I have a data frame like this:
id no age
1 1 7 23
2 1 2 23
3 2 1 25
4 2 4 25
5 3 6 23
6 3 1 23
and I hope to aggregate the date frame by id to a form like this: (just sum the no if they share the same id, but keep age there)
id no age
1 1 9 23
2 2 5 25
3 3 7 23
How to achieve this using R?
Assuming that your data frame is named df.
aggregate(no~id+age, df, sum)
# id age no
# 1 1 23 9
# 2 3 23 7
# 3 2 25 5
Even better, data.table:
library(data.table)
# convert your object to a data.table (by reference) to unlock data.table syntax
setDT(DF)
DF[ , .(sum_no = sum(no), unq_age = unique(age)), by = id]
Alternatively, you could use ddply from plyr package:
require(plyr)
ddply(df,.(id,age),summarise,no = sum(no))
In this particular example the results are identical. However, this is not always the case, the difference between the both functions is outlined here. Both functions have their uses and are worth exploring, which is why I felt this alternative should be mentioned.
Related
I need to clean up the following data frame
df <- data.frame(metric=c(10,20,30,40,NA), cnt=c(1,2,1,2,2))
> df
metric cnt
1 10 1
2 20 2
3 30 1
4 40 2
5 NA 2
I need go back to the original data series (un-pivot ??) which would be like below.
metric
1 10
2 20
3 20
4 30
5 40
6 40
7 NA
8 NA
Is this a use case for tidyr ? If yes, a tidyr based solution would also be helpful.
We can use rep
df1 <- data.frame(metric = rep(df$metric, df$cnt))
There is the function inverse.rle() for inverse RLE. See help("rle"):
df <- data.frame(metric=c(10,20,30,40,NA), cnt=c(1,2,1,2,2))
names(df) <- c("values", "lengths")
inverse.rle(df) # or
data.frame(metric=inverse.rle(df))
I'm having a hard time to describe this so it's best explained with an example (as can probably be seen from the poor question title).
Using dplyr I have the result of a group_by and summarize I have a data frame that I want to do some further manipulation on by factor.
As an example, here's a data frame that looks like the result of my dplyr operations:
> df <- data.frame(run=as.factor(c(rep(1,3), rep(2,3))),
group=as.factor(rep(c("a","b","c"),2)),
sum=c(1,8,34,2,7,33))
> df
run group sum
1 1 a 1
2 1 b 8
3 1 c 34
4 2 a 2
5 2 b 7
6 2 c 33
I want to divide sum by a value that depends on run. For example, if I have:
> total <- data.frame(run=as.factor(c(1,2)),
total=c(45,47))
> total
run total
1 1 45
2 2 47
Then my final data frame will look like this:
> df
run group sum percent
1 1 a 1 1/45
2 1 b 8 8/45
3 1 c 34 34/45
4 2 a 2 2/47
5 2 b 7 7/47
6 2 c 33 33/47
Where I manually inserted the fraction in the percent column by hand to show the operation I want to do.
I know there is probably some dplyr way to do this with mutate but I can't seem to figure it out right now. How would this be accomplished?
(In base R)
You can use total as a look-up table where you get a total for each run of df :
total[df$run,'total']
[1] 45 45 45 47 47 47
And you simply use it to divide the sum and assign the result to a new column:
df$percent <- df$sum / total[df$run,'total']
run group sum percent
1 1 a 1 0.02222222
2 1 b 8 0.17777778
3 1 c 34 0.75555556
4 2 a 2 0.04255319
5 2 b 7 0.14893617
6 2 c 33 0.70212766
If your "run" values are 1,2...n then this will work
divisor <- c(45,47) # c(45,47,...up to n divisors)
df$percent <- df$sum/divisor[df$run]
first you want to merge in the total values into your df:
df2 <- merge(df, total, by = "run")
then you can call mutate:
df2 %<>% mutate(percent = sum / total)
Convert to data.table in-place, then merge and add new column, again in-place:
library(data.table)
setDT(df)[total, on = 'run', percent := sum/total]
df
# run group sum percent
#1: 1 a 1 0.02222222
#2: 1 b 8 0.17777778
#3: 1 c 34 0.75555556
#4: 2 a 2 0.04255319
#5: 2 b 7 0.14893617
#6: 2 c 33 0.70212766
This is something I've been working around for a while just making separate data frames and doing full_join but I think there's an easier way.
Overall, I'm wanting to calculate the differences between an individual ID's value from time 1 to time 2 by type from a long form data frame. This is one of the ways I think I could do it but if other people have other techniques or ideas I'd like to hear them too.
However, I'd also like to know how to address this transposing issue anyway because I'm curious.
Here's my issue.
I have a data frame in long form with 5 different measures for two different time periods. I want to convert this data frame from long form into a wide form so that instead of having a DF look like this (note, not all types are included -- just did 2 for sake of length):
(example df1)
ID Time Value Type
1 1 7 Type1
1 2 8 Type1
2 1 9 Type1
2 2 10 Type1
1 1 13 Type2
1 2 15 Type2
2 1 17 Type2
2 2 19 Type2
I want it to look more like this:
(example df 2)
ID Type1.1 Type1.2 Type2.1 Type2.2
1 7 8 13 15
2 9 10 17 19
I use:
library(dplyr)
library(tidyr)
df.new <- df %>%
spread(Type, Measurement.Value)
and get this from example df 1 which is on the right track:
(example df 3)
ID Time Type1 Type2
1 1 7 13
1 2 8 15
2 1 9 17
2 2 10 19
But now I want to spread the time for each type. When I do something like this on example df3:
newer.df <- df.new %>%
spread(Time, Type1)
to make this:
ID Type1.1 Type1.2
1 7 NA
1 NA 8
2 9 NA
2 NA 10
So, it's producing an NA for each row -- is there a way I can collapse rows on to each other by ID? I think I'm missing something.
