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i want connect column in same dataframe.
for example,
# I have data type is below
region=c("A","B","C")
Q1=c("ads","qwer","zxcv")
Q2=c("poi","lkj","mnb")
temp=data.frame(region, Q1, Q2)
### i want chaged below
region1=c("A","B","C")
Q=c("ads,poi","qwer,lkj","zxcv,mnb")
temp2=data.frame(region1, Q)
How to do it... ?
temp$Q <- apply(temp[-1], 1, toString)
temp[c("Q1", "Q2")] <- NULL
temp
region Q
1 A ads, poi
2 B qwer, lkj
3 C zxcv, mnb
Using base R you can do:
temp$Q <- paste(temp$Q1, temp$Q2, sep=",")
temp <- temp[,c("region", "Q")]
temp
region Q
1 A ads,poi
2 B qwer,lkj
3 C zxcv,mnb
This would be a solution using the mutate function from the dplyr package to create the new column Q by using paste0 to concatenate the columns Q1 and Q2. In the end I just removed the columns Q1 and Q2 by using select with -:
library(dplyr)
temp %>% mutate(Q = paste0(Q1,", ",Q2)) %>% select(-Q1,-Q2)
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I have a DataFrame with three columns:region, year, grdp.
How do I group data with the same name in 'region' column.
Here's the code to create a sample dataset:
Here's the desired result:
store data of values with the same name in the 'region' column
ex) 'region' column has three "서울특별시" data. I want to group the three "서울특별시" data in three columns and assign it to a variable
I'm not completely understanding the question, but I think one of these two might solve what you're looking for?
library(dplyr)
df <- data.frame(region=sample(c('x','y','z'),100,replace=TRUE),
year=sample(c(2017,2018,2019),100,replace=TRUE),
GRDP=sample(200000000:400000000,100))
regions <- unique(df$region)[order(unique(df$region))]
#OPTION 1
for(i in 1:length(regions)){
assign(tolower(LETTERS[i]),df %>% filter(region==regions[i]))
}
a
b
c
#OPTION 2
ltrs <- tolower(LETTERS[1:length(regions)])
df['ex)'] <- sapply(df$region,FUN=function(x){ltrs[which(regions==x)]})
head(df)
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I want to calculate the logarithmic profitabilities of a stock asset. The formula for this calculation is:
ln(Row t+1/row t)
and I want to do this in R. Is it possible?
This is a dirt example of the concept. I hope all you understand it
Thanks in advance
This could be a very basic solution in base R using a for loop:
Date <- c("01-01-2022", "01-02-2022", "01-03-2022")
Date <- as.Date(Date, format = "%d-%m-%Y")
Price <- c(2, 3, 5)
df <- data.frame(Date, Price)
df$Profitablity <- rep(NA, nrow(df))
for(i in 2:nrow(df)) {
df$Profitablity[i] <- log(df$Price[i]/df$Price[i-1])
}
df
Date Price Profitablity
1 2022-01-01 2 NA
2 2022-02-01 3 0.4054651
3 2022-03-01 5 0.5108256
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Basically I have a list of the form
data<-c("1,2,3,4","5,6,7,8")
I want to convert everything to numeric
so that the output is:
[1] 1 2 3 4
[2] 5 6 7 8
how do I do so?
This is a little unclear, but if you have this object:
d <- c("1,2,3,4","5,6,7,8")
then
d |> stringr::str_split(",") |> unlist() |> as.numeric()
will convert it to a numeric vector. If you really want it as a list then appending |> as.list() will do the trick.
If you want the result to be a list of two numeric vectors then
d |> stringr::str_split(",") |> purrr::map(as.numeric)
will work.
It is not clear, but maybe something like this:
d <- c("1,2,3,4","5,6,7,8")
d <- as.list(d)
as.numeric(unlist(strsplit(d[[1]], ",")))
[1] 1 2 3 4
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I need to write an R program that, given a vector, will return another vector giving the number of elements from each position in the input vector to the occurrence in the input vector of a specified value. For example, if the specified value is 0, then with:
input : x<-c(1,1,0,1,1,1,0)
desired output<- (2,1,0,3,2,1,0)
Thanks in advance :)
We can try
library(data.table)
ave(x, rleid(x), FUN= function(x) rev(seq_along(x)))*x
#[1] 2 1 0 3 2 1 0
We could also use rle from base R
rl <- inverse.rle(within.list(rle(x), values <- seq_along(values)))
ave(x, rl, FUN=function(x) rev(seq_along(x)))*x
#[1] 2 1 0 3 2 1 0
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How is it possible to get rid of all NaN values and replace them by zero (0) in a complex function / whole R file?
I'm not sure about what you mean by a whole file/complex function, but depending on the data type you're storing the file with, it's pretty easy using is.na():
df <- data.frame(A = rep(1, 5), B = rep(1,5))
df$B[1] <- NA
df$A[3] <- NA
df[is.na(df)] <- 0