I have some variables in my current R environment:
ls()
[1] "clt.list" "commands.list" "dirs.list" "eq" "hurs.list" "mlist" "prec.list" "temp.list" "vars"
[10] "vars.list" "wind.list"
where each one of the variables "clt.list", "hurs.list", "prec.list", "temp.list" and "wind.list" is a (huge) list of strings.
For example:
clt.list[1:20]
[1] "clt_Amon_ACCESS1-0_historical_r1i1p1_185001-200512.nc" "clt_Amon_ACCESS1-3_historical_r1i1p1_185001-200512.nc"
[3] "clt_Amon_bcc-csm1-1_historical_r1i1p1_185001-201212.nc" "clt_Amon_bcc-csm1-1-m_historical_r1i1p1_185001-201212.nc"
[5] "clt_Amon_BNU-ESM_historical_r1i1p1_185001-200512.nc" "clt_Amon_CanESM2_historical_r1i1p1_185001-200512.nc"
[7] "clt_Amon_CCSM4_historical_r1i1p1_185001-200512.nc" "clt_Amon_CESM1-BGC_historical_r1i1p1_185001-200512.nc"
[9] "clt_Amon_CESM1-CAM5_historical_r1i1p1_185001-200512.nc" "clt_Amon_CESM1-CAM5-1-FV2_historical_r1i1p1_185001-200512.nc"
[11] "clt_Amon_CESM1-FASTCHEM_historical_r1i1p1_185001-200512.nc" "clt_Amon_CESM1-WACCM_historical_r1i1p1_185001-200512.nc"
[13] "clt_Amon_CMCC-CESM_historical_r1i1p1_190001-190412.nc" "clt_Amon_CMCC-CESM_historical_r1i1p1_190001-200512.nc"
[15] "clt_Amon_CMCC-CESM_historical_r1i1p1_190501-190912.nc" "clt_Amon_CMCC-CESM_historical_r1i1p1_191001-191412.nc"
[17] "clt_Amon_CMCC-CESM_historical_r1i1p1_191501-191912.nc" "clt_Amon_CMCC-CESM_historical_r1i1p1_192001-192412.nc"
[19] "clt_Amon_CMCC-CESM_historical_r1i1p1_192501-192912.nc" "clt_Amon_CMCC-CESM_historical_r1i1p1_193001-193412.nc"
What I need to do is extract the subset of the string that is between "Amon_" and "_historical".
I can do this for a single variable, as shown here:
levels(as.factor(sub(".*?Amon_(.*?)_historical.*", "\\1", clt.list[1:20])))
[1] "ACCESS1-0" "ACCESS1-3" "bcc-csm1-1" "bcc-csm1-1-m" "BNU-ESM" "CanESM2" "CCSM4"
[8] "CESM1-BGC" "CESM1-CAM5" "CESM1-CAM5-1-FV2" "CESM1-FASTCHEM" "CESM1-WACCM" "CMCC-CESM"
However, what I'd like to do is to run the command above for all the five variables at once. Instead of using just "ctl.list" as argument in the command above, I'd like to use all variables "clt.list", "hurs.list", "prec.list", "temp.list" and "wind.list" at once.
How can I do that?
Many thanks in advance!
You can put your operation into a function and then iterate over it:
get_my_substr <- function(vecname)
levels(as.factor(sub(".*?Amon_(.*?)_historical.*", "\\1", get(vecname))))
lapply(my_vecnames,get_my_substr)
lapply acts like a loop. You can create your list of vector names with
my_vecnames <- ls(pattern=".list$")
It is generally good practice to post a reproducible example in your question. Since none was provided here, I tested this approach with...
