I have an object in "R" called p_int. This is a list of 1599 peak intensity numbers.
Within every 8 values of this list is a monoisotopic peak. This peak is the most abundant (largest peak value) compared to the other 7 peaks.
Therefore what I'd like to do is write a loop which processes p_int in batches of 8.
So it will take the first 8 values, find the largest value and add this to a new object called "m_iso".
It will then continue, looking at values 9-16, 17-24, 25-32 etc.
Any advice or code in helping me achieve such a loop would be greatly appreciated.
Thanks,
Stephen.
By 1599 do you actually mean 1600? Because 1599 is not evenly divisible by 8. I'm going to assume this is true and offer the following:
m_iso <- sapply(split(p_int,rep(1:200,each=8)),max)
Or:
m_iso <- apply(matrix(p_int,nrow=8),2,max)
This will give you a vector of maximum values for each set of eight observations.
Related
Apparently if I try this:
# first grab the package
install.packages("stringi")
library(stringi)
# and then try to generate some serious dummy data
my_try <- as.vector(sample(1111111111:99999999999,3000000,replace=T))
R will say NOPE, sorry:
Error: cannot allocate vector of size 736.8 Gb
Should I buy more RAM*?
*this is a joke, but I seriously appreciate any help!
EDIT:
The desired output is a dataframe of 20 variables, and 3x10^6 rows. Some columns/variables should be strings, some integers. All in lengths ranging from 2 to 12.
The error isn't coming from sampling 3 million values, it's from trying to create a population of about 90 billion values 1111111111:99999999999 from which to sample. If you want to sample from that range, sample from the range 1:88888888889 and add 11111111110 using
sample(88888888889, 3000000,replace=TRUE) + 11111111110
There's no need for as.vector at the end, it's already a vector.
P.S. I believe in R-devel the range 1111111111:99999999999 will be stored much more efficiently (basically just the limits), but I don't know if sample() will be modified to work with it that way.
I am trying to count the length of occurrances of a value in a vector such as
q <- c(1,1,1,1,1,1,4,4,4,4,4,4,4,4,4,4,4,4,6,6,6,6,6,6,6,6,6,6,1,1,4,4,4)
Actual vectors are longer than this, and are time based. What I would like would be an output for 4 that tells me it occurred for 12 time steps (before the vector changes to 6) and then 3 time steps. (Not that it occurred 15 times total).
Currently my ideas to do this are pretty inefficient (a loop that looks element by element that I can have stop when it doesn't equal the value I specified). Can anyone recommend a more efficient method?
x <- with(rle(q), data.frame(values, lengths)) will pull the information that you want (courtesy of d.b. in the comments).
From the R Documentation: rle is used to "Compute the lengths and values of runs of equal values in a vector – or the reverse operation."
y <- x[x$values == 4, ] will subset the data frame to include only the value of interest (4). You can then see clearly that 4 ran for 12 times and then later for 3.
Modifying the code will let you check whatever value you want.
I think I have a rather simple problem but I can't figure out the best approach. I have a vector with 30 different values. Now I need to divide the vector into 10 groups in such a way that the mean within group variance is as small as possible. the size of the groups is not important, it can anything between one and 21.
Example. Let's say I have vector of six values, that I have to split into three groups:
Myvector <- c(0.88,0.79,0.78,0.62,0.60,0.58)
Obviously the solution would be:
Group1 <-c(0.88)
Group2 <-c(0.79,0.78)
Group3 <-c(0.62,0.60,0.58)
Is there a function that gives the same outcome as the example and that I can use for my vector withe 30 values?
Many thanks in advance.
It sounds like you want to do k-means clustering. Something like this would work
kmeans(Myvector,3, algo="Lloyd")
Note that I changed the default algorithm to match your desired output. If you read the ?kmeans help page you will see that there are different algorithms to calculate the different clusters because it's not a trivial computational problem. They might necessarily guarantee optimality.
