This question already has answers here:
Paste multiple columns together
(11 answers)
Closed 4 years ago.
I have a nested list
l1 <- letters
l2 <- 1:26
l3 <- LETTERS
list <- list(l1,l2,l3)
Is there an elegant way to concatenate all the elements in inner vectors to form one character vector (possibly using paste), the assumption is that all the inner vectors are of the same length.
I would like my final result to be
[1] "a1A"
[2] "b2B"
[3] "c3C"
[4] "d4D"
....
[26] "z26Z"
Try:
apply(sapply(list,paste0),1,paste0,collapse="")
[1] "a1A" "b2B" "c3C" "d4D" "e5E" "f6F" "g7G" "h8H" "i9I" "j10J" "k11K" "l12L" "m13M" "n14N" "o15O" [16] "p16P" "q17Q" "r18R" "s19S" "t20T" "u21U" "v22V" "w23W" "x24X" "y25Y" "z26Z"
user20650's solution is probably as elegant as you are going to get. But for what it's worth, here's a quick hack in dplyr:
library(dplyr)
ll <- list(l1,l2,l3) # I try not to use "list" as a name. Gets confusing sometimes.
as.data.frame(ll) %>%
mutate(x = paste0(.[[1]], .[[2]], .[[3]])) %>%
.$x
# returns
[1] "a1A" "b2B" "c3C" "d4D" "e5E" "f6F" "g7G" "h8H" "i9I" "j10J" "k11K" "l12L"
[13] "m13M" "n14N" "o15O" "p16P" "q17Q" "r18R" "s19S" "t20T" "u21U" "v22V" "w23W" "x24X"
[25] "y25Y" "z26Z"
Related
This question already has answers here:
Alternate, interweave or interlace two vectors
(2 answers)
Closed 1 year ago.
I want to write a R program that creates the vector 0.1^3, 0.2^1, 0.1^6, 0.2^4, ..., 0.1^36, 0.2^34.
v=c(seq(3,36,3))
w=c(seq(1,34,3))
x=c(0.1^v)
y=c(0.2^w)
z=c(x,y)
Please help.
rbind to a matrix and convert to vector again:
c(rbind(x, y))
Or more directly:
rep(c(0.1, 0.2), 12)^c(rbind(seq(3,36,3), seq(1,34,3)))
You can use matrix to create the desired vector.
c(matrix(z, 2, byrow=TRUE))
# [1] 1.000000e-03 2.000000e-01 1.000000e-06 1.600000e-03 1.000000e-09
# [6] 1.280000e-05 1.000000e-12 1.024000e-07 1.000000e-15 8.192000e-10
#[11] 1.000000e-18 6.553600e-12 1.000000e-21 5.242880e-14 1.000000e-24
#[16] 4.194304e-16 1.000000e-27 3.355443e-18 1.000000e-30 2.684355e-20
#[21] 1.000000e-33 2.147484e-22 1.000000e-36 1.717987e-24
This question already has answers here:
Pasting two vectors with combinations of all vectors' elements
(8 answers)
Closed 2 years ago.
I have two vectors, one that contains a list of variables, and one that contains dates, such as
Variables_Pays <- c("PIB", "ConsommationPrivee","ConsommationPubliques",
"FBCF","ProductionIndustrielle","Inflation","InflationSousJacente",
"PrixProductionIndustrielle","CoutHoraireTravail")
Annee_Pays <- c("2000","2001")
I want to merge them to have a vector with each variable indexed by my date, that is my desired output is
> Colonnes_Pays_Principaux
[1] "PIB_2020" "PIB_2021" "ConsommationPrivee_2020"
[4] "ConsommationPrivee_2021" "ConsommationPubliques_2020" "ConsommationPubliques_2021"
[7] "FBCF_2020" "FBCF_2021" "ProductionIndustrielle_2020"
[10] "ProductionIndustrielle_2021" "Inflation_2020" "Inflation_2021"
[13] "InflationSousJacente_2020" "InflationSousJacente_2021" "PrixProductionIndustrielle_2020"
[16] "PrixProductionIndustrielle_2021" "CoutHoraireTravail_2020" "CoutHoraireTravail_2021"
Is there a simpler / more readabl way than a double for loop as I have tried and succeeded below ?
Colonnes_Pays_Principaux <- vector()
for (Variable in (1:length(Variables_Pays))){
for (Annee in (1:length(Annee_Pays))){
Colonnes_Pays_Principaux=
append(Colonnes_Pays_Principaux,
paste(Variables_Pays[Variable],Annee_Pays[Annee],sep="_")
)
}
}
expand.grid will create a data frame with all combinations of the two vectors.
