Im trying to find a way to substitute out the "|" symbol in this dataset for the "/" symbol
df<-c(2|4,5|6,3|4,4|7,5|8)
So that way the final would look like
df<-c(2/4,5/6,3/4,4/7,5/8)
Any help would be great thankyou
If it is a single string (or a vector of elements) use chartr from base R
chartr("|", "/", df)
[1] "2/4,5/6,3/4,4/7,5/8"
df2$col1 <- chartr("|", "/", df2$col1)
> df2
col1
1 2/4
2 5/6
3 3/4
4 4/7
5 5/8
data
df <- "2|4,5|6,3|4,4|7,5|8"
df2 <- data.frame(col1 = c("2|4", "5|6", "3|4", "4|7", "5|8"))
Related
Consider the following dataframe:
status
1 file-status-done-bad
2 file-status-maybe-good
3 file-status-underreview-good
4 file-status-complete-final-bad
We want to extract the last part of status, wherein part is delimited by -. Such:
status status_extract
1 file-status-done-bad done
2 file-status-maybe-good maybe
3 file-status-ok-underreview-good underreview
4 file-status-complete-final-bad final
In SQL this is easy, select split_part(status, '-', -2).
However, the solutions I've seen with R either operate on vectors or are messy to extract particular elements (they return ALL elements). How is this done in a mutate chain? The below is a failed attempt.
df %>%
mutate(status_extract = str_split_fixed(status, pattern = '-')[[-2]])
Found the a really simple answer.
library(tidyverse)
df %>%
mutate(status_extract = word(status, -1, sep = "-"))
In base R you can combine the functions sapply and strsplit
df$status_extract <- sapply(strsplit(df$status, "-"), function(x) x[length(x) - 1])
# status status_extract
# 1 file-status-done-bad done
# 2 file-status-maybe-good maybe
# 3 file-status-underreview-good underreview
# 4 file-status-complete-final-bad final
You can use map() and nth() to extract the nth value from a vector.
library(tidyverse)
df %>%
mutate(status_extract = map_chr(str_split(status, "-"), nth, -2))
# status status_extract
# 1 file-status-done-bad done
# 2 file-status-maybe-good maybe
# 3 file-status-underreview-good underreview
# 4 file-status-complete-final-bad final
which is equivalent to a base version like
sapply(strsplit(df$status, "-"), function(x) rev(x)[2])
# [1] "done" "maybe" "underreview" "final"
You can use regex to get what you want without splitting the string.
sub('.*-(\\w+)-.*$', '\\1', df$status)
#[1] "done" "maybe" "underreview" "final"
I am trying to train a data that's converted from a document term matrix to a dataframe. There are separate fields for the positive and negative comments, so I wanted to add a string to the column names to serve as a "tag", to differentiate the same word coming from the different fields - for example, the word hello can appear both in the positive and negative comment fields (and thus, represented as a column in my dataframe), so in my model, I want to differentiate these by making the column names positive_hello and negative_hello.
I am looking for a way to rename columns in such a way that a specific string will be appended to all columns in the dataframe. Say, for mtcars, I want to rename all of the columns to have "_sample" at the end, so that the column names would become mpg_sample, cyl_sample, disp_sample and so on, which were originally mpg, cyl, and disp.
I'm considering using sapplyor lapply, but I haven't had any progress on it. Any help would be greatly appreciated.
Use colnames and paste0 functions:
df = data.frame(x = 1:2, y = 2:1)
colnames(df)
[1] "x" "y"
colnames(df) <- paste0('tag_', colnames(df))
colnames(df)
[1] "tag_x" "tag_y"
If you want to prefix each item in a column with a string, you can use paste():
# Generate sample data
df <- data.frame(good=letters, bad=LETTERS)
# Use the paste() function to append the same word to each item in a column
df$good2 <- paste('positive', df$good, sep='_')
df$bad2 <- paste('negative', df$bad, sep='_')
# Look at the results
head(df)
good bad good2 bad2
1 a A positive_a negative_A
2 b B positive_b negative_B
3 c C positive_c negative_C
4 d D positive_d negative_D
5 e E positive_e negative_E
6 f F positive_f negative_F
Edit:
Looks like I misunderstood the question. But you can rename columns in a similar way:
colnames(df) <- paste(colnames(df), 'sample', sep='_')
colnames(df)
[1] "good_sample" "bad_sample" "good2_sample" "bad2_sample"
Or to rename one specific column (column one, in this case):
colnames(df)[1] <- paste('prefix', colnames(df)[1], sep='_')
colnames(df)
[1] "prefix_good_sample" "bad_sample" "good2_sample" "bad2_sample"
You can use setnames from the data.table package, it doesn't create any copy of your data.
library(data.table)
df <- data.frame(a=c(1,2),b=c(3,4))
# a b
# 1 1 3
# 2 2 4
setnames(df,paste0(names(df),"_tag"))
print(df)
# a_tag b_tag
# 1 1 3
# 2 2 4
In R I'd like to take a collection of file names in the format below and return the number to the right of the second underscore (this will always be a number) and the text string to the right of the third underscore (this will be combinations of letters and numbers).
