I need to standardize how subgroups are referred to in a data set. To do this I need to identify when a variable matches one of several strings and then set a new variable with the standardized name. I am trying to do that with the following:
df <- data.frame(a = c(1,2,3,4), b = c(depression_male, depression_female, depression_hsgrad, depression_collgrad))
TestVector <- "male"
for (i in TestVector) {
df$grpl <- grepl(paste0(i), df$b)
df[ which(df$grpl == TRUE),]$standard <- "male"
}
The test vector will frequently have multiple elements. The grepl works (I was going to deal with the male/female match confusion later but I'll take suggestions on that) but the subsetting and setting a new variable doesn't. It would be better (and work) if I could transform the grepl output directly into the standard name variable.
Your only real issue is that you need to initialize the standard column. But we can simplify your code a bit:
df <- data.frame(a = c(1,2,3,4), b = c("depression_male", "depression_female", "depression_hsgrad", "depression_collgrad"))
TestVector <- "male"
df$standard <- NA
for (i in TestVector) {
df[ grepl(i, df$b), "standard"] <- "male"
}
df
# a b standard
# 1 1 depression_male male
# 2 2 depression_female male
# 3 3 depression_hsgrad <NA>
# 4 4 depression_collgrad <NA>
Then you've got the issue that the "male" pattern matches "female" as well.
Perhaps you're looking for sub instead? It works like find/replace:
df$standard = sub(pattern = "depression_", replacement = "", df$b)
df
# a b standard
# 1 1 depression_male male
# 2 2 depression_female female
# 3 3 depression_hsgrad hsgrad
# 4 4 depression_collgrad collgrad
It's hard to generalize what will be best in your case without more example input/output pairs. If all your data is of the form "depression_" this will work well. Or maybe the standard name is always after an underscore, so you could use pattern = ".*_" to replace everything before the last underscore. Or maybe something else... Hopefully these ideas give you a good start.
Related
HEADLINE: Is there a way to get R to recognize data.frame column names contained within lists in the same way that it can recognize free-floating vectors?
SETUP: Say I have a vector named varA:
(varA <- 1:6)
# [1] 1 2 3 4 5 6
To get the length of varA, I could do:
length(varA)
#[1] 6
and if the variable was contained within a larger list, the variable and its length could still be found by doing:
list <- list(vars = "varA")
length(get(list$vars[1]))
#[1] 6
PROBLEM:
This is not the case when I substitute the vector for a dataframe column and I don't know how to work around this:
rows <- 1:6
cols <- c("colA")
(df <- data.frame(matrix(NA,
nrow = length(rows),
ncol = length(cols),
dimnames = list(rows, cols))))
# colA
# 1 NA
# 2 NA
# 3 NA
# 4 NA
# 5 NA
# 6 NA
list <- list(vars = "varA",
cols = "df$colA")
length(get(list$vars[1]))
#[1] 6
length(get(list$cols[1]))
#Error in get(list$cols[1]) : object 'df$colA' not found
Though this contrived example seems inane, because I could always use the simple length(variable) approach, I'm actually interested in writing data from hundreds of variables varying in lengths onto respective dataframe columns, and so keeping them in a list that I could iterate through would be very helpful. I've tried everything I could think of, but it may be the case that it's just not possible in R, especially given that I cannot find any posts with solutions to the issue.
You could try:
> length(eval(parse(text = list$cols[1])))
[1] 6
Or:
list <- list(vars = "varA",
cols = "colA")
length(df[, list$cols[1]])
[1] 6
Or with regex:
list <- list(vars = "varA",
cols = "df$colA")
length(df[, sub(".*\\$", "", list$cols[1])])
[1] 6
If you are truly working with a data frame d, then nrow(d) is the length of all of the variables in d. There should be no reason to use length in this case.
If you are actually working with a list x containing variables of potentially different lengths, then you should use the [[ operator to extract those variables by name (see ?Extract):
x <- list(a = 1:10, b = rnorm(20L))
l <- list(vars = "a")
length(d[[l$vars[1L]]]) # 10
If you insist on using get (you shouldn't), then you need to supply a second argument telling it where to look for the variable (see ?get):
length(get(l$vars[1L], x)) # 10
Just a quick question: how can I replace some values with others if these values are present in all the dataframe's column? Functions like mapvalues and recode work only if the column is specified, but in my case the dataframe has 89 columns so that would be time-consuming.