Remember, in my example code I'm only using 2 types but in reality I have 5 types -- just wanted to give simplified code.
We can use dcast() from reshape2 package.
library(reshape2)
dcast(df, ID ~ Type + Time, value.var = "Value")
# ID Type1_1 Type1_2 Type2_1 Type2_2
#1 1 7 8 13 15
#2 2 9 10 17 19
Or using the original tidyr package, we could do this:
library(tidyr)
df$Type <- paste(df$Type, df$Time, sep="_")
df$Time <- NULL
spread(df, key=Type, value=Value)
ID Type1_1 Type1_2 Type2_1 Type2_2
1 7 8 13 15
2 9 10 17 19
Nulling the time column did the trick for me. It seems that spread considers all columns not used otherwise as what dcast would call id.vars. There might be a more elegant solution using tidyr, though.
Here is the data.
set.seed(23) data<-data.frame(ID=rep(1:12), group=rep(1:3,times=4), value=(rnorm(12,mean=0.5, sd=0.3)))
ID group value
1 1 1 0.4133934
2 2 2 0.6444651
3 3 3 0.1350871
4 4 1 0.5924411
5 5 2 0.3439465
6 6 3 0.3673059
7 7 1 0.3202062
8 8 2 0.8883733
9 9 3 0.7506174
10 10 1 0.3301955
11 11 2 0.7365258
12 12 3 0.1502212
I want to get z-standardized scores within each group. so I try
library(weights)
data_split<-split(data, data$group) #split the dataframe
stan<-lapply(data_split, function(x) stdz(x$value)) #compute z-scores within group
However, It looks wrong because I want to add a new variable following 'value'
How can I do that? Kindly provide some suggestions(sample code). Any help is greatly appreciated .
Use this instead:
within(data, stan <- ave(value, group, FUN=stdz))
No need to call split nor lapply.
One way using data.table package:
library(data.table)
library(weights)
set.seed(23)
data <- data.table(ID=rep(1:12), group=rep(1:3,times=4), value=(rnorm(12,mean=0.5, sd=0.3)))
setkey(data, ID)
dataNew <- data[, list(ID, stan = stdz(value)), by = 'group']
the result is:
group ID stan
1: 1 1 -0.6159312
2: 1 4 0.9538398
3: 1 7 -1.0782747
4: 1 10 0.7403661
5: 2 2 -1.2683237
6: 2 5 0.7839781
7: 2 8 0.8163844
8: 2 11 -0.3320388
9: 3 3 0.6698418
10: 3 6 0.8674548
11: 3 9 -0.2131335
12: 3 12 -1.3241632
I tried Ferdinand.Kraft's solution but it didn't work for me. I think the stdz function isn't included in the basic R install. Moreover, the within part troubled me in a large dataset with many variables. I think the easiest way is:
data$value.s <- ave(data$value, data$group, FUN=scale)
Add the new column while in your function, and have the function return the whole data frame.
stanL<-lapply(data_split, function(x) {
x$stan <- stdz(x$value)
x
})
stan <- do.call(rbind, stanL)
I am essentially trying to get disorganized data into long form for linear modeling.
I have 2 data.frames "rec" and "book"
Each row in "book" needs to be pasted onto the end of several of the rows of "rec" according to two variables in the row: "MRN" and "COURSE" which match.
I have tried the following and variations thereon to no avail:
i=1
newlist=list()
colnames(newlist)=colnames(book)
for ( i in 1:dim(rec)[1]) {
mrn=as.numeric(as.vector(rec$MRN[i]));
course=as.character(rec$COURSE[i]);
get.vector<-as.vector(((as.numeric(as.vector(book$MRN))==mrn) & (as.character(book$COURSE)==course)))
newlist[i]<-book[get.vector,]
i=i+1;
}
If anyone has any suggestions on
1)getting this to work
2) making it more elegant (or perhaps just less clumsy)
If I have been unclear in any way I beg your pardons.
I do understand I haven't combined any data above, I think if I can generate a long-format data.frame I can combine them all on my own
Sounds like you need to merge the two data-frames. Try this:
merge(rec, book, by = c('MRN', 'COURSE'))
and do read the help for merge (by doing ?merge at the R console) for more options on how to merge these.
I've created a simple example that may help you. In my case i wanted to paste the 'value' column from df1 in each row of df2, according to variables x1 and x2:
df1 <- read.table(textConnection("
x1 x2 value
1 2 12
1 3 56
2 1 35
2 2 68
"),header=T)
df2 <- read.table(textConnection("
test x1 x2
1 1 2
2 1 3
3 2 1
4 2 2
5 1 2
6 1 3
7 2 1
"),header=T)
library(sqldf)
sqldf("select df2.*, df1.value from df2 join df1 using(x1,x2)")
test x1 x2 value
1 1 1 2 12
2 2 1 3 56
3 3 2 1 35
4 4 2 2 68
5 5 1 2 12
6 6 1 3 56
7 7 2 1 35