# example-maker
prestr <- "grr_Amon_"
posstr <- "_historical_zzz"
make_ex <- function()
replicate(
sample(10,1),
paste0(prestr,paste0(sample(LETTERS,sample(5,1)),collapse=""),posstr)
)
# make a couple examples
set.seed(1)
m01 <- make_ex()
m02 <- make_ex()
# test result
lapply(ls(pattern="^m[0-9][0-9]$"),get_my_substr)
One solution would be to create a vector containing the variable names that you want extract the data from, for example:
var.names <- c("clt.list", "commands.list", "dirs.list")
Then to access the value of each variable from the name:
for (var.name in var.names) {
var.value <- as.list(environment())[[var.name]]
# Do something with var.value
}
Related
I got a list with a weird format:
[[1]]
[1] "Freq.2432.40862794099" "Freq.2792.87280096993" "Freq.2955.16577598796"
[4] "Freq.3161.12982491516" "Freq.3194.19720315405" "Freq.3218.83311568825"
[7] "Freq.3265.37951283662" "Freq.3317.86908506493" "Freq.3900.50408838719"
[10] "Freq.4073.33935633108" "Freq.4302.8830598659" "Freq.4404.80065271461"
[13] "Freq.4469.12305573234" "Freq.4567.90688886175" "Freq.4965.4984006347"
[16] "Freq.5854.45161215455" "Freq.5905.64933878776" "Freq.6175.68130655941"
[19] "Freq.6433.22411185796" "Freq.6631.46775487994" "Freq.6958.20015968149"
[22] "Freq.7469.83422424355" "Freq.8602.43342069553" "Freq.8766.14436081853"
[25] "Freq.8811.22677706485" "Freq.8915.90029255773" "Freq.9131.39810096"
[28] "Freq.9378.82122607608"
Never saw that [[1]] in a list before, and the problem is that I can't append things to this list.
How can I solve this?
This is a list in a list. Normally this can be referred to as a nested list.
a <- c(1,2,3)
b <- c(4,5,6)
list <- list(a,b)
In this code snippet we are creating two vectors and put them into a list. Now you can access the nested vectors/lists using the double brackets. Like so:
list[[1]]
> [1] 1 2 3
Now, if you want to change the value (or append it, see comment) you can use the normal syntax but solely assign it to the nested object.
list[[1]] <- c(7,8,9)
list[[1]]
> [1] 7 8 9
I'm reading data data and trying to convert it to data frame to save it into readable format. However no clue about converting the dat data. A bit beginner to R. Any help will be highly appreciated.
Code so Far:
data <- readLines("Day8.dat")
print(data)
Output So Far:
[1] "<d2lm:d2LogicalModel extensionVersion=\"2.0\" extensionName=\"NTIS Published Services\"
modelBaseVersion=\"2\" xmlns:ns4=\"http://www.thalesgroup.com/NTIS/Datex2Extensions/1.0Beta1\"
xmlns:ns3=\"http://datex2.eu/schema/2/2_0/inrix\" xmlns:d2lm=\"http://datex2.eu/schema/2/2_0\">
<d2lm:exchange><d2lm:supplierIdentification><d2lm:country>gb</d2lm:country>
<d2lm:nationalIdentifier>NTIS</d2lm:nationalIdentifier></d2lm:supplierIdentification></d2lm:exchange>
<d2lm:payloadPublication xsi:type=\"d2lm:SituationPublication\" lang=\"en\"
xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\"><d2lm:feedType>Event Data</d2lm:feedType>
<d2lm:publicationTime>2020-05-10T00:00:44.778+01:00</d2lm:publicationTime><d2lm:publicationCreator>
<d2lm:country>gb</d2lm:country><d2lm:nationalIdentifier>NTIS</d2lm:nationalIdentifier>
</d2lm:publicationCreator><d2lm:situation version=\"\" id=\"2922904\"><d2lm:headerInformation>
<d2lm:areaOfInterest>national</d2lm:areaOfInterest>
....
Thanks
It all depends on what you want to do with the data, i.e., how you want to process it.
