Divide rows by its number of columns without an NA value - r

I have a dataframe containing a single NA in the first row of column b:
a <- c(16.54868281, 47.64097026, 51.0773201)
b <- c(NA, 39.40217391, 13.04347826)
c <- c(17.80821918, 42.92237443, 36.75799087)
d <- c(22.90809328, 56.37860082, 61.04252401)
data <- data.frame(cbind(a, b, c, d))
data
a b c d
1 16.54868 NA 17.80822 22.90809
2 47.64097 39.40217 42.92237 56.37860
3 51.07732 13.04348 36.75799 61.04252
Here, I am trying to acquire an average score of each row. But, because of the NA, the first row returns an NA for its mean.
safety <- data.frame(
(data$a + data$b + data$c + data$d) / 4
)
names(safety)[1] <- "safety"
safety
safety
1 NA
2 46.58603
3 40.48033
To resolve this, I have replaced NA to 0. Unfortunately, the computer is treating the missing value as a number, and I am dividing every row by 4. Therefore, I am getting a wrong mean for the first row.
a <- c(16.54868281, 47.64097026, 51.0773201)
b <- c(NA, 39.40217391, 13.04347826)
c <- c(17.80821918, 42.92237443, 36.75799087)
d <- c(22.90809328, 56.37860082, 61.04252401)
data <- data.frame(cbind(a, b, c, d))
data[is.na(data)] <- 0
safety <- data.frame(
(data$a + data$b + data$c + data$d) / 4
)
names(safety)[1] <- "safety"
safety
safety
1 14.31625
2 46.58603
3 40.48033
I need the first row to read 19.08833 instead of 14.31625. Is there a function in R that allows me to divide each row by the number of columns in its equation? I can probably create a long way to solving this issue. But, as the dataset grows bigger, my primitive shortcomings would soon meet its end.

Use rowMeans with na.rm = TRUE:
rowMeans(data, na.rm = TRUE)
# [1] 19.08833 46.58603 40.48033

Related

Create dataframe from smallest vector available

I want to create a dataframe from a list of dataframes, specifically from a certain column of those dataframes. However each dataframe contains a different number of observations, so the following code gives me an error.
diffs <- data.frame(sensor1 = sensores[[1]]$Diff,
sensor2 = sensores[[2]]$Diff,
sensor3 = sensores[[3]]$Diff,
sensor4 = sensores[[4]]$Diff,
sensor5 = sensores[[5]]$Diff)
The error:
Error in data.frame(sensor1 = sensores[[1]]$Diff, sensor2 = sensores[[2]]$Diff, :
arguments imply differing number of rows: 29, 19, 36, 26
Is there some way to force data.frame() to take the minimal number or rows available from each one of the columns, in this case 19?
Maybe there is a built-in function in R that can do this, any solution is appreciated but I'd love to get something as general and as clear as possible.
Thank you in advance.
I can think of two approaches:
Example data:
df1 <- data.frame(A = 1:3)
df2 <- data.frame(B = 1:4)
df3 <- data.frame(C = 1:5)
Compute the number of rows of the smallest dataframe:
min_rows <- min(sapply(list(df1, df2, df3), nrow))
Use subsetting when combining:
diffs <- data.frame(a = df1[1:min_rows,], b = df2[1:min_rows,], c = df3[1:min_rows,] )
diffs
a b c
1 1 1 1
2 2 2 2
3 3 3 3
Alternatively, use merge:
rowmerge <- function(x,y){
# create row indicators for the merge:
x$ind <- 1:nrow(x)
y$ind <- 1:nrow(y)
out <- merge(x,y, all = T, by = "ind")
out["ind"] <- NULL
return(out)
}
Reduce(rowmerge, list(df1, df2, df3))
A B C
1 1 1 1
2 2 2 2
3 3 3 3
4 NA 4 4
5 NA NA 5
To get rid of the rows with NAs, remove the all = T.
For your particular case, you would probably call Reduce(rowmerge, sensores), assuming that sensores is a list of dataframes.
Note: if you already have an index somewhere (e.g. a timestamp of some sort), then it would be advisable to simply merge on that index instead of creating ind.

