It seems that when we have duplicated data, most of the time we want to remove the duplicated data.
Lets say, we do not want to exclude it, but instead assign it with a new variable.
Taking the following data as a example
b <- c(1:100,1:99,1:104,1:105,1:105)
So we see that between the values for 1-99 are repeated 5 times, the number 100 repeated 4 times, the number 101 repeated 4 times etc.....
How can one search through b (ideally in sequential order), find a repeated/duplicate number and then assign it a new value
Try this if you're interested in assigning one (universal) new value
b <- c(1:100,1:99,1:104,1:105,1:105)
b[duplicated(b)] = 888 # new value
The duplicated command helps you spot the positions of all values that are duplicates in b.
Related
I have a data frame that looks like this:
df <- data.frame(Set = c("A","A","A","B","B","B","B"), Values=c(1,1,2,1,1,2,2))
I want to collapse the data frame so I have one row for A and one for B. I want the Values column for those two rows to reflect the most common Values from the whole dataset.
I could do this as described here (How to find the statistical mode?), but notably when there's a tie (two values that each occur once, therefore no "true" mode) it simply takes the first value.
I'd prefer to use my own hierarchy to determine which value is selected in the case of a tie.
Create a data frame that defines the hierarchy, and assigns each possibility a numeric score.
hi <- data.frame(Poss = unique(df$Set), Nums =c(105,104))
In this case, A gets a numerical value of 105, B gets a numerical score of 104 (so A would be preferred over B in the case of a tie).
Join the hierarchy to the original data frame.
require(dplyr)
matched <- left_join(df, hi, by = c("Set"="Poss"))
Then, add a frequency column to your original data frame that lists the number of times each unique Set-Value combination occurs.
setDT(matched)[, freq := .N, by = c("Set", "Value")]
Now that those frequencies have been recorded, we only need row of each Set-Value combo, so get rid of the rest.
multiplied <- distinct(matched, Set, Value, .keep_all = TRUE)
Now, multiply frequency by the numeric scores.
multiplied$mult <- multiplied$Nums * multiplied$freq
Lastly, sort by Set first (ascending), then mult (descending), and use distinct() to take the highest numerical score for each Value within each Set.
check <- multiplied[with(multiplied, order(Set, -mult)), ]
final <- distinct(check, Set, .keep_all = TRUE)
This works because multiple instances of B (numerical score = 104) will be added together (3 instances would give B a total score in the mult column of 312) but whenever A and B occur at the same frequency, A will win out (105 > 104, 210 > 208, etc.).
If using different numeric scores than the ones provided here, make sure they are spaced out enough for the dataset at hand. For example, using 2 for A and 1 for B doesn't work because it requires 3 instances of B to trump A, instead of only 2. Likewise, if you anticipate large differences in the frequencies of A and B, use 1005 and 1004, since A will eventually catch up to B with the scores I used above (200 * 104 is less than 199 * 205).
first I can't understand aggregate function and cbind I need explanation really simple words, second I have data
permno number mean std
1 10107 120 0.0117174000 0.06802718
2 11850 120 0.0024398083 0.04594591
3 12060 120 0.0005072167 0.08544500
4 12490 120 0.0063569167 0.05325215
5 14593 120 0.0200060583 0.08865493
6 19561 120 0.0154743500 0.07771348
7 25785 120 0.0184815583 0.16510082
8 27983 120 0.0025951333 0.09538822
9 55976 120 0.0092889000 0.04812975
10 59328 120 0.0098526167 0.07135423
I NEED TO process this by
data_processed2 <- aggregate(cbind(return)~permno, Data_summary, median)
I cant understand this command please explain me very simple THANK YOU!
cbind takes two or more tables (dataframes), puts them side by side, and then makes them into one big table. So for example, if you have one table with columns A, B and C, and another with column D and E, after you cbind them, you'll have one table with five columns: A, B, C, D and E. for the rows, cbind assumes all tables are in the same order.
As noted by Rui, in your example cbind doesn't do anything, because return is not a table, and even if it was, it's only one thing.
aggregate takes a table, divides it by some variable, and the calculates a statistic on a variable within each group. For example, if I have data for sales by month and day of month, I can aggregate by month, and calculate the average sales per day for each of the months.
The command you provided uses the following syntax:
aggregate(VARIABLES~GROUPING, DATA, FUNCTION)
Variables (cbind(return) - which doesn't make sense, really) is the list of all the variables for which your statistic will be calculated
Grouping (pernmo) is the variable by which you will break the data into groups (in the sample data you provided each row has a unique number for this variable, so that doesn't really make sense either).
