I would like to understand how to subset multiple columns from same data frame by matching the first 5 letters of the column names with each other and if they are equal then subset it and store it in a new variable.
Here is a small explanation of my required output. It is described below,
Lets say the data frame is eatable
fruits_area fruits_production vegetable_area vegetable_production
12 100 26 324
33 250 40 580
66 510 43 581
eatable <- data.frame(c(12,33,660),c(100,250,510),c(26,40,43),c(324,580,581))
names(eatable) <- c("fruits_area", "fruits_production", "vegetables_area",
"vegetable_production")
I was trying to write a function which will match the strings in a loop and will store the subset columns after matching first 5 letters from the column names.
checkExpression <- function(dataset,str){
dataset[grepl((str),names(dataset),ignore.case = TRUE)]
}
checkExpression(eatable,"your_string")
The above function checks the string correctly but I am confused how to do matching among the column names in the dataset.
Edit:- I think regular expressions would work here.
You could try:
v <- unique(substr(names(eatable), 0, 5))
lapply(v, function(x) eatable[grepl(x, names(eatable))])
Or using map() + select_()
library(tidyverse)
map(v, ~select_(eatable, ~matches(.)))
Which gives:
#[[1]]
# fruits_area fruits_production
#1 12 100
#2 33 250
#3 660 510
#
#[[2]]
# vegetables_area vegetable_production
#1 26 324
#2 40 580
#3 43 581
Should you want to make it into a function:
checkExpression <- function(df, l = 5) {
v <- unique(substr(names(df), 0, l))
lapply(v, function(x) df[grepl(x, names(df))])
}
Then simply use:
checkExpression(eatable, 5)
I believe this may address your needs:
checkExpression <- function(dataset,str){
cols <- grepl(paste0("^",str),colnames(dataset),ignore.case = TRUE)
subset(dataset,select=colnames(dataset)[cols])
}
Note the addition of "^" to the pattern used in grepl.
Using your data:
checkExpression(eatable,"fruit")
## fruits_area fruits_production
##1 12 100
##2 33 250
##3 660 510
checkExpression(eatable,"veget")
## vegetables_area vegetable_production
##1 26 324
##2 40 580
##3 43 581
Your function does exactly what you want but there was a small error:
checkExpression <- function(dataset,str){
dataset[grepl((str),names(dataset),ignore.case = TRUE)]
}
Change the name of the object from which your subsetting from obje to dataset.
checkExpression(eatable,"fr")
# fruits_area fruits_production
#1 12 100
#2 33 250
#3 660 510
checkExpression(eatable,"veg")
# vegetables_area vegetable_production
#1 26 324
#2 40 580
#3 43 581
Related
Instead of looking at the first n rows of a data frame, as head(mydf) does, or the last n as tail(mydf) does, it occurs to me that I would often rather see n evenly-spaced rows, including the first and the last row. For example, if a data frame had 601 rows, this hypothetical function would display row 1, 101, 201, 301, 401, 501, and 601, assuming that 6 is the default number, as it is for head() and tail().
Is there a built-in function of some package that does this, and if not what would be the best way to implement?
For example, for the data frame mydf <- data.frame(name=letters, value=101:126), I would want the output of an alternative to head() called myview() to be something like:
> myview(mydf)
name value
1 a 101
6 f 106
11 k 111
16 p 116
21 u 121
26 z 126
You can directly do this in seq :
looksee <- function(df, n = 6) df[seq(1, nrow(df), length.out = n),]
looksee(mydf)
# name value
#1 a 101
#6 f 106
#11 k 111
#16 p 116
#21 u 121
#26 z 126
looksee(mydf, 10)
# name value
#1 a 101
#3 c 103
#6 f 106
#9 i 109
#12 l 112
#14 n 114
#17 q 117
#20 t 120
#23 w 123
#26 z 126
This is my try at implementing, but it is probably not very robust compared to head()--it will only work for things that nrow() works for, for one thing.
