This may be a very simple question, but I don't see how to answer it.
I have the following reproducible code, where I have two small dataframes that I use to calculate a percentage value based on each column total:
#dataframe x
x <- structure(list(PROV = structure(c(1L, 1L), .Label = "AG", class = "factor"),
APT = structure(1:2, .Label = c("AAA", "BBB"), class = "factor"),
PAX.2013 = c(5L, 4L), PAX.2014 = c(4L, 2L), PAX.2015 = c(4L,0L)),
.Names = c("PROV", "APT", "PAX.2013", "PAX.2014", "PAX.2015"),
row.names = 1:2, class = "data.frame")
#dataframe y
y <- structure(list(PROV = structure(c(1L, 1L), .Label = "AQ", class = "factor"),
APT = structure(1:2, .Label = c("CCC", "AAA"), class = "factor"),
PAX.2013 = c(3L, 7L), PAX.2014 = c(2L, 1L), PAX.2015 = c(0L,3L)),
.Names = c("PROV", "APT", "PAX.2013", "PAX.2014", "PAX.2015"),
row.names = 1:2, class = "data.frame")
#list z (with x and y)
z <- list(x,y)
#percentage value of x and y based on columns total
round(prop.table(as.matrix(z[[1]][3:5]), margin = 2)*100,1)
round(prop.table(as.matrix(z[[2]][3:5]), margin = 2)*100,1)
as you can see, it works just fine.
Now I want to automate for all the list, but I can't figure out how to get the results. This is my simple code:
#for-loop that is not working
for (i in length(z))
{round(prop.table(as.matrix(z[[i]][3:5]), margin = 2)*100,1)}
You have two problems.
First, you have not put a range into your for loop so you are just trying to iterate over a single number and second, you are not assigning your result anywhere on each iteration.
Use 1:length(z) to define a range. Then assign the results to a variable.
This would work:
my_list <- list()
for (i in 1:length(z)){
my_list[[i]] <- round(prop.table(as.matrix(z[[i]][3:5]),
margin = 2)*100,1)
}
my_list
But it would be more efficient and idiomatic to use lapply:
lapply(1:length(z),
function(x) round(prop.table(as.matrix(z[[x]][3:5]), margin = 2)*100,1))
Barring discussions whether for-loops is the best approach, you had two issues. One, your for loop only iterates over 2 (which is length(z)) instead of 1:2. Two, you need to do something with the round(....) statement. In this solution, I added a print statement.
for (i in 1:length(z)){
print(round(prop.table(as.matrix(z[[i]][3:5]), margin = 2)*100,1))
}
Related
This thread follows on from this answered qestion: Matching strings loop over multiple columns
I opened a new thread as I would like to make an update to flag for exact matches only..
I have a table of key words in separate colums as follows:
#codes table
codes <- structure(
list(
Support = structure(
c(2L, 3L, NA),
.Label = c("",
"help", "questions"),
class = "factor"
),
Online = structure(
c(1L,
3L, 2L),
.Label = c("activities", "discussion board", "quiz", "sy"),
class = "factor"
),
Resources = structure(
c(3L, 2L, NA),
.Label = c("", "pdf",
"textbook"),
class = "factor"
)
),
row.names = c(NA,-3L),
class = "data.frame"
)
I also have a comments table structured as follows:
#comments table
comments <- structure(
list(
SurveyID = structure(
1:5,
.Label = c("ID_1", "ID_2",
"ID_3", "ID_4", "ID_5"),
class = "factor"
),
Open_comments = structure(
c(2L,
4L, 3L, 5L, 1L),
.Label = c(
"I could never get the pdf to download",
"I could never get the system to work",
"I didn’t get the help I needed on time",
"my questions went unanswered",
"staying motivated to get through the textbook",
"there wasn’t enough engagement in the discussion board"
),
class = "factor"
)
),
class = "data.frame",
row.names = c(NA,-5L)
)
What I am trying to do:
Search for an exact match keyword. The following working code has been provided by #Len Greski and #Ronak Shah from the previous thread (with huge thanks to both):
resultsList <- lapply(1:ncol(codes),function(x){
y <- stri_detect_regex(comments$Open_comments,paste(codes[[x]],collapse = "|"))
ifelse(y == TRUE,1,0)
})
results <- as.data.frame(do.call(cbind,resultsList))
colnames(results) <- colnames(codes)
mergedData <- cbind(comments,results)
mergedData
and
comments[names(codes)] <- lapply(codes, function(x)
+(grepl(paste0(na.omit(x), collapse = "|"), comments$Open_comments)))
Both work great but I have come across a snag and now need to match the keywords exactly. As per the example tables above, if I have a keyword "sy", the code will flag any comment with the word "system". I would modify either of the above pieces of code to flag the comment where only "sy" exact match is present.