Data is the dataframe you're using.
Function is median.
So this call will break Data_summery into groups that have the same pernmo, and calculate the median for each of the columns.
With the data you provided, you'll basically get the same table back, since you're grouping the data by groups of one row each... -- Actually, since your variables are an empty group, as far as I can tell, you'll get nothing back.
I have a presence absence database with a bunch of zeroes and ones, but when i use rowsums, it seems to only count a portion of the data and then stop. Here's my code
site_matrix=read.csv("TriassicMatrix1.csv", header=T) # create object called site_matrix
summary(site_matrix) # get summary
head(site_matrix) # check out first few columns
tail(site_matrix) # check out last few columns
View(site_matrix) # take a look at whole dataset in new window
# here's the problematic line
spp_rich=rowSums(site_matrix [,2:25]) # generate richness for sites
There are 25 rows of data, and it gives me incorrect output, such as suggesting the first row only has 4 occurances when it has 7.
I tried changing it to [,1,25] and it won't work since row 1 is my title row, so I know it's not that. When I view the data within R i can very easily go to row 2 and count out the data, since there is only a few hundred columns.
It appears to be 'cutting off' at about the halfway point, column-wise.
I have a datatable, DT, with columns A, B and C. I want only one A per unique B, and I want to choose that A based on the value of C (choose the largest C).
Based on this (incredibly helpful) SO page, Use data.table to get first of subgroup based on a variable, I tried something like this:
test <- data.table(A=c(1:3,1:2),B=c(1:5),C=c(11:15))
setkey(test,A,C)
test[,.SD[.N],by="A"]
In my test case, this gives me an answer that seems right:
# A B C
# 1: 1 6 16
# 2: 2 7 17
# 3: 3 8 18
# 4: 4 4 14
# 5: 5 5 15
And, as expected, the number of rows matches the number of unique entries for "A" in my DT:
length(unique(test$A))
# 5
However, when I apply this to my actual dataset, I am missing approximately 20% of my initially ~2 million rows.
I cannot seem to put together a test dataset that will recreate this type of a loss. There are no null values in the actual dataset. What else could be a factor in a dataset that would cause a discrepancy between the number of results from something like test[,.SD[.N],by="A"] and length(unique(test$A))?
Thanks to #Eddi's debugging coaching, here's the answer, at least for my dataset: differential handling of numbers in scientific notation.
In particular: In my actual dataset, columns A and B were very long numbers that, upon import from SQL to R, had been imported in scientific notation. It turns out the test[,.SD[.N],by="A"] and length(unique(test$A)) commands were handling this differently: length(unique(test$A)) was preserving the difference between two values that differed only in a small digit that is not visible in the collapsed scientific notation format printed as visual output, but test[,.SD[.N],by="A"] was, in essence, rounding the values and thus collapsing some of them together.
(I feel foolish that I didn't catch this myself before posting, but much appreciate the help - I hope somehow this spares someone else the same confusion, perhaps!)
Super short version: I'm trying to use a user-defined function to populate a new column in a dataframe with the command:
TestDF$ELN<-EmployeeLocationNumber(TestDF$Location)
However, when I run the command, it seems to just apply EmployeeLocationNumber to the first row's value of Location rather than using each row's value to determine the new column's value for that row individually.
Please note: I'm trying to understand R, not just perform this particular task. I was actually able to get the output I was looking for using the Apply() function, but that's irrelevant. My understanding is that the above line should work on a row-by-row basis, but it isn't.
Here are the specifics for testing:
TestDF<-data.frame(Employee=c(1,1,1,1,2,2,3,3,3),
Month=c(1,5,6,11,4,10,1,5,10),
Location=c(1,5,6,7,10,3,4,2,8))
This testDF keeps track of where each of 3 employees was over the course of the year among several locations.
(You can think of "Location" as unique to each Employee...it is eseentially a unique ID for that row.)
The the function EmployeeLocationNumber takes a location and outputs a number indicating the order that employee visited that location. For example EmployeeLocationNumber(8) = 2 because it was the second location visited by the employee who visited it.
