Altering a large distance matrix to be just three columns - r

I have a large data frame/.csv that is a matrix with 42 columns and 110,357,407. It was derived from the x and y coordinates for two datasets of points, one with 41 and another with 110,357,407 and the values of the rows represent the distances between these two sets of points (the distance of each point on list 1 to every single point on list 2). The first column is a list of points (from 1 to 110,357,407). An excerpt from the matrix is below.
V1 V2 V3 V4 V5 V6 V7
1 38517.05 38717.8 38840.16 38961.37 39281.06 88551.03 88422.62
2 38514.05 38714.79 38837.15 38958.34 39278 88545.48 88417.09
3 38511.05 38711.79 38834.14 38955.3 39274.94 88539.92 88411.56
4 38508.05 38708.78 38831.13 38952.27 39271.88 88534.37 88406.03
5 38505.06 38705.78 38828.12 38949.24 39268.83 88528.82 88400.5
6 38502.07 38702.78 38825.12 38946.21 39265.78 88523.27 88394.97
7 38499.08 38699.78 38822.12 38943.18 39262.73 88517.72 88389.44
8 38496.09 38696.79 38819.12 38940.15 39259.68 88512.17 88383.91
9 38493.1 38693.8 38816.12 38937.13 39256.63 88506.62 88378.38
10 38490.12 38690.8 38813.12 38934.11 39253.58 88501.07 88372.85
11 38487.14 38687.81 38810.13 38931.09 39250.54 88495.52 88367.33
12 38484.16 38684.83 38807.14 38928.07 39247.5 88489.98 88361.8
13 38481.18 38681.84 38804.15 38925.06 39244.46 88484.43 88356.28
14 38478.21 38678.86 38801.16 38922.04 39241.43 88478.88 88350.75
15 38475.23 38675.88 38798.17 38919.03 39238.39 88473.34 88345.23
16 38472.26 38672.9 38795.19 38916.03 39235.36 88467.8 88339.71
My issue is that I would like to change this matrix into just 3 columns, the first column would be similar to the first column of the matrix with the 110,357,407 rows, the second would be the 41 data points (each matched up with a distance each of the first points to all of the others) and the third would be the distance between those points. So it would look something like this
Back Pres Dist
1 1 3486
2 1 3456
3 1 3483
4 1 3456
5 1 3429
6 1 3438
7 1 3422
8 1 3427
9 1 3428
(After the distances between the back and all of the first value of pres are complete, pres will change to 2 and will eventually work its way up to 41)
I realize that this will output a hugely ridiculous number of rows, but this is the format that I need to run some processes that are outside of R.
I tried using this code
cols.Output <- data.frame(col = rep(colnames(output3), each = nrow(output3)),
row = rep(rownames(output3), ncol(output3)),
value = as.vector(output3))
But there won’t be the same number of rows for each column, so I received an error (and I don’t think it would have really worked with my pres column needs). I tried experimenting with some of the rbind.fill and cbind.fill functions (the one in plyr and ones that others have come up with in the forum). I also looked into some of the melting and reshaping but I was very confused about the functions and couldn’t figure out how to implement them appropriately (or if they even are appropriate for what I need). I would really appreciate any help on this as I’ve been struggling with it for a long time.
Edit: Just to be a little more clear about what I need. Take these two smaller data sets
back <- 1 dataset with 5 sets of x, y points
pres <- 1 dataset with 3 sets of x, y points
Calculating distances between these two data frames generates the initial matrix:
Back 1 2 3
1 3427 3444 3451
2 3432 3486 3476
3 3486 3479 3486
4 3449 3438 3484
5 3483 3486 3486
And my desired output would look like this:
Back Pres Dist
1 1 3427
2 1 3432
3 1 3486
4 1 3449
5 1 3483
1 2 3444
2 2 3486
3 2 3479
4 2 3438
5 2 3486
1 3 3451
2 3 3476
3 3 3486
4 3 3484
5 3 3486

Yes, it looks this is the kind of problem generally solved with some combination of melt and cast in the reshape2 package. That said, with 100+ million rows, I'm not sure that that's the most efficient way to go in this case.
You could do it all manually as follows. I'll assume your data frame is called df, and the distances are in columns 2 to 42. See if this works.
d <- unlist(df[-1]) # put all the distances into a vector
newdf <- cbind(expand.grid(back=seq_len(nrow(df)), pres=seq_len(ncol(df) - 1)), d)
This will probably die unless you have tons of memory. The same holds for any simple solution though, since you have > 4.2 billion elements in the vector of distances. You can work on subsets of the full dataset at a time to get around this problem.

