I have a R data frame which looks like
data.1 data.character
a **str1**,str2,str2,str3,str4,str5,str6
b str3,str4,str5
c **str1**,str6
I am currently using grepl to identify if the column data.character has my search string "<str>" and if so I want all the row values in data.1 to be concatenated into one string with a separator
eg. if I use grepl(str1,data.character) it will return two rows of df$data.1 and I want an output like
a,c ( rows which contain str1 in data.character)
I am currently using two for loops but i know this is not an efficient method. I was wondering if someone could suggest a more elegant and less time consuming method.
You were almost there - (now my long-winded answer)
# Data
df <- read.table(text="data.1 data.character
a **str1**,str2,str2,str3,str4,str5,str6
b str3,str4,str5
c **str1**,str6",header=T,stringsAsFactors=F)
Match string
# In your question you used grepl which produces a logical vector (TRUE if
#string is present)
grepl("str1" , df$data.character)
#[1] TRUE FALSE TRUE
# In my comment I used grep which produces an positional index of the vector if
# string is present (this was due to me not reading your grepl properly rather
# than because of any property)
grep("str1" , df$data.character)
# [1] 1 3
Then subset the vector that you want at these positions resulting from grep (or grepl)
(s <- df$data.1[grepl("str1" , df$data.character)])
# [1] "a" "c" first and third elements are selected
Paste these together into the required format (collapse argument is used to define the separator between the elements)
paste(s,collapse=",")
# [1] "a,c"
So more succinctly
paste(df$data.1[grep("str1" , df$data.character)],collapse=",")
Related
I have 6 vector values
vec<-c("Col1","Col2islonger","Col3isabitlonger","Col4isless","Col5willbelongest")
I run
nchar(vec)
my results are
4,12,16,10,17
Based on those values I would like to run a conditional statement or for loop whichever is better that would rename the columns based on their lengths and current values. For example
If nchar(vec) is less or equal to 10 keep the name as is. If the it is greater than 10 make sure the renamed element takes the first 9 values and skips over to the last.
newvec<- c("Col1","Col2islonr","Col3isabir","Col4isless","Col5willbt")
Try this:
#Data
vec<-c("Col1","Col2islonger","Col3isabitlonger","Col4isless","Col5willbelongest")
#Rename
vec2 <- ifelse(nchar(vec)<=10,vec,paste0(substr(vec,1,9),substr(vec,nchar(vec),nchar(vec))))
Output:
[1] "Col1" "Col2islonr" "Col3isabir" "Col4isless" "Col5willbt"
We can use sub to trim the vec. We create a logical vector with the number of character of the 'vec' ('i1'). Using that, we update for those elements that are greater than 10 characters to remove the characters between the 10th and the last character with sub and update it
i1 <- nchar(vec) > 10
vec[i1] <- sub("^(.{1,9}).*(.)", "\\1\\2", vec[i1])
-output
vec
#[1] "Col1" "Col2islonr" "Col3isabir" "Col4isless" "Col5willbt"
Trying to remove columns from a large data frame. Using grep and it works fine when actually there are matching columns. But when there is zero matching columns it drops all the columns.
s <- s[, -grep("^Test", colnames(s))]
To confirm that there are no columns that match Test
> y <- grep("^Test", colnames(s))
> y
integer(0)
What is exactly going on here?
You need to use grepl and ! instead.
df2 <- data.frame(ID =c(1,2,3), T = c("words", "stuff","things"))
df2[,!grepl("^Test", colnames(df2))]
ID T
1 1 words
2 2 stuff
3 3 things
-grep() or -grepl() return integer(0) when there isn't a match.
-TRUE == -1 where as !TRUE == FALSE
Using !grepl() returns the full logical vector (TRUE TRUE) for each column header, allowing you to correctly subset when no columns meet the condition. In other words for colname(df)[i], grepl(..., colnames(df))[i] returns TRUE where your pattern is matched, then using ! you invert to keep the values that don't match, and remove the ones that do.
I am having trouble subsetting my data. I want the data subsetted on column x, where the first 3 characters begin G45.