Many thanks
I have a functions which yields 2 dataframes. As functions can only return one object, I combined these dataframes as a list. However, I need to work with both dataframes separately. Is there a way to automatically split the list into the component dataframes, or to write the function in a way that both objects are returned separately?
The function:
install.packages("plyr")
require(plyr)
fun.docmerge <- function(x, y, z, crit, typ, doc = checkmerge) {
mergedat <- paste(deparse(substitute(x)), "+",
deparse(substitute(y)), "=", z)
countdat <- nrow(x)
check_t1 <- data.frame(mergedat, countdat)
z1 <- join(x, y, by = crit, type = typ)
countdat <- nrow(z1)
check_t2 <- data.frame(mergedat, countdat)
doc <- rbind(doc, check_t1, check_t2)
t1<-list()
t1[["checkmerge"]]<-doc
t1[[z]]<-z1
return(t1)
}
This is the call to the function, saving the result list to the new object results.
results <- fun.docmerge(x = df1, y = df2, z = "df3", crit = c("id"), typ = "left")
In the following sample data to replicate the problem:
df1 <- structure(list(id = c("XXX1", "XXX2", "XXX3",
"XXX4"), tr.isincode = c("ISIN1", "ISIN2",
"ISIN3", "ISIN4")), .Names = c("id", "isin"
), row.names = c(NA, 4L), class = "data.frame")
df2 <- structure(list(id= c("XXX1", "XXX5"), wrong= c(1L,
1L)), .Names = c("id", "wrong"), row.names = 1:2, class = "data.frame")
checkmerge <- structure(list(mergedat = structure(integer(0), .Label = character(0), class = "factor"),
countdat = numeric(0)), .Names = c("mergedat", "countdat"
), row.names = integer(0), class = "data.frame")
In the example, a list with the dataframes df3 and checkmerge are returned. I would need both dataframes separately. I know that I could do it via manual assignment (e.g., checkmerge <- results$checkmerge) but I want to eliminate manual changes as much as possible and am therefore looking for an automated way.
I am learning the use of the ifelse function from Zuur et al (2009) A Beginners guide to R. In one exercise, there is a data frame called Owls which contains data about about 27 nests and two night of observations.
structure(list(Nest = structure(c(1L, 1L, 1L, 1L), .Label = "AutavauxTV", class = "factor"),
FoodTreatment = structure(c(1L, 2L, 1L, 1L), .Label = c("Deprived",
"Satiated"), class = "factor"), SexParent = structure(c(1L,
1L, 1L, 1L), .Label = "Male", class = "factor"), ArrivalTime = c(22.25,
22.38, 22.53, 22.56), SiblingNegotiation = c(4L, 0L, 2L,
2L), BroodSize = c(5L, 5L, 5L, 5L), NegPerChick = c(0.8,
0, 0.4, 0.4)), .Names = c("Nest", "FoodTreatment", "SexParent",
"ArrivalTime", "SiblingNegotiation", "BroodSize", "NegPerChick"
), row.names = c(NA, 4L), class = "data.frame")
The two nights differed as to the feeding regime (satiated or deprived) and are indicated in the Foodregime variable. The task is to use ifelse and past functions that make a new categorical variable that defines observations from a single night at a particular nest.