with(
expand.grid(Variables_Pays, Annee_Pays),
paste0(Var1, "_", Var2)
)
#> [1] "PIB_2000" "ConsommationPrivee_2000"
#> [3] "ConsommationPubliques_2000" "FBCF_2000"
#> [5] "ProductionIndustrielle_2000" "Inflation_2000"
#> [7] "InflationSousJacente_2000" "PrixProductionIndustrielle_2000"
#> [9] "CoutHoraireTravail_2000" "PIB_2001"
#> [11] "ConsommationPrivee_2001" "ConsommationPubliques_2001"
#> [13] "FBCF_2001" "ProductionIndustrielle_2001"
#> [15] "Inflation_2001" "InflationSousJacente_2001"
#> [17] "PrixProductionIndustrielle_2001" "CoutHoraireTravail_2001"
We can use outer :
c(t(outer(Variables_Pays, Annee_Pays, paste, sep = '_')))
# [1] "PIB_2000" "PIB_2001"
# [3] "ConsommationPrivee_2000" "ConsommationPrivee_2001"
# [5] "ConsommationPubliques_2000" "ConsommationPubliques_2001"
# [7] "FBCF_2000" "FBCF_2001"
# [9] "ProductionIndustrielle_2000" "ProductionIndustrielle_2001"
#[11] "Inflation_2000" "Inflation_2001"
#[13] "InflationSousJacente_2000" "InflationSousJacente_2001"
#[15] "PrixProductionIndustrielle_2000" "PrixProductionIndustrielle_2001"
#[17] "CoutHoraireTravail_2000" "CoutHoraireTravail_2001"
No real need to go beyond the basics here! Use paste for pasting the strings and rep to repeat either Annee_Pays och Variables_Pays to get all combinations:
Variables_Pays <- c("PIB", "ConsommationPrivee","ConsommationPubliques",
"FBCF","ProductionIndustrielle","Inflation","InflationSousJacente",
"PrixProductionIndustrielle","CoutHoraireTravail")
Annee_Pays <- c("2000","2001")
# To get this is the same order as in your example:
paste(rep(Variables_Pays, rep(2, length(Variables_Pays))), Annee_Pays, sep = "_")
# Alternative order:
paste(Variables_Pays, rep(Annee_Pays, rep(length(Variables_Pays), 2)), sep = "_")
# Or, if order doesn't matter too much:
paste(Variables_Pays, rep(Annee_Pays, length(Variables_Pays)), sep = "_")
In base R:
Variables_Pays <- c("PIB", "ConsommationPrivee","ConsommationPubliques",
"FBCF","ProductionIndustrielle","Inflation","InflationSousJacente",
"PrixProductionIndustrielle","CoutHoraireTravail")
Annee_Pays <- c("2000","2001")
cbind(paste(Variables_Pays, Annee_Pays,sep="_"),paste(Variables_Pays, rev(Annee_Pays),sep="_")
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 have some codes like this example, if you run these codes
library(hurricaneexposure)
library(hurricaneexposuredata)
data("hurr_tracks")
storms <- unique(hurr_tracks$storm_id)
storms
then you will see that "storms" has a long string list with "stormname-year" structure.
[1] "Alberto-1988" "Beryl-1988" "Chris-1988" "Florence-1988" "Gilbert-1988" "Keith-1988" "Allison-1989" "Chantal-1989"
[9] "Hugo-1989" "Jerry-1989" "Bertha-1990" "Marco-1990" "Ana-1991" "Bob-1991" "Fabian-1991" "Notnamed-1991"
[17] "Andrew-1992" "Danielle-1992" "Earl-1992" "Arlene-1993" "Emily-1993" "Alberto-1994" "Beryl-1994" "Gordon-1994"
[25] "Allison-1995" "Dean-1995" "Erin-1995" "Gabrielle-1995" "Jerry-1995" "Opal-1995" "Arthur-1996" "Bertha-1996"
[33] "Edouard-1996" "Fran-1996" "Josephine-1996" "Subtrop-1997" "Ana-1997" "Danny-1997" "Bonnie-1998" "Charley-1998"
[41] "Earl-1998" "Frances-1998" "Georges-1998" "Hermine-1998" "Mitch-1998" "Bret-1999" "Dennis-1999" "Floyd-1999"
[49] "Harvey-1999" "Irene-1999" "Beryl-2000" "Gordon-2000" "Helene-2000" "Leslie-2000" "Allison-2001" "Barry-2001"
My question is how to split these elements based on same year. For example, I want to create a new variable "y1988" which is a list that has all storms in 1998. If I run y1988, it will output:
y1988
[1] "Alberto-1988" "Beryl-1988" "Chris-1988" "Florence-1988" "Gilbert-1988" "Keith-1988"
So as for y1989 until 2001. I am guessing it might use gsub() and a for-loop，however, I am a rookie in R, so really hope you could give me some suggestion.
We can use split with grouping variable created by removing the prefix substring including the - with sub.
lst <- split(storms, sub(".*-", "", storms))
lst$`1988`
#[1] "Alberto-1988" "Beryl-1988" "Chris-1988" "Florence-1988"
#[5] "Gilbert-1988" "Keith-1988"
data
storms <- c("Alberto-1988", "Beryl-1988", "Chris-1988", "Florence-1988",
"Gilbert-1988", "Keith-1988", "Allison-1989", "Chantal-1989",
"Hugo-1989", "Jerry-1989", "Bertha-1990", "Marco-1990", "Ana-1991",
"Bob-1991", "Fabian-1991", "Notnamed-1991", "Andrew-1992", "Danielle-1992",
"Earl-1992", "Arlene-1993", "Emily-1993", "Alberto-1994", "Beryl-1994",
"Gordon-1994", "Allison-1995", "Dean-1995", "Erin-1995", "Gabrielle-1995",
"Jerry-1995", "Opal-1995", "Arthur-1996", "Bertha-1996", "Edouard-1996",
"Fran-1996", "Josephine-1996", "Subtrop-1997", "Ana-1997", "Danny-1997",
"Bonnie-1998", "Charley-1998", "Earl-1998", "Frances-1998", "Georges-1998",
"Hermine-1998", "Mitch-1998", "Bret-1999", "Dennis-1999", "Floyd-1999",
"Harvey-1999", "Irene-1999", "Beryl-2000", "Gordon-2000", "Helene-2000",
"Leslie-2000", "Allison-2001", "Barry-2001")
Why don't you extract the year directly within your original dataframe? libraries dplyr and tidyr are well suited for problems like this.
I suggest the following:
library(dplyr)
library(tidyr)
hurr_tracks %>%
extract(storm_id, c("storm", "year"),"(.+)-(.+)")
Alternative way using stringr
split(storms,str_extract(storms,"[0-9]+"))
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)