looksee <- function(df, n=6){
q <- seq(0, 1, length.out=n)
n = nrow(df)
rows <- round(quantile(1:n, probs=q))
return(df[rows,])
}
Example usage:
> mydf <- data.frame(name=letters, value=101:126)
> looksee(mydf)
name value
1 a 101
6 f 106
11 k 111
16 p 116
21 u 121
26 z 126
I have a list of lists, like so:
x <-list()
x[[1]] <- c('97', '342', '333')
x[[2]] <- c('97','555','556','742','888')
x[[3]] <- c ('100', '442', '443', '444', '445','446')
The first number in each list (97, 97, 100) refers to a node in a tree and the following numbers refer to traits associated with that node.
My goal is to create a dataframe that looks like this:
df= data.frame(node = c('97','97','97','97','97','97','100','100','100','100','100'),
trait = c('342','333','555','556','742','888','442','443','444','445','446'))
where each trait has its corresponding node.
I think the first thing I need to do is convert the list of lists into a single dataframe. I've tried doing so using:
do.call(rbind,x)
but that repeats the values in x[[1]] and x[[2]] to match the length of x[[3]]. I've also tried using:
dt_list <- map(x, as.data.table)
dt <- rbindlist(dt_list, fill = TRUE, idcol = T)
Which I think gets me closer, but I'm still unsure of how to assign the first node value to the corresponding trait values. I know this is probably a simple task but it's stumping me today!
Maybe you can try the code below
h <- sapply(x, `[`,1)
d <- lapply(x, `[`,-1)
df <- data.frame(node = rep(h,lengths(d)), trait = unlist(d))
such that
> df
node trait
1 97 342
2 97 333
3 97 555
4 97 556
5 97 742
6 97 888
7 100 442
8 100 443
9 100 444
10 100 445
11 100 446
You can create a data frame with the first value from the vector in column 'node' and the rest of the values in column 'trait'. This strategy can be applied to all entries in the list using the map_df() function from purrr package, giving the output you describe.
library(purrr)
library(dplyr)
x %>%
map_df(., function(vec) data.frame(node = vec[1],
trait = vec[-1],
stringsAsFactors = F))
An option with base R is
stack(setNames(lapply(x, `[`, -1), sapply(x, `[`, 1)))[2:1]
# ind values
#1 97 342
#2 97 333
#3 97 555
#4 97 556
#5 97 742
#6 97 888
#7 100 442
#8 100 443
#9 100 444
#10 100 445
#11 100 446
Another solution
library(tidyverse)
library(purrr)
node <- map(x, ~rep(.x[1], length(.x)-1)) %>% flatten_chr()
trait <- map(x, ~.x[2:length(.x)]) %>% flatten_chr()
out <- tibble(node, trait)
node trait
<chr> <chr>
1 97 342
2 97 333
3 97 555
4 97 556
5 97 742
6 97 888
7 100 442
8 100 443
9 100 444
10 100 445
11 100 446
I have a binomail dataset that looks like this:
df <- data.frame(replicate(4,sample(1:200,1000,rep=TRUE)))
addme <- data.frame(replicate(1,sample(0:1,1000,rep=TRUE)))
df <- cbind(df,addme)
df <-df[order(df$replicate.1..sample.0.1..1000..rep...TRUE..),]
The data is currently soreted in a way to show the instances belonging to 0 group then the ones belonging to the 1 group. Is there a way I can sort the data in a 0-1-0-1-0... fashion? I mean to show a row that belongs to the 0 group, the row after belonging to the 1 group then the zero group and so on...
All I can think about is complex functions. I hope there's a simple way around it.