For example, let's assume your interest is in parsing all XML tags as separate strings, then you can extract the tags using regular expression and the function str_extract:
library(stringr)
str_extract_all(dat, "<(d2lm:[^>]*)>.*</\\1>|<d2lm:[^>]*>")
This regex works even if the XML element names are variable:
str_extract_all(dat, "<([^>]*)>.*</\\1>|<[^>]*>")
The result is a list:
[[1]]
[1] "<d2lm:d2LogicalModel extensionVersion=\"2.0\" extensionName=\"NTIS Published Services\" \nmodelBaseVersion=\"2\" xmlns:ns4=\"http://www.thalesgroup.com/NTIS/Datex2Extensions/1.0Beta1\" \nxmlns:ns3=\"http://datex2.eu/schema/2/2_0/inrix\" xmlns:d2lm=\"http://datex2.eu/schema/2/2_0\">"
[2] "<d2lm:exchange>"
[3] "<d2lm:supplierIdentification>"
[4] "<d2lm:country>gb</d2lm:country>"
[5] "<d2lm:nationalIdentifier>NTIS</d2lm:nationalIdentifier>"
[6] "<d2lm:payloadPublication xsi:type=\"d2lm:SituationPublication\" lang=\"en\" \nxmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\">"
[7] "<d2lm:feedType>Event Data</d2lm:feedType>"
[8] "<d2lm:publicationTime>2020-05-10T00:00:44.778+01:00</d2lm:publicationTime>"
[9] "<d2lm:publicationCreator>"
[10] "<d2lm:country>gb</d2lm:country>"
[11] "<d2lm:nationalIdentifier>NTIS</d2lm:nationalIdentifier>"
[12] "<d2lm:situation version=\"\" id=\"2922904\">"
[13] "<d2lm:headerInformation>"
[14] "<d2lm:areaOfInterest>national</d2lm:areaOfInterest>"
To turn the list into a dataframe:
datDF <- data.frame(tags = unlist(str_extract_all(dat, "<(d2lm:[^>]*)>.*</\\1>|<d2lm:[^>]*>")))
EDIT:
If you want to have a dataframe with the text values between XML start tag and XML end tag, you can extract these tags and values along these lines:
datDF <- data.frame(
tags = unlist(str_extract_all(dat, "<([^>]*)>(?=[^>]*</\\1>)")),
values = unlist(str_extract_all(dat, "(?<=<([^>]{1,100})>).*(?=</\\1>)"))
)
datDF
tags values
1 <d2lm:country> gb
2 <d2lm:nationalIdentifier> NTIS
3 <d2lm:feedType> Event Data
4 <d2lm:publicationTime> 2020-05-10T00:00:44.778+01:00
5 <d2lm:country> gb
6 <d2lm:nationalIdentifier> NTIS
7 <d2lm:areaOfInterest> national
Is this--roughly--what you had in mind?
DATA:
dat <- '<d2lm:d2LogicalModel extensionVersion=\"2.0\" extensionName=\"NTIS Published Services\"
modelBaseVersion=\"2\" xmlns:ns4=\"http://www.thalesgroup.com/NTIS/Datex2Extensions/1.0Beta1\"
xmlns:ns3=\"http://datex2.eu/schema/2/2_0/inrix\" xmlns:d2lm=\"http://datex2.eu/schema/2/2_0\">
<d2lm:exchange><d2lm:supplierIdentification><d2lm:country>gb</d2lm:country>
<d2lm:nationalIdentifier>NTIS</d2lm:nationalIdentifier></d2lm:supplierIdentification></d2lm:exchange>
<d2lm:payloadPublication xsi:type=\"d2lm:SituationPublication\" lang=\"en\"
xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\"><d2lm:feedType>Event Data</d2lm:feedType>
<d2lm:publicationTime>2020-05-10T00:00:44.778+01:00</d2lm:publicationTime><d2lm:publicationCreator>
<d2lm:country>gb</d2lm:country><d2lm:nationalIdentifier>NTIS</d2lm:nationalIdentifier>
</d2lm:publicationCreator><d2lm:situation version=\"\" id=\"2922904\"><d2lm:headerInformation>
<d2lm:areaOfInterest>national</d2lm:areaOfInterest>'
Would like to efficiently replace all partial match strings over a single column by supplying a vector of strings which will be searched (and matched) and also be used as replacement. i.e. for each vector in df below, it will partially match for vectors in vec_string. Where matches is found, it will simply replace the entire string with vec_string. i.e. turning 'subscriber manager' to 'manager'. By supplying more vectors into vec_string, it will search through the whole df until all is complete.
