It seems a silly question, but I have searched on line, but still did not find any sufficient reply.
My question is: suppose we have a matrix M, then we use the scale() function, how can we extract the center and scale of each column by writing a line of code (I know we can see the centers and scales..), but my matrix has lots of columns, it is cumbersome to do it manually.
Any ideas? Many thanks!
you are looking for the attributes function:
set.seed(1)
mat = matrix(rnorm(1000),,10) # Suppose you have 10 columns
s = scale(mat) # scale your data
attributes(s)#This gives you the means and the standard deviations:
$`dim`
[1] 100 10
$`scaled:center`
[1] 0.1088873669 -0.0378080766 0.0296735350 0.0516018586 -0.0391342406 -0.0445193567 -0.1995797418
[8] 0.0002549694 0.0100772648 0.0040650015
$`scaled:scale`
[1] 0.8981994 0.9578791 1.0342655 0.9916751 1.1696122 0.9661804 1.0808358 1.0973012 1.0883612 1.0548091
These values can also be obtained as:
colMeans(mat)
[1] 0.1088873669 -0.0378080766 0.0296735350 0.0516018586 -0.0391342406 -0.0445193567 -0.1995797418
[8] 0.0002549694 0.0100772648 0.0040650015
sqrt(diag(var(mat)))
[1] 0.8981994 0.9578791 1.0342655 0.9916751 1.1696122 0.9661804 1.0808358 1.0973012 1.0883612 1.0548091
you get a list that you can subset the way you want:
or you can do
attr(s,"scaled:center")
[1] 0.1088873669 -0.0378080766 0.0296735350 0.0516018586 -0.0391342406 -0.0445193567 -0.1995797418
[8] 0.0002549694 0.0100772648 0.0040650015
attr(s,"scaled:scale")
[1] 0.8981994 0.9578791 1.0342655 0.9916751 1.1696122 0.9661804 1.0808358 1.0973012 1.0883612 1.0548091
I have a vector such as this; (1X2406)
head(lnreturn)
[1] NA 0.004002188 0.003262646 -0.009454616 0.001460387
[6] 0.004005103
I would like to insert an NA as a first element so that I could reach a vector like this:
[1] NA NA 0.004002188 0.003262646 -0.009454616
[6] 0.001460387
Hence, I would get a vector in (1X2407) dimension.
Just use c()
x<-rnorm(10)
x<-c(NA,x)
x
[1] NA -0.004620768 0.760242168 0.038990913 0.735072142 -0.146472627
[7] -0.057887335 0.482369466 0.992943637 -1.246395498 -0.033487525
its easy (like etienne posted)
if you want a vector with same length as a result (like in your example) you can use length().
x<-rnorm(10)
x<-c(NA,x)[1:length(x)]
I want to create a vector of names that act as variable names so I can then use themlater on in a loop.
years=1950:2012
for(i in 1:length(years))
{
varname[i]=paste("mydata",years[i],sep="")
}
this gives:
> [1] "mydata1950" "mydata1951" "mydata1952" "mydata1953" "mydata1954" "mydata1955" "mydata1956" "mydata1957" "mydata1958"
[10] "mydata1959" "mydata1960" "mydata1961" "mydata1962" "mydata1963" "mydata1964" "mydata1965" "mydata1966" "mydata1967"
[19] "mydata1968" "mydata1969" "mydata1970" "mydata1971" "mydata1972" "mydata1973" "mydata1974" "mydata1975" "mydata1976"
[28] "mydata1977" "mydata1978" "mydata1979" "mydata1980" "mydata1981" "mydata1982" "mydata1983" "mydata1984" "mydata1985"
[37] "mydata1986" "mydata1987" "mydata1988" "mydata1989" "mydata1990" "mydata1991" "mydata1992" "mydata1993" "mydata1994"
[46] "mydata1995" "mydata1996" "mydata1997" "mydata1998" "mydata1999" "mydata2000" "mydata2001" "mydata2002" "mydata2003"
[55] "mydata2004" "mydata2005" "mydata2006" "mydata2007" "mydata2008" "mydata2009" "mydata2010" "mydata2011" "mydata2012"
All I want to do is remove the quotes and be able to call each value individually.
I want:
>[1] mydata1950 mydata1951 mydata1952 mydata1953, #etc...
stored as a variable such that
varname[1]
> mydata1950
varname[2]
> mydata1951
and so on.
I have played around with
cat(varname[i],"\n")
but this just prints values as one line and I can't call each individual string. And
gsub("'",'',varname)
but this doesn't seem to do anything.
Suggestions? Is this possible in R? Thank you.
