R Minimum Value from Datatable Not Equal to a Particular Value - r

How do I find the minimum value from an R data table other than a particular value?
For example, there could be zeroes in the data table and the goal would be to find the minimum non zero value.
I tried using the sapply with min, but am not sure how to specify the extra criteria that we have so that the minimum is not equal to a certain value.
More generally, How do we find the minimum from a data table not equal to any element from a list of possible values?

If you want to find the minimum value from a vector while excluding certain values from that vector, then you can use %in%:
v <- c(1:10) # values 1 .. 10
v.exclude <- c(1, 2) # exclude the values 1 and 2 from consideration
min.exclude <- min(v[!v %in% v.exclude])
The logic won't change much if you are using a column from a data table/frame. In this case you can just replace the vector v with the apropriate column. If you have your excluded values in a list, then you can flatten it to produce your v.exclude vector.

This can be done with data.table (as the OP mentioned about data table in the post) after setting the key
library(data.table)
setDT(df, key='a')[!.(exclude)]
# a b
#1: 4 40
#2: 5 50
#3: 6 60
If we need the min value of 'a'
min(setDT(df, key='a')[!.(exclude)]$a)
#[1] 4
For finding the min in all the columns (using the setkey method), we loop over the columns of the dataset, set the key as each of the column, subset the dataset, get the min value in a previously created list object.
setDT(df)
MinVal <- vector('list', length(df))
for(j in seq_along(df)){
setkeyv(df, names(df)[j])
MinVal[[j]] <- min(df[!.(exclude)][[j]])
}
MinVal
#[[1]]
#[1] 4
#[[2]]
#[1] 10
data
df <- data.frame(a = c(0,2,3,2,1,2,3,4,5,6),
b = c(10,10,20,20,30,30,40,40,50,60))
exclude <- c(0,1,2,3)

Assuming you are working with a data.frame
Data
df <- data.frame(a = c(0,2,3,2,1,2,3,4,5,6),
b = c(10,10,20,20,30,30,40,40,50,60))
Values to exlude from our minimum search
exclude <- c(0,1,2,3)
we can find the minimum value from column a excluding our exclude vector
## minimum from column a
min(df[!df$a %in% exclude,]$a)
# [1] 4
Or from b
exclude <- c(10, 20, 30, 40)
min(df[!df$b %in% exclude,]$b)
# [1] 50
To return the row that corresponds to the minimum value
df[df$b == min( df[ !df$b %in% exclude, ]$b ),]
# a b
# 9 5 50
Update
To find the minimum across multiple rows we can do it this way:
## values to exclude
exclude_a <- c(0,1)
exclude_b <- c(10)
## exclude rows/values from each column we don't want
df2 <- df[!(df$a %in% exclude_a) & !(df$b %in% exclude_b),]
## order the data
df3 <- df2[with(df2, order(a,b)),]
## take the first row
df3[1,]
# > df3[1,]
# a b
#4 2 20
Update 2
To select from multiple columns we can iterate over them as #akrun has shown, or alternatively we can construct our subsetting formula using an expression and evaluate it inside our [ operation
exclude <- c(0,1,2, 10)
## construct a formula/expression using the column names
n <- names(df)
expr <- paste0("(", paste0(" !(df$", n, " %in% exclude) ", collapse = "&") ,")")
# [1] "( !(df$a %in% exclude) & !(df$b %in% exclude) )"
expr <- parse(text=expr)
df2 <- df[eval(expr),]
## order and select first row as before
df2 <- df2[with(df2, order(a,b)),]
df2 <- df2[1,]
And if we wanted to use data.table for this:
library(data.table)
setDT(df)[ eval(expr) ][order(a, b),][1,]
comparison of methods
library(microbenchmark)
fun_1 <- function(x){
df2 <- x[eval(expr),]
## order and select first row as before
df2 <- df2[with(df2, order(a,b)),]
df2 <- df2[1,]
return(df2)
}
fun_2 <- function(x){
df2 <- setDT(x)[ eval(expr) ][order(a, b),][1,]
return(df2)
}
## including #akrun's solution
fun_3 <- function(x){
setDT(df)
MinVal <- vector('list', length(df))
for(j in seq_along(df)){
setkeyv(df, names(df)[j])
MinVal[[j]] <- min(df[!.(exclude)][[j]])
}
return(MinVal)
}
microbenchmark(fun_1(df), fun_2(df), fun_3(df) , times=1000)
# Unit: microseconds
# expr min lq mean median uq max neval
# fun_1(df) 770.376 804.5715 866.3499 833.071 869.2195 2728.740 1000
# fun_2(df) 854.862 893.1220 952.1207 925.200 962.6820 3115.119 1000
# fun_3(df) 1108.316 1148.3340 1233.1268 1186.938 1234.3570 5400.544 1000