Here's how to use melt on a small example:
require(reshape2)
a <- matrix(rnorm(9), nrow = 3)
a[, 1] <- 1:3 ## Pretending these are one set of points
rownames(a) <- a[, 1] ## We'll put them as rownames instead of a column
melt(a[, -1]) ## And omit that column when melting
If you have memory issues, you could write a for loop and do it in pieces, writing each to a file when they're completed.

Related

Referencing different coloumn as ranges between two data frames

I have one data frame/ list that gives and ID and a number
1. 25
2. 36
3. 10
4. 18
5. 12
This first list is effectively a list of objects with the number of objects contained in each eg. bricks in a wall, so a a list or walls with the number of bricks in each.
I have a second that contains a a full list of the objects being referred to in that above list and a second attribute for each.
1. 3
2. 4
3. 2
4. 8
5. 5
etc.
in the weak example I'm stringing together this would be a list of the weight of each brick in all walls.
so my first list give me the ranges i would like to average in the second list, or I would like as an end result a list of walls with the average weight of each brick per wall.
ie average the attributes of 1-25, 26-62 ... 89-101
my idea so far was to create a data frame with two coloumns
1. 1 25
2. 26 62
3. n
4. n
5. 89 101
and then attempt to create a third column that uses the first two as x and y in a mean(table2$coloumn1[x:y]) type formula, but I can't get anything to work.
the end result could probably looks something like this
1. 3.2
2. 6.5
3. 3
4. 7.9
5. 8.5
is there a way to do it like this or does anyone have a more elegant solution.
You could do something like this... set the low and high limits of your ranges and then use mapply to work out the mean over the appropriate rows of df2.
df1 <- data.frame(id=c(1,2,3,4,5),no=c(25,36,10,18,12))
df2 <- data.frame(obj=1:100,att=sample(1:10,100,replace=TRUE))
df1$low <- cumsum(c(1,df1$no[-nrow(df1)]))
df1$high <- pmin(cumsum(df1$no),nrow(df2))
df1$meanatt <- mapply(function(l,h) mean(df2$att[l:h]), df1$low, df1$high)
df1
id no low high meanatt
1 1 25 1 25 4.760000
2 2 36 26 61 5.527778
3 3 10 62 71 5.800000
4 4 18 72 89 5.111111
5 5 12 90 100 4.454545