My data frame:
x <- c("G448", "G459", "G479", "G406")
y <- c(1:4)
My.Data <- data.frame (x,y)
I have tried:
subset (My.Data, x=="G45*")
But I am unsure how to use wildcards. I have also tried grep() to find the indicies:
grep ("G45*", My.Data$x)
but it returns all 4 rows, rather than just those beginning G45, probably also as I am unsure how to use wildcards.
It's pretty straightforward using [ to extract:
grep will give you the position in which it matched your search pattern (unless you use value = TRUE).
grep("^G45", My.Data$x)
# [1] 2
Since you're searching within the values of a single column, that actually corresponds to the row index. So, use that with [ (where you would use My.Data[rows, cols] to get specific rows and columns).
My.Data[grep("^G45", My.Data$x), ]
# x y
# 2 G459 2
The help-page for subset shows how you can use grep and grepl with subset if you prefer using this function over [. Here's an example.
subset(My.Data, grepl("^G45", My.Data$x))
# x y
# 2 G459 2
As of R 3.3, there's now also the startsWith function, which you can again use with subset (or with any of the other approaches above). According to the help page for the function, it's considerably faster than using substring or grepl.
subset(My.Data, startsWith(as.character(x), "G45"))
# x y
# 2 G459 2
You may also use the stringr package
library(dplyr)
library(stringr)
My.Data %>% filter(str_detect(x, '^G45'))
You may not use '^' (starts with) in this case, to obtain the results you need
I would like to take a data frame with characters and numbers, and concatenate all of the elements of the each row into a single string, which would be stored as a single element in a vector. As an example, I make a data frame of letters and numbers, and then I would like to concatenate the first row via the paste function, and hopefully return the value "A1"
df <- data.frame(letters = LETTERS[1:5], numbers = 1:5)
df
## letters numbers
## 1 A 1
## 2 B 2
## 3 C 3
## 4 D 4
## 5 E 5
paste(df[1,], sep =".")
## [1] "1" "1"
So paste is converting each element of the row into an integer that corresponds to the 'index of the corresponding level' as if it were a factor, and it keeps it a vector of length two. (I know/believe that factors that are coerced to be characters behave in this way, but as R is not storing df[1,] as a factor at all (tested by is.factor(), I can't verify that it is actually an index for a level)
is.factor(df[1,])
## [1] FALSE
is.vector(df[1,])
## [1] FALSE
So if it is not a vector then it makes sense that it is behaving oddly, but I can't coerce it into a vector
> is.vector(as.vector(df[1,]))
[1] FALSE
Using as.character did not seem to help in my attempts
Can anyone explain this behavior?
While others have focused on why your code isn't working and how to improve it, I'm going to try and focus more on getting the result you want. From your description, it seems you can readily achieve what you want using paste:
df <- data.frame(letters = LETTERS[1:5], numbers = 1:5, stringsAsFactors=FALSE)
paste(df$letters, df$numbers, sep=""))
## [1] "A1" "B2" "C3" "D4" "E5"
You can change df$letters to character using df$letters <- as.character(df$letters) if you don't want to use the stringsAsFactors argument.
But let's assume that's not what you want. Let's assume you have hundreds of columns and you want to paste them all together. We can do that with your minimal example too:
df_args <- c(df, sep="")
do.call(paste, df_args)
## [1] "A1" "B2" "C3" "D4" "E5"
EDIT: Alternative method and explanation:
I realised the problem you're having is a combination of the fact that you're using a factor and that you're using the sep argument instead of collapse (as #adibender picked up). The difference is that sep gives the separator between two separate vectors and collapse gives separators within a vector. When you use df[1,], you supply a single vector to paste and hence you must use the collapse argument. Using your idea of getting every row and concatenating them, the following line of code will do exactly what you want:
apply(df, 1, paste, collapse="")
Ok, now for the explanations:
Why won't as.list work?
as.list converts an object to a list. So it does work. It will convert your dataframe to a list and subsequently ignore the sep="" argument. c combines objects together. Technically, a dataframe is just a list where every column is an element and all elements have to have the same length. So when I combine it with sep="", it just becomes a regular list with the columns of the dataframe as elements.
Why use do.call?
do.call allows you to call a function using a named list as its arguments. You can't just throw the list straight into paste, because it doesn't like dataframes. It's designed for concatenating vectors. So remember that dfargs is a list containing a vector of letters, a vector of numbers and sep which is a length 1 vector containing only "". When I use do.call, the resulting paste function is essentially paste(letters, numbers, sep).