Related

reference x's column in R's apply function

I have a df like this:
a <- c(4,5,3,5,1)
b <- c(8,9,7,3,5)
c <- c(6,7,5,4,3)
df <- data.frame(rbind(a,b,c))
I want a new df, df2, containing the difference between the values in each cell in rows a and b and the value in row c in their respective columns.
df2 would look like this:
a <- c(-2,-2,-2,1,-2)
b <- c(2,2,2,-1,2)
df2 <- data.frame(rbind(a,b))
Here is where I'm getting stuck:
df2 <- data.frame(apply(df,c(1,2),function(x) x - df[nrow(df),the col index of x]))
How do I reference the column index of x? Is there something like JavaScript's this?
We can do this easily by replicating the 3rd row to make the lengths equal before subtracting with the first two rows
out <- df[c("a", "b"),] - df["c",][col(df[c("a", "b"),])]
identical(df2, out)
#[1] TRUE
Or explicitly using rep
df[c("a", "b"),] - rep(unlist(df["c",]), each = 2)

extract highest and lowest values for columns in R, as well as row identifiers

Say I have some data of the following kind:
df<-as.data.frame(matrix(rnorm(10*10000, 1, .5), ncol=10))
I want a new dataframe that keeps the 10 original columns, but for every column retains only the highest 10 and lowest 10 values. Importantly, the rows have names corresponding to id values that need to be kept in the new data frame.
Thus, the end result data.frame is gonna be of dimensions m by 10, where m is very likely to be more than 20. But for every column, I want only 20 valid values.
The only way I can think of doing this is doing it manually per column, using dplyr and arrange, grabbing the top and bottom rows, and then creating a matrix from all the individual vectors. Clearly this is inefficient. Help?
Assuming you want to keep all the rows from the original dataset, where there is at least one value satisfying your condition (value among ten largest or ten smallest in the given column), you could do it like this:
# create a data frame
df<-as.data.frame(matrix(rnorm(10*10000, 1, .5), ncol=10))
# function to find lowes 10 and highest 10 values
lowHigh <- function(x)
{
test <- x
test[!(order(x) <= 10 | order(x) >= (length(x)- 10))] <- NA
test
}
# apply the function defined above
test2 <- apply(df, 2, lowHigh)
# use the original rownames
rownames(test2) <- rownames(df)
# keep only rows where there is value of interest
finalData <- test2[apply(apply(test2, 2, is.na), 1, sum) < 10, ]
Please note that there is definitely some smarter way of doing it...
Here is the data matrix with 10 highest and 10 lowest in each column,
x<-apply(df,2,function(k) k[order(k,decreasing=T)[c(1:10,(length(k)-9):length(k))]])
x is your 20 by 10 matrix.
Your requirement of rownames is conflicting column by column, altogether you only have 20 rownames in this matrix and it can not be same for all 10 columns. Instead, here is your order matrix,
x_roworder<-apply(df,2,function(k) order(k,decreasing=T)[c(1:10,(length(k)-9):length(k))])
This will give you corresponding rows in original data matrix within each column.
I offer a couple of answers to this.
A base R implementation ( I have used %>% to make it easier to read)
ix = lapply(df, function(x) order(x)[-(1:(length(x)-20)+10)]) %>%
unlist %>% unique %>% sort
df[ix,]
This abuses the fact that data frames are lists, finds the row id satisfying the condition for each column, then takes the unique ones in order as the row indices you want to keep. This should retain any row names attached to df
An alternative using dplyr (since you mentioned it) which if I remember correctly doesn't particular like row names
# add id as a variable
df$id = 1:nrow(df) # or row names
df %>%
gather("col",value,-id) %>%
group_by(col) %>%
filter(min_rank(value) <= 10 | min_rank(desc(value)) <= 10) %>%
ungroup %>%
select(id) %>%
left_join(df)
Edited: To fix code alignment and make a neater filter
I'm not entirely sure what you're expecting for your return / output. But this will get you the appropriate indices
# example data
set.seed(41234L)
N <- 1000
df<-data.frame(id= 1:N, matrix(rnorm(10*N, 1, .5), ncol=10))
# for each column, extract ID's for top 10 and bottom 10 values
l1 <- lapply(df[,2:11], function(x,y, n) {
xy <- data.frame(x,y)
xy <- xy[order(xy[,1]),]
return(xy[c(1:10, (n-9):n),2])
}, y= df[,1], n = N)
# check:
xx <- sort(df[,2])
all.equal(sort(df[l1[[1]], 2]), xx[c(1:10, 991:1000)])
[1] TRUE
If you want an m * 10 matrix with these unique values, where m is the number of unique indices, you could do:
l2 <- do.call("c", l1)
l2 <- unique(l2)
df2 <- df[l2,] # in this case, m == 189
This doesn't 0 / NA the columns which you're not searching on for each row. But it's unclear what your question is trying to do.
Note
This isn't as efficient as using data.table since you're going to get a copy of the data in xy <- data.frame(x,y)
Benchmark
library(microbenchmark)
microbenchmark(ira= {
test2 <- apply(df[,2:11], 2, lowHigh);
rownames(test2) <- rownames(df);
finalData <- test2[apply(apply(test2, 2, is.na), 1, sum) < 10, ]
},
alex= {
l1 <- lapply(df[,2:11], function(x,y, n) {
xy <- data.frame(x,y)
xy <- xy[order(xy[,1]),]
return(xy[c(1:10, (n-9):n),2])
}, y= df[,1], n = N);
l2 <- unique(do.call("c", l1));
df2 <- df[l2,]
}, times= 50L)
Unit: milliseconds
expr min lq mean median uq max neval cld
ira 4.360452 4.522082 5.328403 5.140874 5.560295 8.369525 50 b
alex 3.771111 3.854477 4.054388 3.936716 4.158801 5.654280 50 a