Search for value within a range of values in two separate vectors

This is my first time posting to Stack Exchange, my apologies as I'm certain I will make a few mistakes. I am trying to assess false detections in a dataset.
I have one data frame with "true" detections
truth=
ID Start Stop SNR
1 213466 213468 10.08
2 32238 32240 10.28
3 218934 218936 12.02
4 222774 222776 11.4
5 68137 68139 10.99
And another data frame with a list of times, that represent possible 'real' detections
possible=
ID Times
1 32239.76
2 32241.14
3 68138.72
4 111233.93
5 128395.28
6 146180.31
7 188433.35
8 198714.7
I am trying to see if the values in my 'possible' data frame lies between the start and stop values. If so I'd like to create a third column in possible called "between" and a column in the "truth" data frame called "match. For every value from possible that falls between I'd like a 1, otherwise a 0. For all of the rows in "truth" that find a match I'd like a 1, otherwise a 0.
Neither ID, not SNR are important. I'm not looking to match on ID. Instead I wand to run through the data frame entirely. Output should look something like:
ID Times Between
1 32239.76 0
2 32241.14 1
3 68138.72 0
4 111233.93 0
5 128395.28 0
6 146180.31 1
7 188433.35 0
8 198714.7 0
Alternatively, knowing if any of my 'possible' time values fall within 2 seconds of start or end times would also do the trick (also with 1/0 outputs)
(Thanks for the feedback on the original post)
Thanks in advance for your patience with me as I navigate this system.
I think this can be conceptulised as a rolling join in data.table. Take this simplified example:
truth
# id start stop
#1: 1 1 5
#2: 2 7 10
#3: 3 12 15
#4: 4 17 20
#5: 5 22 26
possible
# id times
#1: 1 3
#2: 2 11
#3: 3 13
#4: 4 28
setDT(truth)
setDT(possible)
melt(truth, measure.vars=c("start","stop"), value.name="times")[
possible, on="times", roll=TRUE
][, .(id=i.id, truthid=id, times, status=factor(variable, labels=c("in","out")))]
# id truthid times status
#1: 1 1 3 in
#2: 2 2 11 out
#3: 3 3 13 in
#4: 4 5 28 out
The source datasets were:
truth <- read.table(text="id start stop
1 1 5
2 7 10
3 12 15
4 17 20
5 22 26", header=TRUE)
possible <- read.table(text="id times
1 3
2 11
3 13
4 28", header=TRUE)
I'll post a solution that I'm pretty sure works like you want it to in order to get you started. Maybe someone else can post a more efficient answer.
Anyway, first I needed to generate some example data - next time please provide this from your own data set in your post using the function dput(head(truth, n = 25)) and dput(head(possible, n = 25)). I used:
#generate random test data
set.seed(7)
truth <- data.frame(c(1:100),
c(sample(5:20, size = 100, replace = T)),
c(sample(21:50, size = 100, replace = T)))
possible <- data.frame(c(sample(1:15, size = 15, replace = F)))
colnames(possible) <- "Times"
After getting sample data to work with; the following solution provides what I believe you are asking for. This should scale directly to your own dataset as it seems to be laid out. Respond below if the comments are unclear.
#need the %between% operator
library(data.table)
#initialize vectors - 0 or false by default
truth.match <- c(rep(0, times = nrow(truth)))
possible.between <- c(rep(0, times = nrow(possible)))
#iterate through 'possible' dataframe
for (i in 1:nrow(possible)){
#get boolean vector to show if any of the 'truth' rows are a 'match'
match.vec <- apply(truth[, 2:3],
MARGIN = 1,
FUN = function(x) {possible$Times[i] %between% x})
#if any are true then update the match and between vectors
if(any(match.vec)){
truth.match[match.vec] <- 1
possible.between[i] <- 1
}
}
#i think this should be called anyMatch for clarity
truth$anyMatch <- truth.match
#similarly; betweenAny
possible$betweenAny <- possible.between

Replace values in one data frame from values in another data frame

I need to change individual identifiers that are currently alphabetical to numerical. I have created a data frame where each alphabetical identifier is associated with a number
individuals num.individuals (g4)
1 ZYO 64
2 KAO 24
3 MKU 32
4 SAG 42
What I need to replace ZYO with the number 64 in my main data frame (g3) and like wise for all the other codes.
My main data frame (g3) looks like this
SAG YOG GOG BES ATR ALI COC CEL DUN EVA END GAR HAR HUX ISH INO JUL
1 2
2 2 EVA
3 SAG 2 EVA
4 2
5 SAG 2
6 2
Now on a small scale I can write a code to change it like I did with ATR
g3$ATR <- as.character(g3$ATR)
g3[g3$target == "ATR" | g3$ATR == "ATR","ATR"] <- 2
But this is time consuming and increased chance of human error.
I know there are ways to do this on a broad scale with NAs
I think maybe we could do a for loop for this, but I am not good enough to write one myself.
I have also been trying to use this function which I feel like may work but I am not sure how to logically build this argument, it was posted on the questions board here
Fast replacing values in dataframe in R
df <- as.data.frame(lapply(df, function(x){replace(x, x <0,0)})
I have tried to work my data into this by
df <- as.data.frame(lapply(g4, function(g3){replace(x, x <0,0)})
Here is one approach using the data.table package:
First, create a reproducible example similar to your data:
require(data.table)
ref <- data.table(individuals=1:4,num.individuals=c("ZYO","KAO","MKU","SAG"),g4=c(64,24,32,42))
g3 <- data.table(SAG=c("","SAG","","SAG"),KAO=c("KAO","KAO","",""))
Here is the ref table:
individuals num.individuals g4
1: 1 ZYO 64
2: 2 KAO 24
3: 3 MKU 32
4: 4 SAG 42
And here is your g3 table:
SAG KAO
1: KAO
2: SAG KAO
3:
4: SAG
And now we do our find and replacing:
g3[ , lapply(.SD,function(x) ref$g4[chmatch(x,ref$num.individuals)])]
And the final result:
SAG KAO
1: NA 24
2: 42 24
3: NA NA
4: 42 NA
And if you need more speed, the fastmatch package might help with their fmatch function:
require(fastmatch)
g3[ , lapply(.SD,function(x) ref$g4[fmatch(x,ref$num.individuals)])]
SAG KAO
1: NA 24
2: 42 24
3: NA NA
4: 42 NA