I have started the function, but can't seem to finish it off by replacing the vectors in df with vec_string. Appreciate your help
df <- c(
'solicitor'
,'subscriber manager'
,'licensed conveyancer'
,'paralegal'
,'property assistant'
,'secretary'
,'conveyancing paralegal'
,'licensee'
,'conveyancer'
,'principal'
,'assistant'
,'senior conveyancer'
,'law clerk'
,'lawyer'
,'legal practice director'
,'legal secretary'
,'personal assistant'
,'legal assistant'
,'conveyancing clerk')
vec_string <- c('manager','law')
#function to search and replace
replace_func <-
function(vec,str_vec) {
repl_str <- list()
for(i in 1:length(str_vec)) {
repl_str[[i]] <- grep(str_vec[i],unique(tolower(vec)))
}
names(repl_str) <- vec_string
return(repl_str)
}
replace_func(df,vec_string)
$`manager`
[1] 2
$law
[1] 13 14
As you can see, the function returns a named list with elements to which the replacement will
This should do the trick
res = sapply(df,function(x){
match = which(sapply(vec_string,function(y) grepl(y,x)))
if (length(match)){x=vec_string[match[1]]}else{x}
})
res
[1] "solicitor" "manager" "licensed conveyancer"
[4] "paralegal" "property assistant" "secretary"
[7] "conveyancing paralegal" "licensee" "conveyancer"
[10] "principal" "assistant" "senior conveyancer"
[13] "law" "law" "legal practice director"
[16] "legal secretary" "personal assistant" "legal assistant"
[19] "conveyancing clerk"
We compare each part of df with each part of vec_string. If there is a match, the vec_string part is returned, else it is left as it is. Watch out as if there are more than 1 matches it will keep the first one.
I have several vectors to combine into a named list ("my_list"). The names of the vectors are already stored in the vector ("zI").
> zI
[1] "Chemokines" "Cell_Cycle" "Regulation"
[4] "Senescence" "B_cell_Functions" "T_Cell_Functions"
[7] "Cell_Functions" "Adhesion" "Transporter_Functions"
[10] "Complement" "Pathogen_Defense" "Cytokines"
[13] "Antigen_Processing" "Leukocyte_Functions" "TNF_Superfamily"
[16] "Macrophage_Functions" "Microglial_Functions" "Interleukins"
[19] "Cytotoxicity" "NK_Cell_Functions" "TLR"
If it's a small number of vectors, I'd simply do
my_list <- setNames(list(Chemokines, Adhesion), c("Chemokines", "Adhesion"))
I'd like to find a smarter way, other than to combine the vector names into a long string and then copying/pasting.
> toString(zI)
[1] "Chemokines, Cell_Cycle, Regulation, Senescence, B_cell_Functions, T_Cell_Functions, Cell_Functions, Adhesion, Transporter_Functions, Complement, Pathogen_Defense, Cytokines, Antigen_Processing, Leukocyte_Functions, TNF_Superfamily, Macrophage_Functions, Microglial_Functions, Interleukins, Cytotoxicity, NK_Cell_Functions, TLR"
> my_lists <- list(Chemokines, Cell_Cycle, Regulation, Senescence, B_cell_Functions, T_Cell_Functions, Cell_Functions, Adhesion, Transporter_Functions, Complement, Pathogen_Defense, Cytokines, Antigen_Processing, Leukocyte_Functions, TNF_Superfamily, Macrophage_Functions, Microglial_Functions, Interleukins, Cytotoxicity, NK_Cell_Functions, TLR)
> my_lists <- setNames(my_lists, zI)
This is probably a really fundamental question, but I've searched and read about 10 separate threads and still can't figure it out. Much thanks for any help!
We can use mget to get the values of the character strings.
mget(zI)
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