Passing vector with multiple values into R function to generate data frame

I have a table, called table_wo_nas, with multiple columns, one of which is titled ID. For each value of ID there are many rows. I want to write a function that for input x will output a data frame containing the number of rows for each ID, with column headers ID and nobs respectively as below for x <- c(2,4,8).
## id nobs
## 1 2 1041
## 2 4 474
## 3 8 192
This is what I have. It works when x is a single value (ex. 3), but not when it contains multiple values, for example 1:10 or c(2,5,7). I receive the warning "In ID[counter] <- x : number of items to replace is not a multiple of replacement length". I've just started learning R and have been struggling with this for a week and have searched manuals, this site, Google, everything. Can someone help please?
counter <- 1
ID <- vector("numeric") ## contain x
nobs <- vector("numeric") ## contain nrow
for (i in x) {
r <- subset(table_wo_nas, ID %in% x) ## create subset for rows of ID=x
ID[counter] <- x ## add x to ID
nobs[counter] <- nrow(r) ## add nrow to nobs
counter <- counter + 1 } ## loop
result <- data.frame(ID, nobs) ## create data frame
In base R,
# To make a named vector, either:
tmp <- sapply(split(table_wo_nas, table_wo_nas$ID), nrow)
# OR just:
tmp <- table(table_wo_nas$ID)
# AND
# arrange into data.frame
nobs_df <- data.frame(ID = names(tmp), nobs = tmp)
Alternately, coerce the table into a data.frame directly, and rename:
nobs_df <- data.frame(table(table_wo_nas$ID))
names(nobs_df) <- c('ID', 'nobs')
If you only want certain rows, subset:
nobs_df[c(2, 4, 8), ]
There are many, many more options; these are just a few.
With dplyr,
library(dplyr)
table_wo_nas %>% group_by(ID) %>% summarise(nobs = n())
If you only want certain IDs, add on a filter:
table_wo_nas %>% group_by(ID) %>% summarise(nobs = n()) %>% filter(ID %in% c(2, 4, 8))
Seems pretty straightforward if you just use table again:
tbl <- table( table_wo_nas[ , 'ID'] )
data.frame( IDs = names(tbl), nobs= tbl)
Could also get a quick answer although with different column names using:
as.data.frame(table( table_wo_nas[ , 'ID'] ))
Try this.
x=c(2,4,8)
count_of_id=0
#df is your data frame table_wo_nas
count_of<-function(x)
{for(i in 1 : length(x))
{count_of_id[i]<-length(which(df$id==x[i])) #find out the n of rows for each unique value of x
}
df_1<-cbind(id,count_of_id)
return(df_1)
}