In the solutions the following code is suggested:
Owls <- read.table(file = "Owls.txt", header = TRUE, dec = ".")
ifelse(Owls$FoodTreatment == "Satiated", Owls$NestNight <- paste(Owls$Nest, "1",sep = "_"), Owls$NestNight <- paste(Owls$Nest, "2",sep = "_"))
and apparently it creates a new variable with values the endings of which vary ("-1" or "-2")
however when I call the original dataframe, all "-1" endings in the NestNight variable disappears and are turned to "-2."
Why does this happen? Did the authors miss something from the code or it's me who is not getting it?
Many thanks
EDIT: Sorry, I wanted to give a reproducible example by copying my data using dput but it did not work. If you can let me know how I can correct it so that it appears properly, I'd be grateful too!
Solution
If you do the assignment outside the ifelse structure, it works:
Owls$NestNight <- ifelse(Owls$FoodTreatment == "Satiated",
paste(Owls$Nest, "1",sep = ""),
paste(Owls$Nest, "2",sep = ""))
Explanation
What happens in your case is simply if you would execute the following two lines:
Owls$NestNight <- paste(Owls$Nest, "1",sep = "")
Owls$NestNight <- paste(Owls$Nest, "2",sep = "")
You first assign paste(Owls$Nest, "1",sep = "") to Owls$NestNight and then you reassign paste(Owls$Nest, "2",sep = "") to it. The ifelse is not affected by this, but you don't assign it's result to any variable.
Maybe it is more clear if you test this simple code:
c(a <- 1:5, a <- 6:10) #c is your ifelse, a is your Owls$NestNight
a #[1] 6 7 8 9 10
I have a problem connecting two points with the same y value. My dataset looks like this (I hope the formatting is ok):
attackerip,min,max
125.88.146.123,2016-03-29 17:38:17.949778,2016-03-30 07:28:47.912983
58.218.205.101,2016-04-05 15:53:20.69986,2016-05-12 17:32:08.583255
183.3.202.195,2016-04-05 15:58:27.862509,2016-04-15 18:15:13.117774
58.218.199.166,2016-04-05 16:09:34.448588,2016-04-24 06:02:12.237922
58.218.204.107,2016-04-05 16:57:17.624509,2016-05-31 00:52:44.007908
What I have so far is the following:
mydata = read.csv("timeline.csv", sep=',')
mydata$min <- strptime(as.character(mydata$min), format='%Y-%m-%d %H:%M:%S')
mydata$max <- strptime(as.character(mydata$max), format='%Y-%m-%d %H:%M:%S')
plot(mydata$min, mydata$attackerip, col="red")
points(mydata$max, mydata$attackerip, col="blue")
Which results in:
Now I want to connect the points with the same y-axis value. And can not get lines or abline to work. Thanks in Advance!
EDIT: dput of data
dput(mydata)
structure(list(attackerip = structure(c(1L, 5L, 2L, 3L, 4L), .Label = c("125.88.146.123",
"183.3.202.195", "58.218.199.166", "58.218.204.107", "58.218.205.101"
), class = "factor"), min = structure(1:5, .Label = c("2016-03-29 17:38:17.949778",
"2016-04-05 15:53:20.69986", "2016-04-05 15:58:27.862509", "2016-04-05 16:09:34.448588",
"2016-04-05 16:57:17.624509"), class = "factor"), max = structure(c(1L,
4L, 2L, 3L, 5L), .Label = c("2016-03-30 07:28:47.912983", "2016-04-15 18:15:13.117774",
"2016-04-24 06:02:12.237922", "2016-05-12 17:32:08.583255", "2016-05-31 00:52:44.007908"
), class = "factor")), .Names = c("attackerip", "min", "max"), class = "data.frame", row.names = c(NA,
-5L))
Final Edit:
The reason why plotting lines did not work was, that the datatype of min and max was timestamps. Casting those to numeric values yielded the expected result. Thanks for your help everyone
The lines function should work just fine. However, you will need to call it for every pair (or set) of points that share the same y value. Here is a reproducible example:
# get sets of observations with the same y value
dupeVals <- unique(y[duplicated(y) | duplicated(y, fromLast=T)])
# put the corresponding indices into a list
dupesList <- lapply(dupeVals, function(i) which(y == i))
# scatter plot
plot(x, y)
# plot the lines using sapply
sapply(dupesList, function(i) lines(x[i], y[i]))
This returns
data
set.seed(1234)
x <- sort(5* runif(30))
y <- sample(25, 30, replace=T)
As it appears that you have two separate groups for which you would like draw these lines, the following would be the algorithm:
for each group, (min and max, I believe)
calculate the duplicate values of the y variable
put the indicies of these duplicates into a dupesList (maybe dupesListMin and dupesListMax).
plot the points
run one sapply function on each dupesList.