Thank you,
Here's an attempt, which will add any extra 1's at the end:
First make some example data:
set.seed(2)
df <- data.frame(replicate(4,sample(1:200,10,rep=TRUE)),
addme=sample(0:1,10,rep=TRUE))
Then order:
with(df, df[unique(as.vector(rbind(which(addme==0),which(addme==1)))),])
# X1 X2 X3 X4 addme
#2 141 48 78 33 0
#1 37 111 133 3 1
#3 115 153 168 163 0
#5 189 82 70 103 1
#4 34 37 31 174 0
#6 189 171 98 126 1
#8 167 46 72 57 0
#7 26 196 30 169 1
#9 94 89 193 134 1
#10 110 15 27 31 1
#Warning message:
#In rbind(which(addme == 0), which(addme == 1)) :
# number of columns of result is not a multiple of vector length (arg 1)
Here's another way using dplyr, which would make it suitable for within-group ordering. It's also probably pretty quick. If there's unbalanced numbers of 0's and 1's, it will leave them at the end.
library(dplyr)
df %>%
arrange(addme) %>%
mutate(n0 = sum(addme == 0),
orderme = seq_along(addme) - (n0 * addme) + (0.5 * addme)) %>%
arrange(orderme) %>%
select(-n0, -orderme)
I have written a function that will compare the similarity of IP addresses, and will let the user select the level of detail in the octet. for example, in the address 255.255.255.0 and 255.255.255.1, a user could specify that they only want to compare the first, first and second, first second third etc. octets.
the function is below:
did.change.ip=function(vec, detail){
counter=2
result.vec=FALSE
r.list=strsplit(vec, '.', fixed=TRUE)
for(i in vec){
if(counter>length(vec)){
break
}
first=as.numeric(r.list[[counter-1]][1:detail])
second=as.numeric(r.list[[counter]][1:detail])
if(sum(first==second)==detail){
result.vec=append(result.vec,FALSE)
}
else{
result.vec=append(result.vec,TRUE)
}
counter=counter+1
}
return(result.vec)
}
and it's really slow once the data starts getting larger. for a dataset of 500,000 rows, the system.time() results are:
user system elapsed
208.36 0.59 209.84
are there any R power users who have insight on how to write this more efficiently? I know lapply() is the preferred method for looping over vectors/dataframes, but I'm stumped as to how to access the previous element in a vector for this purpose. I've tried to sketch something out quickly, but It returns a syntax error:
test=function(vec, detail){
rlist=strsplit(vec, '.', fixed=TRUE)
r.value=vapply(rlist, function(x,detail) ifelse(x[1:detail]==x[1:detail] TRUE, FALSE))
}
I've created some sample data for testing purposes below:
stack.data=structure(list(V1 = c("247.116.209.66", "195.121.47.105", "182.136.49.12",
"237.123.100.50", "120.30.174.18", "29.85.72.70", "18.186.76.177",
"33.248.142.26", "109.97.92.50", "217.138.155.145", "20.203.156.2",
"71.1.51.190", "31.225.208.60", "55.25.129.73", "211.204.249.244",
"198.137.15.53", "234.106.102.196", "244.3.87.9", "205.242.10.22",
"243.61.212.19", "32.165.79.86", "190.207.159.147", "157.153.136.100",
"36.151.152.15", "2.254.210.246", "3.42.1.208", "30.11.229.18",
"72.187.36.103", "98.114.189.34", "67.93.180.224")), .Names = "V1", class = "data.frame", row.names = c(NA,
-30L))
Here's another solution just using base R.
did.change.ip <- function(vec, detail=4){
ipv <- scan(text=paste(vec, collapse="\n"),
what=c(replicate(detail, integer()), replicate(4-detail,NULL)),
sep=".", quiet=TRUE)
c(FALSE, rowSums(vapply(ipv[!sapply(ipv, is.null)],
diff, integer(length(vec)-1))!=0)>0)
}
Here we use scan() to break up the ip address into numbers. Then we we look down each octet for differences using diff. It seems this is faster than the original proposal, but slightly slower than #josilber's stringr solution (using microbenchmark with 3,000 ip addresses)
Unit: milliseconds
expr min lq median uq max neval
orig 35.251886 35.716921 36.019354 36.700550 90.159992 100
scan 2.062189 2.116391 2.170110 2.236658 3.563771 100
strngr 2.027232 2.075018 2.136114 2.200096 3.535227 100
The simplest way I can think of to do this is to build a transformed vector that only includes the parts of the IP you want. Then it's a one-liner to check if each element is equal to the one before it:
library(stringr)
did.change.josilber <- function(vec, detail) {
s <- str_extract(vec, paste0("^(\\d+\\.){", detail, "}"))
return(s != c(s[1], s[1:(length(s)-1)]))
}
This seems reasonably efficient for 500,000 rows:
set.seed(144)
big.vec <- sample(stack.data[,1], 500000, replace=T)
system.time(did.change.josilber(big.vec, 3))
# user system elapsed
# 0.527 0.030 0.554
The biggest issue with your code is that you call append each iteration, which requires reallocation of your vector 500,000 times. You can read more about this in the second circle of the R inferno.