But what if my original dataframe had columns "letters", "numbers", "squigs", "blargs" after which I added the separator like I did before? Then the paste function through do.call would look like:
paste(letters, numbers, squigs, blargs, sep)
So you see it works for any number of columns.
For those using library(tidyverse), you can simply use the unite function.
new.df <- df%>%
unite(together, letters, numbers, sep="")
This will give you a new column called together with A1, B2, etc.
This is indeed a little weird, but this is also what is supposed to happen.
When you create the data.frame as you did, column letters is stored as factor. Naturally factors have no ordering, therefore when as.numeric() is applied to a factor it returns the ordering of of the factor. For example:
> df[, 1]
[1] A B C D E
Levels: A B C D E
> as.numeric(df[, 1])
[1] 1 2 3 4 5
A is the first level of the factor df[, 1] therefore A gets converted to the value 1, when as.numeric is applied. This is what happens when you call paste(df[1, ]). Since columns 1 and 2 are of different class, paste first transforms both elements of row 1 to numeric then to characters.
When you want to concatenate both columns, you first need to transform the first row to character:
df[, 1] <- as.character(df[, 1])
paste(df[1,], collapse = "")
As #sebastian-c pointed out, you can also use stringsAsFactors = FALSE in the creation of the data.frame, then you can omit the as.character() step.
if you want to start with
df <- data.frame(letters = LETTERS[1:5], numbers = 1:5, stringsAsFactors=TRUE)
.. then there is no general rule about how df$letters will be interpreted by any given function. It's a factor for modelling functions, character for some and integer for some others. Even the same function such as paste may interpret it differently, depending on how you use it:
paste(df[1,], collapse="") # "11"
apply(df, 1, paste, collapse="") # "A1" "B2" "C3" "D4" "E5"
No logic in it except that it will probably make sense once you know the internals of every function.
The factors seem to be converted to integers when an argument is converted to vector (as you know, data frames are lists of vectors of equal length, so the first row of a data frame is also a list, and when it is forced to be a vector, something like this happens:)
df[1,]
# letters numbers
# 1 A 1
unlist(df[1,])
# letters numbers
# 1 1
I don't know how apply achieves what it does (i.e., factors are represented by character values) -- if you're interested, look at its source code. It may be useful to know, though, that you can trust (in this specific sense) apply (in this specific occasion). More generally, it is useful to store every piece of data in a sensible format, that includes storing strings as strings, i.e., using stringsAsFactors=FALSE.
Btw, every introductory R book should have this idea in a subtitle. For example, my plan for retirement is to write "A (not so) gentle introduction to the zen of data fishery with R, the stringsAsFactors=FALSE way".
I would like to take a data frame with characters and numbers, and concatenate all of the elements of the each row into a single string, which would be stored as a single element in a vector. As an example, I make a data frame of letters and numbers, and then I would like to concatenate the first row via the paste function, and hopefully return the value "A1"
df <- data.frame(letters = LETTERS[1:5], numbers = 1:5)
df
## letters numbers
## 1 A 1
## 2 B 2
## 3 C 3
## 4 D 4
## 5 E 5
paste(df[1,], sep =".")
## [1] "1" "1"
So paste is converting each element of the row into an integer that corresponds to the 'index of the corresponding level' as if it were a factor, and it keeps it a vector of length two. (I know/believe that factors that are coerced to be characters behave in this way, but as R is not storing df[1,] as a factor at all (tested by is.factor(), I can't verify that it is actually an index for a level)
is.factor(df[1,])
## [1] FALSE
is.vector(df[1,])
## [1] FALSE
So if it is not a vector then it makes sense that it is behaving oddly, but I can't coerce it into a vector
> is.vector(as.vector(df[1,]))
[1] FALSE
Using as.character did not seem to help in my attempts
Can anyone explain this behavior?
While others have focused on why your code isn't working and how to improve it, I'm going to try and focus more on getting the result you want. From your description, it seems you can readily achieve what you want using paste:
df <- data.frame(letters = LETTERS[1:5], numbers = 1:5, stringsAsFactors=FALSE)
paste(df$letters, df$numbers, sep=""))
## [1] "A1" "B2" "C3" "D4" "E5"
You can change df$letters to character using df$letters <- as.character(df$letters) if you don't want to use the stringsAsFactors argument.