I have a list like this:
A B score
B C score
A C score
......
where the first two columns contain the variable name and third column contains the score between both. Total number of variables is 250,000 (A,B,C....). And the score is a float [0,1]. The file is approximately 50 GB. And the pairs of A,B where scores are 1, have been removed as more than half the entries were 1.
I wanted to perform hierarchical clustering on the data.
Should I convert the linear form to a matrix with 250,000 rows and 250,000 columns? Or should I partition the data and do the clustering?
I'm clueless with this. Please help!
Thanks.
Your input data already is the matrix.
However hierarchical clustering usually scales O(n^3). That won't work with your data sets size. Plus, they usually need more than one copy of the matrix. You may need 1TB of RAM then... 2*8*250000*250000is a lot.
Some special cases can run in O(n^2): SLINK does. If your data is nicely sorted, it should be possible to run single-link in a single pass over your file. But you will have to implement this yourself. Don't even think of using R or something fancy.
I have a number of fishing boat tracks, and I'm trying to detect a certain pattern in their movement using R. In doing so I have reached a point where I have discarded all points of the track where the desired pattern is not occurring within a given time window, and I'm left with the remaining georeferenced points. These points have a score value associated, which measures the 'intensity' of the desired pattern.
track_1[1:10,]:
LAT LON SCORE
1 32.34855 -35.49264 80.67
2 31.54764 -35.58691 18.14
3 31.38293 -35.25243 46.70
4 31.21447 -35.25830 22.65
5 30.76365 -35.38881 11.93
6 30.75872 -35.54733 22.97
7 30.60261 -35.95472 35.98
8 30.62818 -36.27024 31.09
9 31.35912 -35.73573 14.97
10 31.15218 -36.38027 37.60
The code bellow provides the same data
data.frame(cbind(
LAT=c(32.34855,31.54764,31.38293,31.21447,30.76365,30.75872,30.60261,30.62818,31.35912,31.15218),
LON=c(-35.49264,-35.58691,-35.25243,-35.25830,-35.38881,-35.54733,-35.95472,-36.27024,-35.73573,-36.38027),
SCORE=c(80.67,18.14,46.70,22.65,11.93,22.97,35.98,31.09,14.97,37.60)))
Because some of these points occur geographically close to each other I need to 'pool' their scores together. Hence, I now need a way to throw this data into some kind of a spatial grid and cumulatively sum the scores of all points that fall in the same cell of the grid. This would allow me to find in what areas a given fishing boat exhibits the pattern I'm after the most (and this is not just about time spent in one place). Ultimately, the preferred output would contain lat and lon for every grid cell (center), and the sum of all scores on each cell. In addition, I would also like to be able to adjust the sizing of the grid cells.
I've looked around and all I can find either does not preserve the georeferenced information, is very inefficient, or performs binning of data. There may already be some answers out there, but it might be the case that I'm not able to recognize them since I'm a bit out of my league on this stuff. Can someone please point me to some direction (package, function, etc.)? Any guidance will be greatly appreciated.
Take your lat/lon coordinates, and multiply them by the inverse of your desired grid cell edge lengths, measured in degrees. The result will be a pair of floating point numbers whose integer part identifies the grid cell in question. Take the floor of these and you have two numbers describing the cell, which you could paste to form a single string. You may add that as a new factor column of your data frame. Then you can perform operations based on that factor, like summarizing values.
Example:
latScale <- 2 # one cell for every 0.5 degrees
lonScale <- 2 # likewise
track_1$cell <- factor(with(track_1,
paste(floor(LAT*latScale), floor(LON*lonScale), sep='.')))
library(plyr)
ddply(track_1, .(cell), summarize,
LAT=mean(LAT), LON=mean(LON), SCORE=sum(SCORE))
If you want to, you can use weighted.mean instead of mean. If you don't like these factors, you can put more effort in making them nice (e.g. by using compass directions instead of signs), or drop them altogether and use a pair of integer columns instead.