Intersecting multiple columns between two data frames

I have two data frames with 2 columns in each. For example:
df.1 = data.frame(col.1 = c("a","a","a","a","b","b","b","c","c","d"), col.2 = c("b","c","d","e","c","d","e","d","e","e"))
df.2 = data.frame(col.1 = c("b","b","b","a","a","e"), col.2 = c("a","c","e","c","e","c"))
and I'm looking for an efficient way to look up the row index in df.2 of every col.1 col.2 row pair of df.1. Note that a row pair in df.1 may appear in df.2 in reverse order (for example df.1[1,], which is "a","b" appears in df.2[1,] as "b","a"). That doesn't matter to me. In other words, as long as a row pair in df.1 appears in any order in df.2 I want its row index in df.2, otherwise it should return NA. One more note, row pairs in both data frames are unique - meaning each row pair appears only once.
So for these two data frames the return vector would be:
c(1,4,NA,5,2,NA,3,NA,6,NA)
Maybe something using dplyr package:
first make the reference frame
use row_number() to number as per row index efficiently.
use select to "flip" the column vars.
two halves:
df_ref_top <- df.2 %>% mutate(n=row_number())
df_ref_btm <- df.2 %>% select(col.1=col.2, col.2=col.1) %>% mutate(n=row_number())
then bind together:
df_ref <- rbind(df_ref_top,df_ref_btm)
Left join and select vector:
gives to get your answer
left_join(df.1,df_ref)$n
# Per #thelatemail's comment, here's a more elegant approach:
match(apply(df.1,1,function(x) paste(sort(x),collapse="")),
apply(df.2,1,function(x) paste(sort(x),collapse="")))
# My original answer, for reference:
# Check for matches with both orderings of df.2's columns
match.tmp = cbind(match(paste(df.1[,1],df.1[,2]), paste(df.2[,1],df.2[,2])),
match(paste(df.1[,1],df.1[,2]), paste(df.2[,2],df.2[,1])))
# Convert to single vector of match indices
match.index = apply(match.tmp, 1,
function(x) ifelse(all(is.na(x)), NA, max(x, na.rm=TRUE)))
[1] 1 4 NA 5 2 NA 3 NA 6 NA
Here's a little function that tests a few of the looping options in R (which was not really intentional, but it happened).
check.rows <- function(data1, data2)
{
df1 <- as.matrix(data1);df2 <- as.matrix(data2);ll <- vector('list', nrow(df1))
for(i in seq(nrow(df1))){
ll[[i]] <- sapply(seq(nrow(df2)), function(j) df2[j,] %in% df1[i,])
}
h <- sapply(ll, function(x) which(apply(x, 2, all)))
sapply(h, function(x) ifelse(is.double(x), NA, x))
}
check.rows(df.1, df.2)
## [1] 1 4 NA 5 2 NA 3 NA 6 NA
And here's a benchmark when row dimensions are increased for both df.1 and df.2. Not too bad I guess, considering the 24 checks on each of 40 rows.
> dim(df.11); dim(df.22)
[1] 40 2
[1] 24 2
> f <- function() check.rows(df.11, df.22)
> microbenchmark(f())
## Unit: milliseconds
## expr min lq median uq max neval
## f() 75.52258 75.94061 76.96523 78.61594 81.00019 100
1) sort/merge First sort df.2 creating df.2.s and append a row number column. Then merge this new data frame with df.1 (which is already sorted in the question):
df.2.s <- replace(df.2, TRUE, t(apply(df.2, 1, sort)))
df.2.s$row <- 1:nrow(df.2.s)
merge(df.1, df.2.s, all.x = TRUE)$row
The result is:
[1] 1 4 NA 5 2 NA 3 NA 6 NA
2) sqldf Since dot is an SQL operator rename the data frames as df1 and df2. Note that for the same reason the column names will be transformed to col_1 and col_2 when df1 and df2 are automatically uploaded to the backend database. We sort df2 using min and max and left join it to df1 (which is already sorted):
df1 <- df.1
df2 <- df.2
library(sqldf)
sqldf("select b.rowid row
from df1
left join
(select min(col_1, col_2) col_1, max(col_1, col_2) col_2 from df2) b
using (col_1, col_2)")$row
REVISED Some code improvements. Added second solution.