There are no quotes in that character vector's values. Use:
cat(varname)
.... if you want to see the unquoted values. The R print mechanism is set to use quotes as a signal to your brain that distinct values are present. You can also use:
print(varname, quote=FALSE)
If there are that many named objects in you workspace, then you need desperately to learn to use lists. There are mechanisms for "promoting" character values to names, but this would be seen as a failure on your part to learn to use the language effectively:
var <- 2
> eval(as.name('var'))
[1] 2
> eval(parse(text="var"))
[1] 2
> get('var')
[1] 2
sampleFiles <- list.files(path="/path",pattern="*.txt");
> sampleFiles
[1] "D104.txt" "D121.txt" "D153.txt" "D155.txt" "D161.txt" "D162.txt" "D167.txt"
[8] "D173.txt" "D176.txt" "D177.txt" "D179.txt" "D204.txt" "D221.txt" "D253.txt"
[15] "D255.txt" "D261.txt" "D262.txt" "D267.txt" "D273.txt" "D276.txt" "D277.txt"
[22] "D279.txt" "N101.txt" "N108.txt" "N113.txt" "N170.txt" "N171.txt" "N172.txt"
[29] "N175.txt" "N181.txt" "N182.txt" "N183.txt" "N186.txt" "N187.txt" "N188.txt"
[36] "N201.txt" "N208.txt" "N213.txt" "N270.txt" "N271.txt" "N272.txt" "N275.txt"
[43] "N281.txt" "N282.txt" "N283.txt" "N286.txt" "N287.txt" "N288.txt"
How can I get all started with "N" first and "D" last? In other words swap them.
If you want to sort by letter (N, D) and number (101, ..) you could -just- swap your elements:
#random vector
vec <- c("D104.txt", "D121.txt", "D279.txt", "N101.txt", "N108.txt", "N113.txt")
#swap places
vec[c(grep("N", vec), grep("D", vec))]
[1] "N101.txt" "N108.txt" "N113.txt" "D104.txt" "D121.txt" "D279.txt"
grep finds what element of the vector has the pattern wanted. So, we move elements with "N" in front and with "D" in the back.
If you just want to sort with letters and numbers decreasing, you just (like Thomas suggested):
sort(vec, decreasing = T)
[1] "N113.txt" "N108.txt" "N101.txt" "D279.txt" "D121.txt" "D104.txt"
Also, since you know the indices of the elements you want to swap, then:
sampleFiles[c(23:48, 1:22)]
In this case it would be as simple as:
sampleFiles[c(23:48, 1:22)]
More general solutions have been suggested including, but sort(sampleFiles) would NOT succeed with "D" < "N". You could have used:
sampleFiles[rev(order(substr(sampleFiles, 1,1)))]
If you just used:
sampleFiles[rev(order(sampleFiles, 1,1))]
.. then the numeric values would get reversed as well. So you could have used chartr to swap them as the argument to order to selectively reverse the values of only "D" and "N":
sampleFiles[ order( chartr(c("DN"), c("ND"), sampleFiles) ) ]
I tried to find two values in the following vector, which are close to 10. The expected value is 10.12099196 and 10.63054170. Your inputs would be appreciated.
[1] 0.98799517 1.09055728 1.20383713 1.32927166 1.46857509 1.62380423 1.79743107 1.99241551 2.21226576 2.46106916 2.74346924 3.06455219 3.42958354 3.84350238 4.31005838
[16] 4.83051356 5.40199462 6.01590035 6.65715769 7.30532785 7.93823621 8.53773241 9.09570538 9.61755743 10.12099196 10.63018180 11.16783243 11.74870531 12.37719092 13.04922392
[31] 13.75661322 14.49087793 15.24414627 16.00601247 16.75709565 17.46236358 18.06882072 18.51050094 18.71908344 18.63563523 18.22123225 17.46709279 16.40246292 15.09417699 13.63404124
[46] 12.11854915 10.63054170 9.22947285 7.95056000 6.80923943 5.80717982 4.93764782 4.18947450 3.54966795 3.00499094 2.54283599 2.15165780 1.82114213 1.54222565 1.30703661
[61] 1.10879707 0.94170986 0.80084308 0.68201911 0.58171175 0.49695298 0.42525021 0.36451350 0.31299262 0.26922281 0.23197860 0.20023468 0.17313291 0.14995459 0.13009730
[76] 0.11305559 0.09840485 0.08578789 0.07490387 0.06549894 0.05735864
Another alternative could be allowing the user to control for the "tolerance" in order to set what "closeness" is, this can be done by using a simple function:
close <- function(x, value, tol=NULL){
if(!is.null(tol)){
x[abs(x-10) <= tol]
} else {
x[order(abs(x-10))]
}
}
Where x is a vector of values, value is the value of comparison for closeness, and tol is logical, if it's NULL it returns all the "close" values ordered by "closeness" to value, otherwise it returns just the values meeting the condition given in tol.