Limiting Duplication of Specified Columns

I'm trying to find a way to add some constraints into a linear programme to force the solution to have a certain level of uniqueness to it. I'll try explain what I mean here. Take the example below, the linear programme returns the max possible Score for a combination of 2 males and 1 female.
Looking at the Team/Grade/Rep columns however we can see that there is a lot of duplication from row to row. In fact Shana and Jason are identical.
Name<-c("Jane","Brad","Harry","Shana","Debra","Jason")
Sex<-c("F","M","M","F","F","M")
Score<-c(25,50,36,40,39,62)
Team<-c("A","A","A","B","B","B")
Grade<-c(1,2,1,2,1,2)
Rep<-c("C","D","C","D","D","D")
df<-data.frame(Name,Sex,Score,Team,Grade,Rep)
df
Name Sex Score Team Grade Rep
1 Jane F 25 A 1 C
2 Brad M 50 A 2 D
3 Harry M 36 A 1 C
4 Shana F 40 B 2 D
5 Debra F 39 B 1 D
6 Jason M 62 B 2 D
library(Rglpk)
num <- length(df$Name)
obj<-df$Score
var.types<-rep("B",num)
matrix <- rbind(as.numeric(df$Sex == "M"),as.numeric(df$Sex == "F"))
direction <- c("==","==")
rhs<-c(2,1)
sol <- Rglpk_solve_LP(obj = obj, mat = matrix, dir = direction, rhs = rhs,types = var.types, max = TRUE)
df[sol$solution==1,]
Name Sex Score Team Grade Rep
2 Brad M 50 A 2 D
4 Shana F 40 B 2 D
6 Jason M 62 B 2 D
What I am trying to work out is how to limit say the level of randomness across those last three columns. For example I would like there to no more than ie 2 columns the same across any two rows. So this would mean that either the Shana row or Jason row would be replaced in the model with an alternative.
I'm not sure if this is something that can be easily added into the Rglpk model? Appreciate any help that can be offered.
It sounds like you're asking how to prevent having a pair of individuals who are "too similar" from being returned by your optimization model. Once you have determined a rule for what makes a pair of people "too similar", you can simply add a constraint for each pair, limiting your solution to have no more than one of those two people.
For instance, if we use your rule of having no more than 2 columns the same, we could easily identify all pairs that we want to block:
pairs <- t(combn(nrow(df), 2))
(blocked <- pairs[rowSums(sapply(df[,c("Team", "Grade", "Rep")], function(x) {
x[pairs[,1]] == x[pairs[,2]]
})) >= 3,])
# [,1] [,2]
# [1,] 1 3
# [2,] 4 6
We want to block the pairs Jane/Harry and Shana/Jason. This is easy to do with linear constraints:
library(Rglpk)
num <- length(df$Name)
obj<-df$Score
var.types<-rep("B",num)
matrix <- rbind(as.numeric(df$Sex == "M"), as.numeric(df$Sex == "F"),
outer(blocked[,1], seq_len(num), "==") + outer(blocked[,2], seq_len(num), "=="))
direction <- rep(c("==", "<="), c(2, nrow(blocked)))
rhs<-c(2, 1, rep(1, nrow(blocked)))
sol <- Rglpk_solve_LP(obj = obj, mat = matrix, dir = direction, rhs = rhs,types = var.types, max = TRUE)
df[sol$solution==1,]
# Name Sex Score Team Grade Rep
# 2 Brad M 50 A 2 D
# 5 Debra F 39 B 1 D
# 6 Jason M 62 B 2 D
The approach of computing every pair to block is attractive because we could have a much more complicated rule for which pairs to block, since we don't need to encode the rule into the linear program. All we need to be able to do is to compute every pair that needs to be blocked.
For each group of rows having the same last 3 columns we construct a constraint such that at most one of those rows may appear. If a is an indictor vector of the rows of such a group then the constraint would look like this:
a'x <= 1
To do that split the row numbers by the last 3 columns into a list of vectors s each of whose components is a vector of row numbers for rows having the same last 3 columns. Only keep those conponents having more than 1 row number giving s1. In this case the first component of s1 is c(1, 3) referring to the Jane and Harry rows and the second component is c(4, 6) referring to the Shana and Jason rows. In this particular data there were 2 rows in each of the groups but in other data there could be more than 2 rows in a group. excl has one row (constraint) for each element of s1.
The data in the question only has groups of size 2 but in general if there were k rows in some group one would need k choose 2 constraint rows to ensure that only one of the k were chosen if this were done pairwise whereas the approach here only requires one constraint row for the entire group. For example, if k = 10 then choose(10, 2) = 45 so this uses 1 constraint in place of 45.
Finally rbind excl to matrix giving matrix2 and adjust the other Rglpk_solve_LP arguments accordingly giving:
nr <- nrow(df)
s <- split(1:nr, df[4:6])
s1 <- s[lengths(s) > 1]
excl <-t(sapply(s1, "%in%", x = 1:nr)) + 0
matrix2 <- rbind(matrix, excl)
direction2 <- c(direction, rep("<=", nrow(excl)))
rhs2 <- c(rhs, rep(1, nrow(excl)))
sol2 <- Rglpk_solve_LP(obj = obj, mat = matrix2,
dir = direction2, rhs = rhs2, types = "B", max = TRUE)
df[ sol2$solution == 1, ]
giving:
Name Sex Score Team Grade Rep
2 Brad M 50 A 2 D
5 Debra F 39 B 1 D
6 Jason M 62 B 2 D