EmployeeLocationNumber <- function(Site){
CurrentEmployee <- subset(TestDF,Location==Site,select=Employee, drop = TRUE)[[1]]
LocationDate<- subset(TestDF,Location==Site,select=Month, drop = TRUE)[[1]]
LocationNumber <- length(subset(TestDF,Employee==CurrentEmployee & Month<=LocationDate,select=Month)[[1]])
return(LocationNumber)
}
I realize I probably could have packed all of that into a single subset command, but I didn't know how referencing worked when you used subset commands inside other subset commands.
So, keeping in mind that I'm really trying to understand how to work in R, I have a few questions:
Why won't TestDF$ELN<-EmployeeLocationNumber(TestDF$Location) work row-by-row like other assignment statements do?
Is there an easier way to reference a particular value in a dataframe based on the value of another one? Perhaps one that does not return a dataframe/list that then must be flattened and extracted from?
I'm sure the function I'm using is laughably un-R-like...what should I have done to essentially emulate an INNER Join type query?
Using logical indexing, the condensed one-liner replacement for your function is:
EmployeeLocationNumber <- function(Site){
with(TestDF[do.call(order, TestDF), ], which(Location[Employee==Employee[which(Location==Site)]] == Site))
}
Of course this isn't the most readable way, but it demonstrates the principles of logical indexing and which() in R. Then, like others have said, just wrap it up with a vectorized *ply function to apply this across your dataset.
A) TestDF$Location is a vector. Your function is not set up to return a vector, so giving it a vector will probably fail.
B) In what sense is Location:8 the "second location visited"?
C) If you want within group ordering then you need to pass you dataframe split up by employee to a funciton that calculates a result.
D) Conditional access of a data.frame typically involves logical indexing and or the use of which()
If you just want the sequence of visits by employee try this:
(Changed first argument to Month since that is what determines the sequence of locations)
with(TestDF, ave(Location, Employee, FUN=seq))
[1] 1 2 3 4 2 1 2 1 3
TestDF$LocOrder <- with(TestDF, ave(Month, Employee, FUN=seq))
If you wanted the second location for EE:3 it would be:
subset(TestDF, LocOrder==2 & Employee==3, select= Location)
# Location
# 8 2
The vectorized nature of R (aka row-by-row) works not by repeatedly calling the function with each next value of the arguments, but by passing the entire vector at once and operating on all of it at one time. But in EmployeeLocationNumber, you only return a single value, so that value gets repeated for the entire data set.
Also, your example for EmployeeLocationNumber does not match your description.
> EmployeeLocationNumber(8)
[1] 3
Now, one way to vectorize a function in the manner you are thinking (repeated calls for each value) is to pass it through Vectorize()
TestDF$ELN<-Vectorize(EmployeeLocationNumber)(TestDF$Location)
which gives
> TestDF
Employee Month Location ELN
1 1 1 1 1
2 1 5 5 2
3 1 6 6 3
4 1 11 7 4
5 2 4 10 1
6 2 10 3 2
7 3 1 4 1
8 3 5 2 2
9 3 10 8 3
As to your other questions, I would just write it as
TestDF$ELN<-ave(TestDF$Month, TestDF$Employee, FUN=rank)
The logic is take the months, looking at groups of the months by employee separately, and give me the rank order of the months (where they fall in order).
Your EmployeeLocationNumber function takes a vector in and returns a single value.
The assignment to create a new data.frame column therefore just gets a single value:
EmployeeLocationNumber(TestDF$Location) # returns 1
TestDF$ELN<-1 # Creates a new column with the single value 1 everywhere
Assignment doesn't do any magic like that. It takes a value and puts it somewhere. In this case the value 1. If the value was a vector of the same length as the number of rows, it would work as you wanted.
I'll get back to you on that :)
Dito.
Update: I finally worked out some code to do it, but by then #DWin has a much better solution :(
TestDF$ELN <- unlist(lapply(split(TestDF, TestDF$Employee), function(x) rank(x$Month)))
...I guess the ave function does pretty much what the code above does. But for the record:
First I split the data.frame into sub-frames, one per employee. Then I rank the months (just in case your months are not in order). You could use order too, but rank can handle ties better. Finally I combine all the results into a vector and put it into the new column ELN.
Update again Regarding question 2, "What is the best way to reference a value in a dataframe?":
This depends a bit on the specific problem, but if you have a value, say Employee=3 and want to find all rows in the data.frame that matches that, then simply:
TestDF$Employee == 3 # Returns logical vector with TRUE for all rows with Employee == 3
which(TestDF$Employee == 3) # Returns a vector of indices instead
TestDF[which(TestDF$Employee == 3), ] # Subsets the data.frame on Employee == 3