Re-sample a data frame with panel dimension

I have a data set consisting of 2000 individuals. For each individual, i:2000 , the data set contains n repeated situations. Letting d denote this data set, each row of dis indexed by i and n. Among other variables, d has a variable pid which takes on identical value for an individual across different (situations) rows.
Taking into consideration the panel nature of the data, I want to re-sample d (as in bootstrap):
with replacement,
store each re-sample data as a data frame
I considered using the sample function but could not make it work. I am a new user of r and have no programming skills.
The data set consists of many variables, but all the variables have numeric values. The data set is as follows.
pid x y z
1 10 2 -5
1 12 3 -4.5
1 14 4 -4
1 16 5 -3.5
1 18 6 -3
1 20 7 -2.5
2 22 8 -2
2 24 9 -1.5
2 26 10 -1
2 28 11 -0.5
2 30 12 0
2 32 13 0.5
The first six rows are for the first person, for which pid=1, and the next sex rows, pid=2 are different observations for the second person.
This should work for you:
z <- replicate(100,
d[d$pid %in% sample(unique(d$pid), 2000, replace=TRUE),],
simplify = FALSE)
The result z will be a list of dataframes you can do whatever with.
EDIT: this is a little wordy, but will deal with duplicated rows. replicate has its obvious use of performing a set operation a given number of times (in the example below, 4). I then sample the unique values of pid (in this case 3 of those values, with replacement) and extract the rows of d corresponding to each sampled value. The combination of a do.call to rbind and lapply deal with the duplicates that are not handled well by the above code. Thus, instead of generating dataframes with potentially different lengths, this code generates a dataframe for each sampled pid and then uses do.call("rbind",...) to stick them back together within each iteration of replicate.
z <- replicate(4, do.call("rbind", lapply(sample(unique(d$pid),3,replace=TRUE),
function(x) d[d$pid==x,])),
simplify=FALSE)

how to select matrix element in R?

Reading the data the following way
data<-read.csv("userStats.csv", sep=",", header=F)
I tried to select an element at the specific position.
The example of the data (first five rows) is the following (V2 is the date and V3 is the day of week):
V1 V2
1 00002781A2ADA816CDB0D138146BD63323CCDAB2 2010-09-04
2 00002D2354C7080C0868CB0E18C46157CA9F0FD4 2010-09-04
3 00002D2354C7080C0868CB0E18C46157CA9F0FD4 2010-09-07
4 00002D2354C7080C0868CB0E18C46157CA9F0FD4 2010-09-08
5 00002D2354C7080C0868CB0E18C46157CA9F0FD4 2010-09-17
V3 V4 V5 V6 V7 V8 V9
1 Saturday 2 2 615 1 1 47
2 Saturday 2 2 77 1 1 43
3 Tuesday 1 3 201 1 1 117
4 Wednesday 1 1 44 1 1 74
5 Friday 1 1 3 1 1 18
I tried to divide 6th column with 9th column in the first row the following way:
data[1,6]/data[1,9]
but it returned an error
[1] NA
Warning message:
In Ops.factor(data[1, 6], data[1, 9]) : / not meaningful for factors
Then I tried to select just one element
> data[2,9]
[1] 43
11685 Levels: 0 1 2 3 ... 55311
but don't know what these Levels are and what causes an error. Does anyone know how to select an element at the specific position data[row, column]?
Thank you!
My favorite tool to check variable class is str().
What you have there is a data frame and at least one of the columns you're trying to work with is a factor. See Dirk's answer on how to change classes of a column.
Command
data[1,6]/data[1,9]
is selecting the value in the first row of sixth column and dividing with the value in first row of the ninth column. Is this what you want? If you want to use values from the entire column (and not just the first row), you would write
data[6] / data[9]
or
data[, 6] / data[, 9]
Both arguments are equivalent for data.frames.
The standard modeling data structure in R is a data.frame.
The data.frame objects can hold various types: numeric, character, factor, ...
Now, when reading data via read.csv() et al, you can get bitten by the default valus of the stringsAsFactors option. I presume that at least a row in your data had text, so R decides to decode it as a factor and presto! you no longer can do direct mathematical operations on the column.
In short, do summary(data) and/or a sweep of class() over all the columns. Convert as necessary, or turn the stringsAsFactors variable to a different value or both.
Once your data is numeric, you can divide, slice, dice, ... as you please.

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