Comparing two columns

I am new to R and I am trouble with a command that I did all the time in Python.
I have two data-frames (database and creditIDs), and what I want to do is compare one column in database and one column in creditIDs. More specifically in a value exists in creditIDs[,1] but doesn't in database[,5], I want to delete that entire row in database.
Here is the code:
for (i in 1:lengthColumns){
if (!(database$credit_id[i] %in% creditosVencidos)){
database[i,]<-database[-i,]
}
}
But I keep on getting this error:
50: In `[<-.data.frame`(`*tmp*`, i, , value = structure(list( ... :
replacement element 50 has 9696 rows to replace 1 rows
Could someone explain why this is happening? Thanks!
the which() command will return the row indices that satisfy a boolean statement, much like numpy.where() in python. Using the $ after a dataframe with a column name gives you a vector of that column... alternatively you could do d[,column_number].
In this example I'm creating an x and y column which share the first five values, and use which() to slice the dataframe on their by-row equality:
L3 <- LETTERS[1:3]
fac <- sample(L3, 10, replace = TRUE)
(d <- data.frame(x = rep(1:5, 2), y = 1:10, fac = fac))
d = d[which(d$x == d$y),]
d
x y fac
1 1 A
2 2 B
3 3 C
4 4 B
5 5 B
You will need to adjust this for your column names/numbers.
# Create two example data.frames
creditID <- data.frame(ID = c("896-19", "895-8", "899-1", "899-5"))
database <- data.frame(ID = c("896-19", "camel", "899-1", "goat", "899-1"))
# Method 1
database[database$ID %in% creditID$ID, ]
# Method 2 (subset() function)
database <- subset(database, ID %in% creditID$ID)

How to label ties when creating a variable capturing the most frequent occurence of a group?

In the following example, how do I ask R to identify a tie as "tie" when I want to determine the most frequent value within a group?
I am basically following on from a previous question, that used which.max or which.is.max and a custom function (Create a variable capturing the most frequent occurence by group), but I want to acknowledge the ties as a tie. Any ideas?
df1 <-data.frame(
id=c(rep(1,3),rep(2,3)),
v1=as.character(c("a","b","b",rep("c",3)))
)
I want to create a third variable freq that contains the most frequent observation in v1 by id, but also creates identifies ties as "tie".
From previous answers, this code works to create the freq variable, but just doesn't deal with the ties:
myFun <- function(x){
tbl <- table(x$v1)
x$freq <- rep(names(tbl)[which.max(tbl)],nrow(x))
x
}
ddply(df1,.(id),.fun=myFun)
You could slightly modify your function by testing if the maximum count occurs more than once. This happens in sum(tbl == max(tbl)). Then proceed accordingly.
df1 <-data.frame(
id=rep(1:2, each=4),
v1=rep(letters[1:4], c(2,2,3,1))
)
myFun <- function(x){
tbl <- table(x$v1)
nmax <- sum(tbl == max(tbl))
if (nmax == 1)
x$freq <- rep(names(tbl)[which.max(tbl)],nrow(x))
else
x$freq <- "tie"
x
}
ddply(df1,.(id),.fun=myFun)
id v1 freq
1 1 a tie
2 1 a tie
3 1 b tie
4 1 b tie
5 2 c c
6 2 c c
7 2 c c
8 2 d c