Not sure if all you want is counts, but this is potentially a solution:
library(dplyr)
library(tidyr)
# split ip addresses into "octets"
octets <- stack.data %>%
separate(V1,c("first","second","third","fourth"))
# how many shared both their first and second octets?
octets %>%
group_by(first,second) %>%
summarize(n = n())
first second n
1 109 97 1
2 120 30 1
3 157 153 1
4 18 186 1
5 182 136 1
6 190 207 1
7 195 121 1
8 198 137 1
9 2 254 1
10 20 203 1
11 205 242 1
12 211 204 1
13 217 138 1
14 234 106 1
15 237 123 1
16 243 61 1
17 244 3 1
18 247 116 1
19 29 85 1
20 3 42 1
21 30 11 1
22 31 225 1
23 32 165 1
24 33 248 1
25 36 151 1
26 55 25 1
27 67 93 1
28 71 1 1
29 72 187 1
30 98 114 1
I have a data frame having 20 columns. I need to filter / remove noise from one column. After filtering using convolve function I get a new vector of values. Many values in the original column become NA due to filtering process. The problem is that I need the whole table (for later analysis) with only those rows where the filtered column has values but I can't bind the filtered column to original table as the number of rows for both are different. Let me illustrate using the 'age' column in 'Orange' data set in R:
> head(Orange)
Tree age circumference
1 1 118 30
2 1 484 58
3 1 664 87
4 1 1004 115
5 1 1231 120
6 1 1372 142
Convolve filter used
smooth <- function (x, D, delta){
z <- exp(-abs(-D:D/delta))
r <- convolve (x, z, type='filter')/convolve(rep(1, length(x)),z,type='filter')
r <- head(tail(r, -D), -D)
r
}
Filtering the 'age' column
age2 <- smooth(Orange$age, 5,10)
data.frame(age2)
The number of rows for age column and age2 column are 35 and 15 respectively. The original dataset has 2 more columns and I like to work with them also. Now, I only need 15 rows of each column corresponding to the 15 rows of age2 column. The filter here removed first and last ten values from age column. How can I apply the filter in a way that I get truncated dataset with all columns and filtered rows?
You would need to figure out how the variables line up. If you can add NA's to age2 and then do Orange$age2 <- age2 followed by na.omit(Orange) you should have what you want. Or, equivalently, perhaps this is what you are looking for?
df <- tail(head(Orange, -10), -10) # chop off the first and last 10 observations
df$age2 <- age2
df
Tree age circumference age2
11 2 1004 156 915.1678
12 2 1231 172 876.1048
13 2 1372 203 841.3156
14 2 1582 203 911.0914
15 3 118 30 948.2045
16 3 484 51 1008.0198
17 3 664 75 955.0961
18 3 1004 108 915.1678
19 3 1231 115 876.1048
20 3 1372 139 841.3156
21 3 1582 140 911.0914
22 4 118 32 948.2045
23 4 484 62 1008.0198
24 4 664 112 955.0961
25 4 1004 167 915.1678
Edit: If you know the first and last x observations will be removed then the following works:
x <- 2
df <- tail(head(Orange, -x), -x) # chop off the first and last x observations
df$age2 <- age2