I have two dataframes of equal dimensions.
One has some value in cells (i.e. 'abc') that i need to index. Other has all different values. And I need to replace the values in other dataframe with the same index as 'abc'.
Examples:
df1 <- data.frame('1'=c('abc','bbb','rweq','dsaf','cxc','rwer','anc','ewr','yuje','gda'),
'2'=c(NA,NA,'bbb','dsaf','rwer','dsaf','ewr','cxc','dsaf','cxc'),
'3'=c(NA,NA,'dsaf','abc','bbb','cxc','yuje',NA,'ewr','anc'),
'4'=c(NA,NA,'cxc',NA,'abc','anc',NA,NA,'yuje','rweq'),
'5'=c(NA,NA,'anc',NA,'abc',NA,NA,NA,'rwer','rwer'),
'6'=c(NA,NA,'rweq',NA,'dsaf',NA,NA,NA,'bbb','bbb'),
'7'=c(NA,NA,'abc',NA,'ewr',NA,NA,NA,'abc','abc'),
'8'=c(NA,NA,'abc',NA,'rweq',NA,NA,NA,'cxc','bbb'),
'9'=c(NA,NA,NA,NA,'abc',NA,NA,NA,'anc',NA),
'10'=c(NA,NA,NA,NA,'abc',NA,NA,NA,'rweq',NA))
df2 <- data.frame('1'=c('green','black','white','yelp','help','green','red','brown','green','crack'),
'2'=c(NA,NA,'black','yelp','green','yelp','brown','help','yelp','help'),
'3'=c(NA,NA,'yelp','green','black','help','green',NA,'brown','red'),
'4'=c(NA,NA,'help',NA,'green','red',NA,NA,'green','white'),
'5'=c(NA,NA,'red',NA,'green',NA,NA,NA,'green','green'),
'6'=c(NA,NA,'white',NA,'yelp',NA,NA,NA,'black','black'),
'7'=c(NA,NA,'green',NA,'brown',NA,NA,NA,'green','green'),
'8'=c(NA,NA,'green',NA,'white',NA,NA,NA,'help','black'),
'9'=c(NA,NA,NA,NA,'green',NA,NA,NA,'red',NA),
'10'=c(NA,NA,NA,NA,'green',NA,NA,NA,'white',NA))
I can find sequential index of 'abc', but it returns one-sized vector
which(df1 == 'abc')
#[1] 1 24 35 45 63 69 70 73 85 95
And i don't know how to replace values using this method
In output expected to view df2 with replaced values 'green' only on the same indexes as values 'abc' in df1.
But note!! that 'green' values in df2 are not only in the same indexes as in df1
I don't think your problem is appropriately approached with the data in a data.frame. That introduces several complications. First, each variable (column) in the data frame is a factor with different levels! Second, your code is making a comparison between a list (data.frame) and a factor (which is coerced into an atomic vector). The help function for the == operator states ..if the other is a list R attempts to coerce it to the type of the atomic vector.. The help function also points out that factors get special handling in comparisons where it first assumes you are comparing factor levels, which your code is doing.
I think you want to convert your data frames of identical dimensions to a matrix first. If you need the results in a data.frame, convert it back after as I show here but realize that the factor levels may have changed.
# Starting with the values assigned to df1 and df2
m1 <- as.matrix(df1)
m2 <- as.matrix(df2)
index <- which(m1 == "abc")
m2[index] <- "abc"
df2 <- as.data.frame(m2)
Here is a way to. Learn about the *apply family in R: I think it is the most useful group of functions in this language, whatever you plan to do ;) Also know that data.frame are of 'list' type.
df1 <- lapply(df1, function(frame, pattern, replace){ # for each frame = column:
matches <- which(pattern %in% frame) # what are the matching indexes of the frame
if(length(matches) > 0) # If there is at least one index matching,
frame[matches] <- replace # give it the value you want
return(frame) # Commit your changes back to df1
}, pattern="abc", replace= "<whatYouWant>") # don't forget this part: the needed arguments !
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 am new to r and rstudio and I need to create a vector that stores the first 100 rows of the csv file the programme reads . However , despite all my attempts my variable v1 ends up becoming a dataframe instead of an int vector . May I know what I can do to solve this? Here's my code:
library(readr)
library(readr)
cup_data <- read_csv("C:/Users/Asus.DESKTOP-BTB81TA/Desktop/STUDY/YEAR 2/
YEAR 2 SEM 2/PREDICTIVE ANALYTICS(1_PA_011763)/Week 1 (Intro to PA)/
Practical/cup98lrn variable subset small.csv")
# Retrieve only the selected columns
cup_data_small <- cup_data[c("AGE", "RAMNTALL", "NGIFTALL", "LASTGIFT",
"GENDER", "TIMELAG", "AVGGIFT", "TARGET_B", "TARGET_D")]
str(cup_data_small)
cup_data_small
#get the number of columns and rows
ncol(cup_data_small)
nrow(cup_data_small)
cat("No of column",ncol(cup_data_small),"\nNo of Row :",nrow(cup_data_small))
#cat
#Concatenate and print
#Outputs the objects, concatenating the representations.
#cat performs much less conversion than print.
#Print the first 10 rows of cup_data_small
head(cup_data_small, n=10)
#Create a vector V1 by selecting first 100 rows of AGE
v1 <- cup_data_small[1:100,"AGE",]
Here's what my environment says:
cup_data_small is a tibble, a slightly modified version of a dataframe that has slightly different rules to try to avoid some common quirks/inconsistencies in standard dataframes. E.g. in a standard dataframe, df[, c("a")] gives you a vector, and df[, c("a", "b")] gives you a dataframe - you're using the same syntax so arguably they should give the same type of result.