Fill in data frame with values from rows above

Say I have a data frame like this:
ID, ID_2, FIRST, VALUE
-----------------------
'a', 'aa', TRUE, 2
'a', 'ab', FALSE, NA
'a', 'ac', FALSE, NA
'b', 'aa', TRUE, 5
'b', 'ab', FALSE, NA
So VALUE is only set for FIRST = TRUE once per ID. ID_2 may be duplicate between IDs, but doesn't have to.
How do I put the numbers from the first rows of each ID into all rows of that ID, such that the VALUE column becomes 2, 2, 2, 5, 5?
I know I could simply loop over all IDs with a for loop, but I am looking for a more efficient way.
The question asks for efficiency compared with a loop. Here is a comparison of four solutions:
zoo::na.locf, which introduces a package dependency, and although it handles many edge cases, requires that the 'blank' values are NA. The other solutions are easily adapted to non-NA blanks.
A simple loop in base R.
A recursive function in base R.
My own vectorised solution in base R.
The new fill() function in tidyr version 0.3.0., which works on data.frames.
Note that most of these solutions are for vectors, not data frames, so they don't check any ID column. If the data frame isn't grouped by ID, with the value to be filled down being at the top of each group, then you could try a windowing function in dplyr or data.table
# A popular solution
f1 <- zoo::na.locf
# A loop, adapted from https://stat.ethz.ch/pipermail/r-help/2008-July/169199.html
f2 <- function(x) {
for(i in seq_along(x)[-1]) if(is.na(x[i])) x[i] <- x[i-1]
x
}
# Recursion, also from https://stat.ethz.ch/pipermail/r-help/2008-July/169199.html
f3 <- function(z) {
y <- c(NA, head(z, -1))
z <- ifelse(is.na(z), y, z)
if (any(is.na(z))) Recall(z) else z }
# My own effort
f4 <- function(x, blank = is.na) {
# Find the values
if (is.function(blank)) {
isnotblank <- !blank(x)
} else {
isnotblank <- x != blank
}
# Fill down
x[which(isnotblank)][cumsum(isnotblank)]
}
# fill() from the `tidyr` version 0.3.0
library(tidyr)
f5 <- function(y) {
fill(y, column)
}
# Test data, 2600 values, ~58% blanks
x <- rep(LETTERS, 100)
set.seed(2015-09-12)
x[sample(1:2600, 1500)] <- NA
x <- c("A", x) # Ensure the first element is not blank
y <- data.frame(column = x, stringsAsFactors = FALSE) # data.frame version of x for tidyr
# Check that they all work (they do)
identical(f1(x), f2(x))
identical(f1(x), f3(x))
identical(f1(x), f4(x))
identical(f1(x), f5(y)$column)
library(microbenchmark)
microbenchmark(f1(x), f2(x), f3(x), f4(x), f5(y))
Results:
Unit: microseconds
expr min lq mean median uq max neval
f1(x) 422.762 466.6355 508.57284 505.6760 527.2540 837.626 100
f2(x) 2118.914 2206.7370 2501.04597 2312.8000 2497.2285 5377.018 100
f3(x) 7800.509 7832.0130 8127.06761 7882.7010 8395.3725 14128.107 100
f4(x) 52.841 58.7645 63.98657 62.1410 65.2655 104.886 100
f5(y) 183.494 225.9380 305.21337 331.0035 350.4040 529.064 100
If you need only to carry forward the values from the VALUE column, then I think you can use na.lofc() function from zoo package. Here is an example:
a<-c(1,NA,NA,2,NA)
na.locf(a)
[1] 1 1 1 2 2
If the VALUE for a specific ID always appears in the first record, which seems to be the case for your data, you can use match to find that record:
df <- read.csv(textConnection("
ID, ID_2, FIRST, VALUE
'a', 'aa', TRUE, 2
'a', 'ab', FALSE, NA
'a', 'ac', FALSE, NA
'b', 'aa', TRUE, 5
'b', 'ab', FALSE, NA
"))
df$VALUE <- df$VALUE[match(df$ID, df$ID)]
df
# ID ID_2 FIRST VALUE
# 1 'a' 'aa' TRUE 2
# 2 'a' 'ab' FALSE 2
# 3 'a' 'ac' FALSE 2
# 4 'b' 'aa' TRUE 5
# 5 'b' 'ab' FALSE 5
+1 for #nacnudus
Handles leading blanks
f4 <- function(x, blank = is.na) {
# Find the values
if (is.function(blank)) {
isnotblank <- !blank(x)
} else {
isnotblank <- x != blank
}
# Fill down
xfill <- cumsum(isnotblank)
xfill[ xfill == 0 ] <- NA
# Replace Blanks
xnew <- x[ which(isnotblank) ][ xfill ]
xnew[is.na(xnew)] <- blank
return(xnew)
}

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