I have file names in this format:
HELP_PLEASE_4_ME
I want to extract the number 4 and the text ME
I'd then like to create a new field within my data frame where these two types of data can be stored. Any suggestions?
Here is an option using regexec and regmatches to pull out the patterns:
matches <- regmatches(df$a, regexec("^.*?_.*?_([0-9]+)_([[:alnum:]]+)$", df$a))
df[c("match.1", "match.2")] <- t(sapply(matches, `[`, -1)) # first result for each match is full regular expression so need to drop that.
Produces:
a match.1 match.2
1 HELP_PLEASE_4_ME 4 ME
2 SOS_WOW_3_Y34OU 3 Y34OU
This will break if any rows don't have the expected structure, but I think that is what you want to happen (i.e. be alerted that your data is not what you think it is). strsplit based approaches will require additional checking to ensure that your data is what you think it is.
And the data:
df <- data.frame(a=c("HELP_PLEASE_4_ME", "SOS_WOW_3_Y34OU"), stringsAsFactors=F)
The obligatory stringr version of #BrodieG's quite spiffy answer:
df[c("match.1", "match.2")] <-
t(sapply(str_match_all(df$a, "^.*?_.*?_([0-9]+)_([[:alnum:]]+)$"), "[", 2:3))
Put here for context only. You should accept BrodieG's answer.
Since you already know that you want the text that comes after the second and third underscore, you could use strsplit and take the third and fourth result.
> x <- "HELP_PLEASE_4_ME"
> spl <- unlist(strsplit(x, "_"))[3:4]
> data.frame(string = x, under2 = spl[1], under3 = spl[2])
## string under2 under3
## 1 HELP_PLEASE_4_ME 4 ME
Then for longer vectors, you could do something like the last two lines here.
## set up some data
> word1 <- c("HELLO", "GOODBYE", "HI", "BYE")
> word2 <- c("ONE", "TWO", "THREE", "FOUR")
> nums <- 20:23
> word3 <- c("ME", "YOU", "THEM", "US")
> XX <-paste0(word1, "_", word2, "_", nums, "_", word3)
> XX
## [1] "HELLO_ONE_20_ME" "GOODBYE_TWO_21_YOU"
## [3] "HI_THREE_22_THEM" "BYE_FOUR_23_US"
## ------------------------------------------------
## process it
> spl <- do.call(rbind, strsplit(XX, "_"))[, 3:4]
> data.frame(cbind(XX, spl))
## XX V2 V3
## 1 HELLO_ONE_20_ME 20 ME
## 2 GOODBYE_TWO_21_YOU 21 YOU
## 3 HI_THREE_22_THEM 22 THEM
## 4 BYE_FOUR_23_US 23 US
I want to change the name of the output of my R function to reflect different strings that are inputted. Here is what I have tried:
kd = c("a","b","d","e","b")
test = function(kd){
return(list(assign(paste(kd,"burst",sep="_"),1:6)))
}
This is just a simple test function. I get the warning (which is just as bad an error for me):
Warning message:
In assign(paste(kd, "burst", sep = "_"), 1:6) :
only the first element is used as variable name
Ideally I would get ouput like a_burst = 1, b_burst = 2 and so on but am not getting close.
I would like split up a dataframe by contents of a vector and be able to name everything according to the name from that vector, similar to
How to split a data frame by rows, and then process the blocks?
but not quite. The naming is imperative.
Something like this, maybe?
kd = c("a","b","d","e","b")
test <- function(x){
l <- as.list(1:5)
names(l) <- paste(x,"burst",sep = "_")
l
}
test(kd)
You could use a vector instead of a list by way of setNames:
t1_6 <- setNames( 1:6, kd)
t1_6
a b d e b <NA>
1 2 3 4 5 6
> t1_6["a"]
a
1
Looking at the question again I wondered if you wnated to assign sequential names to a character vector:
> a1_5 <- setNames(kd, paste0("alpha", 1:5))
> a1_5
alpha1 alpha2 alpha3 alpha4 alpha5
"a" "b" "d" "e" "b"
I know that I can change a data.frame column name by:
colnames(df)[3] <- "newname"
But there might be cases where the column I want to change is not in the 3rd position. Is there a way to look up the column by name and change it? Like this...
colnames(df)[,"oldname"] <- "newname"
BTW, I have tried this code and I keep getting incorrect number of subscripts on matrix.
Thanks.
colnames(df)[colnames(df)=="oldname"] <- "newname"
or just names
names(df)[names(df)=="oldname"] <- "newname"
There are various functions for renaming columns in packages as well.
colnames(df)[colnames(df)=="oldname"] <- "newname"
or
names(df)[names(df)=="oldname"] <- "newname"
(since names and colnames are equivalent for a data frame)
or you might be looking for
library(reshape)
df <- rename(df,c(oldname="newname"))
I was using package data.table today and when I tried to change a column name using my usual method a message appeared recommending this approach:
library(data.table)
df <- read.table(text= "
region state county
1 1 1
1 2 2
1 2 3
2 1 4
2 1 4
", header=TRUE, na.strings=NA)
df
setnames(df, "county", "district")
df
A somewhat more general approach that will replace all of the "old"s at the beginning of any current name with "new" in the same character location:
names(df) <- sub("^old", "new", names(df) )