Improving performance of updating contents of large data frame using contents of similar data frame

I'm looking for a general solution for updating one large data frame with the contents of a second similar data frame. I have dozens of datasets, each with thousands of rows and upwards of 10,000 columns. An "update" dataset will overlap its corresponding "base" dataset by anywhere from a few percent to perhaps 50 percent, rowwise. The datasets have a "key" column and there will be only one row per each unique key value in any given dataset.
The basic rule is: if a non-NA value exists in the update dataset for a given cell, replace the same cell in the base dataset with that value. (The "same cell" means same value of the "key" column and colname.)
Note the update dataset will likely contain new rows ("inserts") which I can handle with an rbind.
So given the base data frame "df1", where column "K" is the unique key column, and "P1" .. "P3" represent the 10,000 columns, whose names will vary from one pair of datasets to the next:
K P1 P2 P3
1 A 1 1 1
2 B 1 1 1
3 C 1 1 1
...and the update data frame "df2":
K P1 P2 P3
1 B 2 NA 2
2 C NA 2 2
3 D 2 2 2
The result I need is as follows, where the 1's for "B" and "C" were overwritten by the 2's but not overwritten by the NA's:
K P1 P2 P3
1 A 1 1 1
2 B 2 1 2
3 C 1 2 2
4 D 2 2 2
This doesn't seem to be a merge candidate as merge gives me either duplicate rows (with respect to the "key" column) or duplicate columns (e.g. P1.x, P1.y), which I have to iterate over to collapse somehow.
I have tried pre-allocating a matrix with the dimensions of the final rows/columns, and populating it with the contents of df1, then iterating over the overlapping rows of df2, but I cannot get better than 20 cells per second performance, requiring hours to complete (compared to minutes for the equivalent DATA step UPDATE functionality in SAS).
I'm sure I'm missing something, but can't find a comparable example.
I see ddply usage that looks close, but not a general solution. The data.table package didn't seem to help as it's not obvious to me that this is a join problem, at least not generally over so many columns.
Also a solution that focuses only on the intersecting rows is adequate as I can identify the others and rbind them in.
Here is some code to fabricate the data frames above:
cat("K,P1,P2,P3", "A,1,1,1", "B,1,1,1", "C,1,1,1", file="f1.dat", sep="\n");
cat("K,P1,P2,P3", "B,2,,2", "C,,2,2", "D,2,2,2", file="f2.dat", sep="\n");
df1 <- read.table("f1.dat", sep=",", header=TRUE, stringsAsFactors=FALSE);
df2 <- read.table("f2.dat", sep=",", header=TRUE, stringsAsFactors=FALSE);
Thanks
This loops by column, setting dt1 by reference and (hopefully) should be quick.
dt1 = as.data.table(df1)
dt2 = as.data.table(df2)
if (!identical(names(dt1),names(dt2)))
stop("Assumed for now. Can relax later if needed.")
w = chmatch(dt2$K, dt1$K)
for (i in 2:ncol(dt2)) {
nna = !is.na(dt2[[i]])
set(dt1,w[nna],i,dt2[[i]][nna])
}
dt1 = rbind(dt1,dt2[is.na(w)])
dt1
K P1 P2 P3
[1,] A 1 1 1
[2,] B 2 1 2
[3,] C 1 2 2
[4,] D 2 2 2
This is likely not the fastest solution but is done entirely in base.
(updated answer per Tommy's comments)
#READING IN YOUR DATA FRAMES
df1 <- read.table(text=" K P1 P2 P3
1 A 1 1 1
2 B 1 1 1
3 C 1 1 1", header=TRUE)
df2 <- read.table(text=" K P1 P2 P3
1 B 2 NA 2
2 C NA 2 2