To get just a vector from a tibble, you have to explicitly pass drop = TRUE, e.g.:
library(dplyr)
# Standard dataframe
iris[, "Species"]
iris_tibble = iris %>%
as_tibble()
# Remains a tibble/dataframe
iris_tibble[, "Species"]
# This gives you just the vector
iris_tibble[, "Species", drop = TRUE]
I want to apply some operations to the values in a number of columns, and then sum the results of each row across columns. I can do this using:
x <- data.frame(sample=1:3, a=4:6, b=7:9)
x$a2 <- x$a^2
x$b2 <- x$b^2
x$result <- x$a2 + x$b2
but this will become arduous with many columns, and I'm wondering if anyone can suggest a simpler way. Note that the dataframe contains other columns that I do not want to include in the calculation (in this example, column sample is not to be included).
Many thanks!
I would simply subset the columns of interest and apply everything directly on the matrix using the rowSums function.
x <- data.frame(sample=1:3, a=4:6, b=7:9)
# put column indices and apply your function
x$result <- rowSums(x[,c(2,3)]^2)
This of course assumes your function is vectorized. If not, you would need to use some apply variation (which you are seeing many of). That said, you can still use rowSums if you find it useful like so. Note, I use sapply which also returns a matrix.
# random custom function
myfun <- function(x){
return(x^2 + 3)
}
rowSums(sapply(x[,c(2,3)], myfun))
I would suggest to convert the data set into the 'long' format, group it by sample, and then calculate the result. Here is the solution using data.table:
library(data.table)
melt(setDT(x),id.vars = 'sample')[,sum(value^2),by=sample]
# sample V1
#1: 1 65
#2: 2 89
#3: 3 117
You can easily replace value^2 by any function you want.
You can use apply function. And get those columns that you need with c(i1,i2,..,etc).
apply(( x[ , c(2, 3) ])^2, 1 ,sum )
If you want to apply a function named somefunction to some of the columns, whose indices or colnames are in the vector col_indices, and then sum the results, you can do :
# if somefunction can be vectorized :
x$results<-apply(x[,col_indices],1,function(x) sum(somefunction(x)))
# if not :
x$results<-apply(x[,col_indices],1,function(x) sum(sapply(x,somefunction)))
I want to come at this one from a "no extensions" R POV.
It's important to remember what kind of data structure you are working with. Data frames are actually lists of vectors--each column is itself a vector. So you can you the handy-dandy lapply function to apply a function to the desired column in the list/data frame.
I'm going to define a function as the square as you have above, but of course this can be any function of any complexity (so long as it takes a vector as an input and returns a vector of the same length. If it doesn't, it won't fit into the original data.frame!
The steps below are extra pedantic to show each little bit, but obviously it can be compressed into one or two steps. Note that I only retain the sum of the squares of each column, given that you might want to save space in memory if you are working with lots and lots of data.
create data; define the function
grab the columns you want as a separate (temporary) data.frame
apply the function to the data.frame/list you just created.
lapply returns a list, so if you intend to retain it seperately make it a temporary data.frame. This is not necessary.
calculate the sums of the rows of the temporary data.frame and append it as a new column in x.
remove the temp data.table.
Code:
x <- data.frame(sample=1:3, a=4:6, b=7:9); square <- function(x) x^2 #step 1
x[2:3] #Step 2
temp <- data.frame(lapply(x[2:3], square)) #step 3 and step 4
x$squareRowSums <- rowSums(temp) #step 5
rm(temp) #step 6
Here is an other apply solution
cols <- c("a", "b")
x <- data.frame(sample=1:3, a=4:6, b=7:9)
x$result <- apply(x[, cols], 1, function(x) sum(x^2))
When I pass a row of a data frame to a function using apply, I lose the class information of the elements of that row. They all turn into 'character'. The following is a simple example. I want to add a couple of years to the 3 stooges ages. When I try to add 2 a value that had been numeric R says "non-numeric argument to binary operator." How do I avoid this?
age = c(20, 30, 50)
who = c("Larry", "Curly", "Mo")
df = data.frame(who, age)
colnames(df) <- c( '_who_', '_age_')
dfunc <- function (er) {
print(er['_age_'])
print(er[2])
print(is.numeric(er[2]))
print(class(er[2]))
return (er[2] + 2)
}
a <- apply(df,1, dfunc)
Output follows:
_age_
"20"
_age_
"20"
[1] FALSE
[1] "character"
Error in er[2] + 2 : non-numeric argument to binary operator
apply only really works on matrices (which have the same type for all elements). When you run it on a data.frame, it simply calls as.matrix first.
The easiest way around this is to work on the numeric columns only:
# skips the first column
a <- apply(df[, -1, drop=FALSE],1, dfunc)
# Or in two steps:
m <- as.matrix(df[, -1, drop=FALSE])
a <- apply(m,1, dfunc)
The drop=FALSE is needed to avoid getting a single column vector.
-1 means all-but-the first column, you could instead explicitly specify the columns you want, for example df[, c('foo', 'bar')]
UPDATE
If you want your function to access one full data.frame row at a time, there are (at least) two options:
# "loop" over the index and extract a row at a time
sapply(seq_len(nrow(df)), function(i) dfunc(df[i,]))
# Use split to produce a list where each element is a row
sapply(split(df, seq_len(nrow(df))), dfunc)
The first option is probably better for large data frames since it doesn't have to create a huge list structure upfront.