I have a data frame with an ID column that includes duplicates. There is a column called type that takes the values "S" or "N." There are two additional date columns - admission date and discharge date. My question is a bit similar to comparing two data frames and isolating rows based on certain date differences, but not quite. If needed, I could separate my data into two data frames, but I'm wondering if I can accomplish what I want without the extra steps.
Here is a small example of what the data for two patients looks like in R:
example <- data.frame(ID = c(22,22,22,52,52,52),
admission_date = c("2013-10-03","2014-03-11","2014-03-16","2012-02-08","2014-06-10","2014-06-20"),
discharge_date = c("2013-10-11","2014-03-16","2014-03-28","2012-02-13","2014-06-12","2014-06-30"),
type = c('S','S','N','S','S','N'))
What I want to do is compare within patients, entries that take the value "N" and entries that take the value "S" in the type variable. Based on the discharge date for entries with the value "S," I would like to find entries with the value "N" that have an admission date within 5 days of the former's discharge date (the discharge date with value "S" should be before the admission date with value "N").
So in the example data frame, the only two entries that should be retained are rows 2 and 3 and not rows 5 and 6 since the difference between admission date and discharge date is greater than 5.
Does anyone have any suggestions of how I can filter this data? Any help is greatly appreciated.
This was an interesting challenge. One reason for this is because iterating over rows is less intuitive than iterating over columns (see this question for lots of suggestions: For each row in an R dataframe).
Now I know vectorized solutions are preferred over for loops, but one of the challenges with this problem was that instead of just performing functions on each row, we're comparing the iterated rows to other rows and deleting some rows as we go along. I expect there's a better solution out there and I hope someone posts a better solution to help me learn.
One minor note before I begin, "example" isn't a great name for an object because it's also a function in base R. Additionally, the solution is much easier if we're only dealing with alternating rows of "S" and "N" - that is if many S's precede an N then only the lowest S might be within 5 days of N. Nonetheless it was worth the effort to attack the more challenging case.
Ultimately I ended up solving this as a 2-stage problem, each solved with a for loop. First, I took out all the S rows which weren't within 5 days of the corresponding N rows. Then I took out those N rows which didn't have any appropriate S companions. All of this is implemented in base R.
So to begin:
example_df <- data.frame(ID = c(22,22,22,52,52,52),
admission_date = c("2013-10-03","2014-03-11","2014-03-16","2012-02-08","2014-06-10","2014-06-20"),
discharge_date = c("2013-10-11","2014-03-16","2014-03-28","2012-02-13","2014-06-12","2014-06-30"),
type = c('S','S','N','S','S','N'))
example_df$admission_date<-as.numeric(as.Date(example_df$admission_date))
example_df$discharge_date<-as.numeric(as.Date(example_df$discharge_date))
The first thing I did was to take the date columns (which were characters) and convert them to numeric based on date. Originally I was doing mathematical operations with date objects, but this became complicated with the subsetting operations I ended up using.
Here's the first for loop:
del_vec <- vector("integer")
for( i in 1:nrow(example_df)) {
if (example_df[i,"type"]== "S") {
next
}
if (example_df[i,"type"] == "N") {
add_on <- which
(
example_df["type"] == "S" &
example_df["ID"]==example_df[i,"ID"] &
example_df["discharge_date"] < (example_df[i,"admission_date"] - 5)
)
}
del_vec<- append(del_vec,add_on)
}
example_df_new <- example_df[-c(del_vec),]
rownames(example_df_new) <- 1:nrow(example_df_new)
example_df_new
What I did here is start by creating a vector which will contain the row numbers that we delete. To get rid of the inappropriate S rows we need to actually work on the N rows, so I have the loop skip the S rows. Then when the loop encounters an N row, we find the rows which meet the following conditions:
have type S
have the same ID as the N row in question
have a discharge date which is more than 5 days from the admission date for the N row in question
Using which()captures the row numbers that meet these criteria. Now I add these rows to the empty vector and remove them from the original df. I also rename the rows of the new df to get the following output for example_df_new
ID admission_date discharge_date type
1 22 16140 16145 S
2 22 16145 16157 N
3 52 16241 16251 N
So we've preserved the 2 rows you wanted to keep, but now we have this bottom row that we want to get rid of. I do this in the second loop which iterates over the rows in the new reduced df:
del_vec2 <- vector()
for(i in 1:nrow(example_df_new)) {
if (example_df_new[i,"type"]=="S") {
next
}
if (example_df_new[i,"type"] == "N") {
add_on_two <- which(example_df_new["type"] == "S" & example_df_new["ID"] == example_df_new[i,"ID"])
}
if(length(add_on_two !=0)) {
next
} else {
del_vec2 <- append(del_vec2,i)
}
}
example_df_3<-example_df_new[-c(del_vec2),]
example_df_3
Again, we tell the loop to skip the S rows — whichever ones made the first cut should stay in. Now when the loop encounters an N row we ask the loop to look for rows that meet the following criteria:
is type S
has the same ID as the N row in question
Again I use which() to save the positions of these rows. If these criteria are met then we skip ahead - we want to keep all the N's that have an appropriate S companion. If not then we add the row number of (i) - that is the row number for the N in question to our vector of rows that we want to delete.
We then delete those rows and end up with the desired output:
ID admission_date discharge_date type
1 22 16140 16145 S
2 22 16145 16157 N
At this point you can change the date columns back to a date format.
Again, while this may be the first, I expect it's not the best solution. I hope to see an improved solution, but the problem is more tricky than it appears at first.
After attempting to filter within the same data frame, I decided to separate the data into two tables: one containing only data of type "S" and the other containing only data of type "N." Then, I did a full join while matching on the ID column. While this creates a greater number of rows than before, I was then able to compare the two date of interest. The resulting data frame contains only one row - the entry of a patient with an admission date with type "N" within 5 days of a discharge date with type "S."