For the sake of clarity, take in consideration the following example. I want to replace [NULL] with another value.
Example:
a <- c("NULL",2,"NULL")
b <- c(3, "NULL", 1)
df <- data.frame(a, b)
df
a b
0 NULL 3
1 2 NULL
2 NULL 1
The difference between the example and my case is that the dataset is [35383 x 89], and the values I want to replace are more than one.
Thank you in advance for your time.
An extension to the comment by Ronak Shah. You can add 0 if you want like that. Or you can replace it with desired values, if you like that.
For example, replace the NULLs with mean of the respective columns:
#Run a loop to convert the characters into numbers because for your case it is all characters
#This will change the NULL to NAs.
for (i in colnames(df)){
df[,i] <- as.numeric(df[,i])
}
#Now replace the NAs with the mean of the column
for (i in colnames(df)){
df[,i][is.na(df[,i])] <- mean(df[,i], na.rm=TRUE)
}
You can similarly do this for median also. Let me know in the comment if you have any doubts.
For starters, I have added a few more rows to your example to better show how the code works
df
# a b
#1 NULL 3
#2 2 NULL
#3 NULL 1
#4 a 14
#5 1 a
#6 14 5
First, create two vectors: one with whe values you want to replace (pattern) and one with replacements in the same order. To make sure you have done it right, put them together in a data frame and take a look at the rows (this will also help in next step)
In this case, I want NULL to be 0, "a" to be "alpha", and so on, as shown below
pattern <- c("NULL", "a", 14, 1)
replacement <- c(0, "alpha", "fourteen", "one")
subs <- data.frame(pattern, replacement)
subs
# pattern replacement
#1 NULL 0
#2 a alpha
#3 14 fourteen
#4 1 one
To finish it, we will make a for tthat each time we will pick a pattern and its replacement from the subs data frame we created, and with these values execute a map_df(). This function iterates over the columns from our original data frame (df) and apply the gsub() function with the pattern and replacement
for (i in 1:nrow(subs)) {
df <- map_df(df, gsub, pattern = subs$pattern[i], replacement = subs$replacement[i])
}
df
# a b
#1 0 3
#2 2 0
#3 0 one
#4 alpha fourteen
#5 one alpha
#6 fourteen 5
I hope this was clear. Let me know if you have any doubts
Another question for me as a beginner. Consider this example here:
n = c(2, 3, 5)
s = c("ABBA", "ABA", "STING")
b = c(TRUE, "STING", "STRING")
df = data.frame(n,s,b)
n s b
1 2 ABBA TRUE
2 3 ABA STING
3 5 STING STRING
How can I search within this dataframe for similar strings, i.e. ABBA and ABA as well as STING and STRING and make them the same (doesn't matter whether ABBA or ABA, either fine) that would not require me knowing any variations? My actual data.frame is very big so that it would not be possible to know all the different variations.
I would want something like this returned:
> n = c(2, 3, 5)
> s = c("ABBA", "ABBA", "STING")
> b = c(TRUE, "STING", "STING")
> df = data.frame(n,s,b)
> print(df)
n s b
1 2 ABBA TRUE
2 3 ABBA STING
3 5 STING STING
I have looked around for agrep, or stringdist, but those refer to two data.frames or are able to name the column which I can't since I have many of those.
Anyone an idea? Many thanks!
Best regards,
Steffi
This worked for me but there might be a better solution
The idea is to use a recursive function, special, that uses agrepl, which is the logical version of approximate grep, https://www.rdocumentation.org/packages/base/versions/3.4.1/topics/agrep. Note that you can specify the 'error tolerance' to group similar strings with agrep. Using agrepl, I split off rows with similar strings into x, mutate the s column to the first-occurring string, and then add a grouping variable grp. The remaining rows that were not included in the ith group are stored in y and recursively passed through the function until y is empty.