I have defined the following function in r:
#A function that compares color and dates to determine if there is a match
getTagColor <- function(color, date){
for (i in (1:nrow(TwistTieFix))){
if ((color == TwistTieFix$color_match[i]) &
(date > TwistTieFix$color_match[i]) &
(date <= TwistTieFix$julian_cut_off_date[i])) {
Data$color_code <- TwistTieFix$color_code[i]
print(Data$color_code)
}
}
}
I then used apply() in an attempt to apply the function to each row.
#Apply the above function to the data set
testData <- apply(Data, 1, getTagColor(Data$tag_color,Data$julian_date))`
The goal of the code is to use two variables in Data and find another value to put into a new column in Data (color_code) that will be based on the information in TwistTieFix. When I run the code, I get a list of warnings saying
In if ((color == TwistTieFix$color_match[i]) & (date > ... :
the condition has length > 1 and only the first element will be used
I cannot determine why the function does not use the date and color from each row and use it in the function (at least that is what I think is going wrong here). Thanks!
Here are examples of the data frames being used:
TwistTieFix
color_name date color_code cut_off_date color_match julian_start julian_cut_off_date
yellow 2013-08-12 y1 2001-07-02 yellow 75 389
blue 2000-09-28 b1 2001-08-12 blue 112 430
Data
coll_date julian_date tag_color
2013-08-13 76 yellow
2013-08-14 76 yellow
2000-09-29 112 blue
Data has a lot more columns of different variables, but I am not allowed to include all of the columns. However, I have included the columns in Data that I am referencing in function. The data sets are loaded into r using read.csv and are from Excel csv files.
To me, it seems like you want to join Data and TwistTieFix where tag_color=color_match and julian_start <= julian_date <= julian_cut_off_date. Here are your sample data.sets in dput form
TwistTieFix <- structure(list(color_name = structure(c(2L, 1L), .Label = c("blue",
"yellow"), class = "factor"), date = structure(c(2L, 1L), .Label = c("2000-09-28",
"2013-08-12"), class = "factor"), color_code = structure(c(2L,
1L), .Label = c("b1", "y1"), class = "factor"), cut_off_date = structure(1:2, .Label = c("2001-07-02",
"2001-08-12"), class = "factor"), color_match = structure(c(2L,
1L), .Label = c("blue", "yellow"), class = "factor"), julian_start = c(75L,
112L), julian_cut_off_date = c(389L, 430L)), .Names = c("color_name",
"date", "color_code", "cut_off_date", "color_match", "julian_start",
"julian_cut_off_date"), class = "data.frame", row.names = c(NA,
-2L))
Data <- structure(list(coll_date = structure(c(2L, 3L, 1L), .Label = c("2000-09-29",
"2013-08-13", "2013-08-14"), class = "factor"), julian_date = c(76L,
76L, 112L), tag_color = structure(c(2L, 2L, 1L), .Label = c("blue",
"yellow"), class = "factor")), .Names = c("coll_date", "julian_date",
"tag_color"), class = "data.frame", row.names = c(NA, -3L))
An easy way to perform this merge would be using the data.table library. You can do
#convert to data.table and set keys
ttf<-setDT(TwistTieFix)
setkey(ttf, color_match, julian_start)
dt<-setDT(Data)
setkey(dt, tag_color, julian_date)
#merge and extract columns
ttf[dt, roll=T][julian_start<julian_cut_off_date,list(coll_date,
julian_date=julian_start, tag_color=color_match, color_code)]
to get
coll_date julian_date tag_color color_code
1: 2000-09-29 112 blue b1
2: 2013-08-13 76 yellow y1
3: 2013-08-14 76 yellow y1