The code in R is as follows:
library(dplyr)
example_df <- data.frame(ID = c(22,22,22,52,52,52),
admission_date = c("2013-10-03","2014-03-11","2014-03-16","2012-02-08","2014-06-10","2014-06-20"),
discharge_date = c("2013-10-11","2014-03-16","2014-03-28","2012-02-13","2014-06-12","2014-06-30"),
type = c('S','S','N','S','S','N'))
N_only <- example_df %>%
filter(type == "N")
S_only <- example_df %>%
filter(type == "S")
example_df_merged <- merge(N_only, S_only, by = "ID")
example_df_merged$admission_date.x <- as.Date(as.character(example_df_merged$admission_date.x), format="%Y-%m-%d")
example_df_merged$discharge_date.y <- as.Date(as.character(example_df_merged$discharge_date.y), format="%Y-%m-%d")
example_df_merged$dateDiff <- example_df_merged$discharge_date.y - example_df_merged$admission_date.x
example_df_final <- example_df_merged %>%
filter(dateDiff <= 5 & dateDiff >= 0)
For clearer variable names, I would have changed the variables ending in ".x" and ".y," but that is not necessary.
I have a dataframe with 3 columns. First two columns are IDs (ID1 and ID2) referring to the same item and the third column is a count of how many times items with these two IDs appear. The dataframe has many rows so I want to use binary search to first find the appropriate row where both IDs match and then add 1 to the cell under the count column in that row.
I have used the which() function to find the index of the correct row and then using the index added 1 to the count column.
For example:
index <- which(DF$ID1 == x & DF$ID1 == y)
DF$Count[index] <- DF$Count[index] + 1
While this works, the which function is very inefficient. Because I have to do this within a for loop for more than a trillion times, it takes a lot of time. Also, there is only one row in the data frame with this ID combination. While the which function goes through all the rows, a function that stops once it finds the correct row should suffice. I have looked into using data.table and setkey for this purpose but do not know how to implement that for my purpose. Thank you in advance.
Indeed you can use data.table and setkeyv (not setkey because you need 2 columns as indexes)
library(data.table)
DF <- data.frame(ID1=sample(1:100,100000,replace=TRUE),ID2=sample(1:100,100000,replace=TRUE))
# convert DF to a data.table
DF <- as.data.table(DF)
# put both ID1 and ID2 as indexes, in that order
setkeyv(DF,c("ID1","ID2"))
# random x and y values
x <- 10
y <- 18
# select value for ID1=x and ID2=y and add 1 in the Count column
DF[.(x,y),"Count"] <- DF[,.(x,y),"Count"]+1
I want to find a series of consecutive rows in a dataset where a condition is met the most often.
I have two columns that I can use for this; Either one with ones and zeros that alternate based on the presence or absence of a condition or a column which increments for the duration across which the desirable condition is present. I envision that I will need to use subset(),filter(), and/or rle() in order to make this happen but am at a loss as to how to get it to work.
In the example, I want to find 6 sequential rows that maximize the instances in which happens occurs.
Given the input:
library(data.frame)
df<-data.frame(time=c(1:10),happens=c(1,1,0,0,1,1,1,0,1,1),count=c(1,2,0,0,1,2,3,0,1,2))
I would like to see as the output the rows 5 through 10, inclusive, as the data subset output, using either the happens or count columns since this sequence of rows would yield the highest output of happens occurrences on 6 consecutive rows.
library(zoo)
which.max( rollapply( df$happens, 6, sum) )
#[1] 5
The fifth window of 6 rows apparently holds the maximum sum of df$happens
So the answer is row 5:10
I have a dataframe with multiple columns and I want to apply different functions on each column.
An example of my dataset -
I want to calculate the count of column pq110a for each country mentioned in qcountry2 column(me-mexico,br-brazil,ar-argentina). The problem I face here is that I have to use filter on these columns for example for sample patients I want-
Count of pq110 when the values are 1 and 2 (for some patients)
Count of pq110 when the value is 3 (for another patients)
Similarly when the value is 6.
For total patient I want-total count of pq110.
Output I am expecting is-Output
Similalry for each country I want this output.
Please suggest how can I do this for other columns also,countrywise.
Thanks !!
I guess what you want to do is count the number of columns of 'pq110' which have the same value within different 'qcountry2'.
So I'll try to use 'tapply' to divide data into several subsets and then use 'table' to count column number for each different value.
tapply(my_data[,"pq110"], INDEX = as.factor(my_data[,"qcountry2"]), function(x)table(x))
My Problem:
I have a dataframe consisting of 86016000 rows of observations:
there are 512000 observations for each hour
there are 24 hours data for seven days
So 24*7*512000 = 86016000
there are 40 columns (variables)
There is no column of date or datetimestamp
Only row numbers are good enough to identify how many obs. for each day, and there are no errors in recording of this data.
Given such a large dataset, what I want to do is create subsets of 12288000 (i.e. 24 * 512000) rows, so that we have 7 each day's subset.
What I tried:
d <- split(PltB_Fold3_1_Data, rep(1:12288000, each=7))
But unfortunately after almost half an hour, I termicated the process as there was no result.
Is there any better solution then the one above?
You're probably looking for seq rather than rep. With seq, you can generate a sequence of numbers from 0 to 86016000 incremented by 12288000.
To save resources, you can then use this sequence to generate temporary data frames and do whatever you want with each.
sequence <- seq(from = 0, to = 86016000, by = 12288000)
for(i in 1:(length(sequence)-1)){
temp <- df[sequence[i]+1:sequence[i+1], ]
# do something here with your temporary data frame
}