3 D 2 2 2", header=TRUE)
all <- c(levels(df1$K), levels(df2$K)) #all cells of key column
dups <- all[duplicated(all)] #the overlapping key cells
ndups <- all[!all %in% dups] #unique key cells
df3 <- rbind(df1[df1$K%in%ndups, ], df2[df2$K%in%ndups, ]) #bind the unique rows
decider <- function(x, y) ifelse(is.na(x), y, x) #function replaces NAs if existing
df4 <- data.frame(mapply(df2[df2$K%in%dups, ], df1[df1$K%in%dups, ],
FUN = decider)) #repalce all NAs of df2 with df1 values if they exist
df5 <- rbind(df3, df4) #bind unique rows of df1 and df2 with NA replaced df4
df5 <- df5[order(df5$K), ] #reorder based on key column
rownames(df5) <- 1:nrow(df5) #give proper non duplicated rownames
df5
This yields:
K P1 P2 P3
1 A 1 1 1
2 B 2 1 2
3 C 1 2 2
4 D 2 2 2
Upon closer reading not all columns have the same name but I am assuming the same order. this may be a more helpful approach:
all <- c(levels(df1$K), levels(df2$K))
dups <- all[duplicated(all)]
ndups <- all[!all %in% dups]
LS <- list(df1, df2)
LS2 <- lapply(seq_along(LS), function(i) {
colnames(LS[[i]]) <- colnames(LS[[2]])
return(LS[[i]])
}
)
LS3 <- lapply(seq_along(LS2), function(i) LS2[[i]][LS2[[i]]$K%in%ndups, ])
LS4 <- lapply(seq_along(LS2), function(i) LS2[[i]][LS2[[i]]$K%in%dups, ])
decider <- function(x, y) ifelse(is.na(x), y, x)
DF <- data.frame(mapply(LS4[[2]], LS4[[1]], FUN = decider))
DF$K <- LS4[[1]]$K
LS3[[3]] <- DF
df5 <- do.call("rbind", LS3)
df5 <- df5[order(df5$K), ]
rownames(df5) <- 1:nrow(df5)
df5
EDIT : Please ignore this answer. Bad idea to loop by row. It works but is very slow. Left for posterity! See my 2nd attempt as separate answer.
require(data.table)
dt1 = as.data.table(df1)
dt2 = as.data.table(df2)
K = dt2[[1]]
for (i in 1:nrow(dt2)) {
k = K[i]
p = unlist(dt2[i,-1,with=FALSE])
p = p[!is.na(p)]
dt1[J(k),names(p):=as.list(p),with=FALSE]
}
or, can you use matrix instead of data.frame? If so it could be a single line using A[B] syntax where B is a 2-column matrix containing the row and column numbers to update.
The following gives the correct answer for the small example data, tries to minimize the number of "copies" of tables, and uses the new fread and (new?) rbindlist. Does it work with your larger actual data set? I didn't quite follow all the comments in the original post about the memory issues you had when trying to flatten/normalize/stack, so apologies if you've already tried this route.
library(data.table)
library(reshape2)
cat("K,P1,P2,P3", "A,1,1,1", "B,1,1,1", "C,1,1,1", file="f1.dat", sep="\n")
cat("K,P1,P2,P3", "B,2,,2", "C,,2,2", "D,2,2,2", file="f2.dat", sep="\n")
dt1s<-data.table(melt(fread("f1.dat"), id.vars="K"), key=c("K","variable")) # read f1.dat, melt to long/stacked format, and convert to data.table
dt2s<-data.table(melt(fread("f2.dat"), id.vars="K", na.rm=T), key=c("K","variable")) # read f2.dat, melt to long/stacked format (removing NAs), and convert to data.table
setnames(dt2s,"value","value.new")
dt1s[dt2s,value:=value.new] # Update new values
dtout<-reshape(rbindlist(list(dt1s,dt1s[dt2s][is.na(value),list(K,variable,value=value.new)])), direction="wide", idvar="K", timevar="variable") # Use rbindlist to insert new records, and then reshape
setkey(dtout,K)
setnames(dtout,colnames(dtout),sub("value.", "", colnames(dtout))) # Clean up the column names

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