But let's assume that's not what you want. Let's assume you have hundreds of columns and you want to paste them all together. We can do that with your minimal example too:
df_args <- c(df, sep="")
do.call(paste, df_args)
## [1] "A1" "B2" "C3" "D4" "E5"
EDIT: Alternative method and explanation:
I realised the problem you're having is a combination of the fact that you're using a factor and that you're using the sep argument instead of collapse (as #adibender picked up). The difference is that sep gives the separator between two separate vectors and collapse gives separators within a vector. When you use df[1,], you supply a single vector to paste and hence you must use the collapse argument. Using your idea of getting every row and concatenating them, the following line of code will do exactly what you want:
apply(df, 1, paste, collapse="")
Ok, now for the explanations:
Why won't as.list work?
as.list converts an object to a list. So it does work. It will convert your dataframe to a list and subsequently ignore the sep="" argument. c combines objects together. Technically, a dataframe is just a list where every column is an element and all elements have to have the same length. So when I combine it with sep="", it just becomes a regular list with the columns of the dataframe as elements.
Why use do.call?
do.call allows you to call a function using a named list as its arguments. You can't just throw the list straight into paste, because it doesn't like dataframes. It's designed for concatenating vectors. So remember that dfargs is a list containing a vector of letters, a vector of numbers and sep which is a length 1 vector containing only "". When I use do.call, the resulting paste function is essentially paste(letters, numbers, sep).
But what if my original dataframe had columns "letters", "numbers", "squigs", "blargs" after which I added the separator like I did before? Then the paste function through do.call would look like:
paste(letters, numbers, squigs, blargs, sep)
So you see it works for any number of columns.
For those using library(tidyverse), you can simply use the unite function.
new.df <- df%>%
unite(together, letters, numbers, sep="")
This will give you a new column called together with A1, B2, etc.
This is indeed a little weird, but this is also what is supposed to happen.
When you create the data.frame as you did, column letters is stored as factor. Naturally factors have no ordering, therefore when as.numeric() is applied to a factor it returns the ordering of of the factor. For example:
> df[, 1]
[1] A B C D E
Levels: A B C D E
> as.numeric(df[, 1])
[1] 1 2 3 4 5
A is the first level of the factor df[, 1] therefore A gets converted to the value 1, when as.numeric is applied. This is what happens when you call paste(df[1, ]). Since columns 1 and 2 are of different class, paste first transforms both elements of row 1 to numeric then to characters.
When you want to concatenate both columns, you first need to transform the first row to character:
df[, 1] <- as.character(df[, 1])
paste(df[1,], collapse = "")
As #sebastian-c pointed out, you can also use stringsAsFactors = FALSE in the creation of the data.frame, then you can omit the as.character() step.
if you want to start with
df <- data.frame(letters = LETTERS[1:5], numbers = 1:5, stringsAsFactors=TRUE)
.. then there is no general rule about how df$letters will be interpreted by any given function. It's a factor for modelling functions, character for some and integer for some others. Even the same function such as paste may interpret it differently, depending on how you use it:
paste(df[1,], collapse="") # "11"
apply(df, 1, paste, collapse="") # "A1" "B2" "C3" "D4" "E5"
No logic in it except that it will probably make sense once you know the internals of every function.
The factors seem to be converted to integers when an argument is converted to vector (as you know, data frames are lists of vectors of equal length, so the first row of a data frame is also a list, and when it is forced to be a vector, something like this happens:)
df[1,]
# letters numbers
# 1 A 1
unlist(df[1,])
# letters numbers
# 1 1
I don't know how apply achieves what it does (i.e., factors are represented by character values) -- if you're interested, look at its source code. It may be useful to know, though, that you can trust (in this specific sense) apply (in this specific occasion). More generally, it is useful to store every piece of data in a sensible format, that includes storing strings as strings, i.e., using stringsAsFactors=FALSE.
Btw, every introductory R book should have this idea in a subtitle. For example, my plan for retirement is to write "A (not so) gentle introduction to the zen of data fishery with R, the stringsAsFactors=FALSE way".