sample (1:22, 100, replace = TRUE)
)
temp1.df$cyl <- as.character(temp1.df$cyl)
temp2.df$Cyl <- as.character(temp2.df$Cyl)
My attempt:
temp1.df <- temp1.df %>% mutate (new_mpg = case_when (
temp1.df$cyl %in% temp2.df$Cyl & temp2.df$Start <= temp1.df$mpg & temp2.df$End >= temp1.df$mpg ~ 1
))
Error:
Error in mutate_impl(.data, dots) :
Column `new_mpg` must be length 32 (the number of rows) or one, not 100
Expected Result:
Compare temp1.df$cyl and temp2.df$Cyl. If they are match then -->
Check if temp1.df$mpg is between temp2.df$Start and temp2.df$End -->
if it is, then create a new variable new_mpg with value of 1.
It's hard to show the exact expected output here.
I realize I could loop this so for each row of temp1.df but the original temp2.df has over 250,000 rows. An efficient solution would be much appreciated.
Thanks
temp1.df$new_mpg<-apply(temp1.df, 1, function(x) {
temp<-temp2.df[temp2.df$Cyl==x[2],]
ifelse(any(apply(temp, 1, function(y) {
dplyr::between(as.numeric(x[1]),as.numeric(y[2]),as.numeric(y[3]))
})),1,0)
})
Note that this makes some assumptions about the organization of your actual data (in particular, I can't call on the column names within apply, so I'm using indexes - which may very well change, so you might want to rearrange your data between receiving it and calling apply, or maybe changing the organization of it within apply, e.g., by apply(temp1.df[,c("mpg","cyl")]....
At any rate, this breaks your data set into lines, and each line is compared to the a subset of the second dataset with the same Cyl count. Within this subset, it checks if any of the mpg for this line falls between (from dplyr) Start and End, and returns 1 if yes (or 0 if no). All these ones and zeros are then returned as a (named) vector, which can be placed into temp1.df$new_mpg.
I'm guessing there's a way to do this with rowwise, but I could never get it to work properly...
I have a large data set I am attempting to sample rows from. Each row has a family ID, and there may be one or multiple rows for each family ID. I want to parse the data set by randomly sampling one row for each family ID. I have attempted to accomplish this by using both tapply() and split() + lapply() functions, but to no avail. Below is code that reproduces my issue - the size and scope of the factor levels and data entries mirror the data set I am working with.
set.seed(63)
f1 <- factor(c(rep(30000:32000, times=1),
rep(30500:31700, times = 2),
rep(30900:31900, times = 3)))
f2 <- factor(rep(sample(1:7, replace = TRUE), times = length(f1)/7))
x1 <- round(matrix(rnorm(length(f1)*300), nrow = length(f1), ncol = 300),3)
df <- data.frame(f1, f2, x1)
Next, I used tapply to sample one row per factor from f1, and then check for repeats. (f2 is a secondary factor that indexes another aspect of the observations, but is [hopefully] irrelevant here; I only include it for full disclosure of the structure of my data set).
s1 <- tapply(1:nrow(df), df$f1, sample, size=1)
any(duplicated(s1))
The output for the second line of code using duplicated is TRUE, which means there are repeats. Stumped, I tried split to see if that was the problem.
df.split <- split(1:nrow(df), df$f1)
any(duplicated(df.split))
The output here for duplicated is FALSE, so the problem is not split. I then used the output df.split with lapply and sample to see if the problem was with tapply.
df.unique <- unlist(lapply(df.split, sample, size = 1, replace = FALSE,
prob = NULL))
any(duplicated(df.unique))
In the first line, I sampled one value from each element of df.split which outputs a list, then I used unlist to convert into a vector. The output for duplicated here is also TRUE.
Somewhere within sample and lapply there is funky stuff going on (since tapply merely calls lapply). I'm not sure how to fix the issue (I searched SO and Google and found nothing related to my issue), so any help would be greatly appreciated!
EDIT: I'm hoping someone could tell me why the above code using tapply and lapply is not working as intended. Arthur has provided a nice answer, and I have coded a loop for sample as well. I'm wondering why the above code is misbehaving.
I would do that:
library(data.table)
data.table(df)[,.SD[sample(.N,1)],by='f1']
... but actually your original approach with tapply is faster if you just want an index and not the actual subset table ; however, you must notice that sample(n) actually samples in 1:n when length(n)==1. See ?sample. This version is error-proof:
s1 <- tapply(1:nrow(df), list(df$f1), function(v) v[sample(1:length(v), 1)])` is error prooff