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I have two data frames, each with 9 columns, and DF2 is a subset of DF1. I'm trying to create a third data frame that contains only the contents of DF1 that are NOT present in DF2.
What is the most efficient way of doing this? I can write a while loop, but I was wondering if there is another way (besides sqldf as for some reason I cannot upload it into my R Studio) that I can do this?
The following can work (directly from Identify records in data frame A not contained in data frame B)
fun.12 <- function(x.1,x.2,...){
x.1p <- do.call("paste", x.1)
x.2p <- do.call("paste", x.2)
x.1[! x.1p %in% x.2p, ]
}
DF1 <- data.frame(a=c(1,2,3,4,5), b=c(1,2,3,4,5))
DF2 <- data.frame(a=c(1,1,2,3,4), b=c(1,1,99,3,4))
fun.12(DF1, DF2)
# a b
# 2 2 2
# 5 5 5
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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$region,FUN=function(x){ltrs[which(regions==x)]})
head(df)
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How can I scale(x) only certain columns of a dataframe? I have a dataframe with 7 columns and I want to scale only column 3 and 6. The rest should stay as it is.
We can do this with lapply. Subset the columns of interest, loop through them with lapply, assign the output back to the subset of data. Here, we are using c because the outpuf of scale is a matrix with a single column. Using c or as.vector, it gets converted to vector
df[c(3,6)] <- lapply(df[c(3, 6), function(x) c(scale(x)))
Or another option is mutate_at from dplyr
library(dplyr)
df %>%
mutate_at(c(3,6), funs(c(scale(.))))
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I have two data frames called df1 and df2, and df1 has 2 columns named poi, score. And the another data frame df2 has only one column called poi_ and it contains a few common data from df1$poi. I would be needing to check which df2$poi_ have their score defined in df1$poi and if score is present then put a new column called score_ in df2 and fill the column with the score found in df1
try this:
res <- merge(df2,df1,by.x="poi_",by.y="poi")
names(res)[2] <- "score_"
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Trying to merge two data frames, using a variable called hash_id. For some reason R does not recognize the hash-id's in one of the data frames, while it does so in the other.
I have checked and I just don't get it. See below how I checked:
> head(df1[46],1) # so I take the first 'hash-id' from df1
# hash_id
# 1 abab123123
> which(df2 == "abab123123", arr.ind=TRUE) # here it shows that row 6847 contains a match
# row col
# [1,] 6847 32`
> which(df1 == "abab123123", arr.ind=TRUE) # and here there is NO matching value!
# row col
#
One possibility is trailing or leading spaces in the concerned columns for one of the datasets. You could do:
library(stringr)
df1[, "hash_id"] <- str_trim(df1[,"hash_id"])
df2[, "hash_id"] <- str_trim(df2[, "hash_id"])
which(df1[, "hash_id"]=="abab123123", arr.ind=TRUE)
which(df2[, "hash_id"]=="abab123123", arr.ind=TRUE)
Another way would be use grep
grepl("\\babab123123\\b", df1[,"hash_id"])
grepl("\\babab123123\\b", df2[,"hash_id"])
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I have a data frame with daily data in R (148 columns by 6230 rows). I want to find the correlations coefficients using sliding windows with length of 600 (days) with windows displacement of 5 (days) and trying to generate 1220 correlation matrices (approx.). All the examples that I saw used only one information vector. There exist an easy way to find those correlation matrices using sliding window? I'll appreciate any suggestion.
If M is the input matrix then each row of out is one correlation matrix strung out column by column:
library(zoo)
out <- rollapply(M, 600, by = 5, function(x) c(cor(x)), by.column = FALSE)
They could be reshaped into a list of correlation matrices, if need be:
L <- lapply(1:nrow(out), function(i) matrix(out[i, ], ncol(M)))
or as an array:
simplify2array(L)