You need the dplyr package, install.packages("dplyr")
library(dplyr)
desired <- NULL
grp <- 1
special <- function(x, y, grp) {
if (nrow(y) < 1) { # if y is empty return data
return(x)
} else {
similar <- agrepl(y$s[1], y$s) # find similar occurring strings
x <- rbind(x, y[similar,] %>% mutate(s=head(s,1)) %>% mutate(grp=grp))
y <- setdiff(y, y[similar,])
special(x, y, grp+1)
}
}
desired <- special(desired,df,grp)
To change the stringency of string similarity, change max.distance like agrepl(x,y,max.distance=0.5)
Output
n s b grp
1 2 ABBA TRUE 1
2 3 ABBA STING 1
3 5 STING STRING 2
To remove the grouping variable
withoutgrp <- desired %>% select(-grp)
I am trying to train a data that's converted from a document term matrix to a dataframe. There are separate fields for the positive and negative comments, so I wanted to add a string to the column names to serve as a "tag", to differentiate the same word coming from the different fields - for example, the word hello can appear both in the positive and negative comment fields (and thus, represented as a column in my dataframe), so in my model, I want to differentiate these by making the column names positive_hello and negative_hello.
I am looking for a way to rename columns in such a way that a specific string will be appended to all columns in the dataframe. Say, for mtcars, I want to rename all of the columns to have "_sample" at the end, so that the column names would become mpg_sample, cyl_sample, disp_sample and so on, which were originally mpg, cyl, and disp.
I'm considering using sapplyor lapply, but I haven't had any progress on it. Any help would be greatly appreciated.
Use colnames and paste0 functions:
df = data.frame(x = 1:2, y = 2:1)
colnames(df)
[1] "x" "y"
colnames(df) <- paste0('tag_', colnames(df))
colnames(df)
[1] "tag_x" "tag_y"
If you want to prefix each item in a column with a string, you can use paste():
# Generate sample data
df <- data.frame(good=letters, bad=LETTERS)
# Use the paste() function to append the same word to each item in a column
df$good2 <- paste('positive', df$good, sep='_')
df$bad2 <- paste('negative', df$bad, sep='_')
# Look at the results
head(df)
good bad good2 bad2
1 a A positive_a negative_A
2 b B positive_b negative_B
3 c C positive_c negative_C
4 d D positive_d negative_D
5 e E positive_e negative_E
6 f F positive_f negative_F
Edit:
Looks like I misunderstood the question. But you can rename columns in a similar way:
colnames(df) <- paste(colnames(df), 'sample', sep='_')
colnames(df)
[1] "good_sample" "bad_sample" "good2_sample" "bad2_sample"
Or to rename one specific column (column one, in this case):
colnames(df)[1] <- paste('prefix', colnames(df)[1], sep='_')
colnames(df)
[1] "prefix_good_sample" "bad_sample" "good2_sample" "bad2_sample"
You can use setnames from the data.table package, it doesn't create any copy of your data.
library(data.table)
df <- data.frame(a=c(1,2),b=c(3,4))
# a b
# 1 1 3
# 2 2 4
setnames(df,paste0(names(df),"_tag"))
print(df)
# a_tag b_tag
# 1 1 3
# 2 2 4
I want to update a dataframe with values from a table of new values where there is a one-to-many relationship between the dataframe and table of new values. This code illustrates the intent:
df = data.frame(x=rep(letters[1:4],5,rep=T), y=1:20)
and new values..
eds = data.frame(x=c('c','d'), val=c(101, 102))
For a one-to-one relationship the following should work:
df$x[match(eds$x, df$x)] = eds$x[match(df$x, eds$x)]
But match only works with first match, so this throws the error number of items to replace is not a multiple of replacement length. Grateful for any tips on the most efficient way to approach this. I'm guessing some sapply wrapper but I can't think of the method.
Thanks in advance.
tmp <- eds$val[match(df$x, eds$x)] # Matching indices (with NAs for no match)
df$y <- ifelse(is.na(tmp), df$y, tmp) # Values at matches (leaving alone for NAs)
head(df, 5)
# x y
# 1 a 1
# 2 b 2
# 3 c 101
# 4 d 102
# 5 a 5
Not that this not a very robust solution. It depends on your exact data structure here (repeating 'c', 'd' pattern) but it works for this case:
df[df[["x"]] %in% eds[["x"]], "y"] = eds[[2]]