> close(x, value=10, tol=.7)
[1] 9.617557 10.120992 10.630182 10.630542
> close(x, value=10)
[1] 10.12099196 9.61755743 10.63018180 10.63054170 9.22947285 9.09570538 11.16783243
[8] 8.53773241 11.74870531 7.95056000 7.93823621 12.11854915 12.37719092 7.30532785
[15] 13.04922392 6.80923943 6.65715769 13.63404124 13.75661322 6.01590035 5.80717982
[22] 14.49087793 5.40199462 4.93764782 15.09417699 4.83051356 15.24414627 4.31005838
[29] 4.18947450 16.00601247 3.84350238 16.40246292 3.54966795 3.42958354 16.75709565
[36] 3.06455219 3.00499094 2.74346924 2.54283599 17.46236358 17.46709279 2.46106916
[43] 2.21226576 2.15165780 1.99241551 18.06882072 1.82114213 1.79743107 18.22123225
[50] 1.62380423 1.54222565 18.51050094 1.46857509 18.63563523 1.32927166 1.30703661
[57] 18.71908344 1.20383713 1.10879707 1.09055728 0.98799517 0.94170986 0.80084308
[64] 0.68201911 0.58171175 0.49695298 0.42525021 0.36451350 0.31299262 0.26922281
[71] 0.23197860 0.20023468 0.17313291 0.14995459 0.13009730 0.11305559 0.09840485
[78] 0.08578789 0.07490387 0.06549894 0.05735864
In the first example I defined "closeness" to be at most a difference of 0.7 between value and each elements in x. In the second example the function close returns a vector of values where the firsts are the closest to the value given in value and the lasts are the farest values from value.
Since my solution does not provide an easy (practical) way to find tol as #Arun pointed out, one way to find the closest values would be seting tol=NULL and asking for the exact number of close values as in:
> close(x, value=10)[1:3]
[1] 10.120992 9.617557 10.630182
This shows the three values in x closest to 10.
I can't think of a way without using sort. However, you can speed it up by using partial sort.
x[abs(x-10) %in% sort(abs(x-10), partial=1:2)[1:2]]
# [1] 9.617557 10.120992
In case the same values are present more than once, you'll get all of them here. So, you can either wrap this with unique or you can use match instead as follows:
x[match(sort(abs(x-10), partial=1:2)[1:2], abs(x-10))]
# [1] 10.120992 9.617557
dput output:
dput(x)
c(0.98799517, 1.09055728, 1.20383713, 1.32927166, 1.46857509,
1.62380423, 1.79743107, 1.99241551, 2.21226576, 2.46106916, 2.74346924,
3.06455219, 3.42958354, 3.84350238, 4.31005838, 4.83051356, 5.40199462,
6.01590035, 6.65715769, 7.30532785, 7.93823621, 8.53773241, 9.09570538,
9.61755743, 10.12099196, 10.6301818, 11.16783243, 11.74870531,
12.37719092, 13.04922392, 13.75661322, 14.49087793, 15.24414627,
16.00601247, 16.75709565, 17.46236358, 18.06882072, 18.51050094,
18.71908344, 18.63563523, 18.22123225, 17.46709279, 16.40246292,
15.09417699, 13.63404124, 12.11854915, 10.6305417, 9.22947285,
7.95056, 6.80923943, 5.80717982, 4.93764782, 4.1894745, 3.54966795,
3.00499094, 2.54283599, 2.1516578, 1.82114213, 1.54222565, 1.30703661,
1.10879707, 0.94170986, 0.80084308, 0.68201911, 0.58171175, 0.49695298,
0.42525021, 0.3645135, 0.31299262, 0.26922281, 0.2319786, 0.20023468,
0.17313291, 0.14995459, 0.1300973, 0.11305559, 0.09840485, 0.08578789,
0.07490387, 0.06549894, 0.05735864)
I'm not sure your question is clear, so here's another approach. To find the value closest to your first desired value, 10.12099196 , subtract that from the vector, take the absolute value, and then find the index of the closest element. Explicit:
delx <- abs( 10.12099196 - x)
min.index <- which.min(delx) #returns index of first minimum if there are duplicates
x[min.index] #gets you the value itself
Apologies if this was not the intent of your question.