This question already has answers here:
Combine Multiple Columns Into Tidy Data [duplicate]
(3 answers)
Closed 5 years ago.
In R, I have data where each person has multiple session dates, and the scores on some tests, but this is all in one row. I would like to change it so I have multiple rows with the persons info, but only one of the session dates and corresponding test scores, and do this for every person. Also, each person may have completed different number of sessions.
Ex:
ID Name Session1Date Score Score Session2Date Score Score
23 sjfd 20150904 2 3 20150908 5 7
28 addf 20150905 3 4 20150910 6 8
To:
ID Name SessionDate Score Score
23 sjfd 20150904 2 3
23 sjfd 20150908 5 7
28 addf 20150905 3 4
28 addf 20150910 6 8
You can use melt from the devel version of data.table ie. v1.9.5. It can take multiple 'measure' columns as a list. Instructions to install are here
library(data.table)#v1.9.5+
melt(setDT(df1), measure = patterns("Date$", "Score(\\.2)*$", "Score\\.[13]"))
# ID Name variable value1 value2 value3
#1: 23 sjfd 1 20150904 2 3
#2: 28 addf 1 20150905 3 4
#3: 23 sjfd 2 20150908 5 7
#4: 28 addf 2 20150910 6 8
Or using reshape from base R, we can specify the direction as 'long' and varying as a list of column index
res <- reshape(df1, idvar=c('ID', 'Name'), varying=list(c(3,6), c(4,7),
c(5,8)), direction='long')
res
# ID Name time Session1Date Score Score.1
#23.sjfd.1 23 sjfd 1 20150904 2 3
#28.addf.1 28 addf 1 20150905 3 4
#23.sjfd.2 23 sjfd 2 20150908 5 7
#28.addf.2 28 addf 2 20150910 6 8
If needed, the rownames can be changed
row.names(res) <- NULL
Update
If the columns follow a specific order i.e. 3rd grouped with 6th, 4th with 7th, 5th with 8th, we can create a matrix of column index and then split to get the list for the varying argument in reshape.
m1 <- matrix(3:8,ncol=2)
lst <- split(m1, row(m1))
reshape(df1, idvar=c('ID', 'Name'), varying=lst, direction='long')
If your data frame name is data
Use this
data1 <- data[1:5]
data2 <- data[c(1,2,6,7,8)]
newdata <- rbind(data1,data2)
This works for the example you've given. You might have to change column names appropriately in data1 and data2 for a proper rbind
Related
This question already has answers here:
Calculate the mean by group
(9 answers)
Aggregate / summarize multiple variables per group (e.g. sum, mean)
(10 answers)
Closed 5 years ago.
Hi I have 3 data set with contains the items and counts. I need to add the all data sets and combine the count based on the item names. He is my input.
Df1 <- data.frame(items =c("Cookies", "Candys","Toys","Games"), Counts = c( 10,20,30,5))
Df2 <- data.frame(items =c( "Candys","Cookies","Toys"), Counts = c( 5,21,20))
Df3 <- data.frame(items =c( "Playdows","Gummies","Candys"), Counts = c(10,15,20))
Df_all <- rbind(Df1,Df2,Df3)
Df_all
items Counts
1 Cookies 10
2 Candys 20
3 Toys 30
4 Games 5
5 Candys 5
6 Cookies 21
7 Toys 20
8 Playdows 10
9 Gummies 15
10 Candys 20
I need to combine the columns based on the item values. Delete the Row after adding the values. My output should be
items Counts
1 Cookies 31
2 Candys 45
3 Toys 50
4 Games 5
5 Playdows 10
6 Gummies 15
Could you help in getting this output in r.
use dplyr:
library(dplyr)
result<-Df_all%>%group_by(items)%>%summarize(sum(Counts))
> result
# A tibble: 6 x 2
items `sum(Counts)`
<fct> <dbl>
1 Candys 45.0
2 Cookies 31.0
3 Games 5.00
4 Toys 50.0
5 Gummies 15.0
6 Playdows 10.0
You can use tapply
tapply(Df_all$Counts, Df_all$items, FUN=sum)
what returns
Candys Cookies Games Toys Gummies Playdows
45 31 5 50 15 10
This question already has answers here:
calculate average over multiple data frames
(5 answers)
Closed 6 years ago.
I am new to R and I need help in this. I have 3 data sets from 3 different years. they have the same columns with different values for each year. I want to find the average for the column values across the three years based on the name field. To be specific:
assume : first data set
Name Age Height Weight
A 4 20 20
B 5 22 22
C 8 25 21
D 10 25 23
second data set
Name Age Height Weight
A 5 22 25
B 6 23 26
Third data set
Name Age Height Weight
A 6 24 24
B 7 24 27
C 10 27 28
I want to find the average height for "A" across the three data sets
We can place them in a list and rbind them, group by 'Name' and get the mean of each column
library(data.table)
rbindlist(list(df1, df2, df3))[, lapply(.SD, mean), by = Name]
Or with dplyr
bind_rows(df1, df2, df3) %>%
group_by(Name) %>%
summarise_each(funs(mean))
I have a dataset which looks something like this:-
Key Days
A 1
A 2
A 3
A 8
A 9
A 36
A 37
B 14
B 15
B 44
B 45
I would like to split the individual keys based on the days in groups of 7. For e.g.:-
Key Days
A 1
A 2
A 3
Key Days
A 8
A 9
Key Days
A 36
A 37
Key Days
B 14
B 15
Key Days
B 44
B 45
I could use ifelse and specify buckets of 1-7, 7-14 etc until 63-70 (max possible value of days). However the issue lies with the days column. There are lots of cases wherein there is an overlap in days - Take days 14-15 as an example which would fall into 2 brackets if split using the ifelse logic (7-14 & 15-21).
The ideal method of splitting this would be to identify a day and add 7 to it and check how many rows of data are actually falling under that category. I think we need to use loops for this. I could do it in excel but i have 20000 rows of data for 2000 keys hence i'm using R. I would need a loop which checks each key value and for each key it further checks the value of days and buckets them in group of 7 by checking the first day value of each range.
We create a grouping variable by applying %/% on the 'Day' column and then split the dataset into a list based on that 'grp'.
grp <- df$Day %/%7
split(df, factor(grp, levels = unique(grp)))
#$`0`
# Key Days
#1 A 1
#2 A 2
#3 A 3
#$`1`
# Key Days
#4 A 8
#5 A 9
#$`5`
# Key Days
#6 A 36
#7 A 37
#$`2`
# Key Days
#8 B 14
#9 B 15
#$`6`
# Key Days
#10 B 44
#11 B 45
Update
If we need to split by 'Key' also
lst <- split(df, list(factor(grp, levels = unique(grp)), df$Key), drop=TRUE)
This question already has an answer here:
R // Sum by based on date range
(1 answer)
Closed 7 years ago.
Suppose I have df1 like this:
Date Var1
01/01/2015 1
01/02/2015 4
....
07/24/2015 1
07/25/2015 6
07/26/2015 23
07/27/2015 15
Q1: Sum of Var1 on previous 3 days of 7/27/2015 (not including 7/27).
Q2: Sum of Var1 on previous 3 days of 7/25/2015 (This is not last row), basically I choose anyday as reference day, and then calculate rolling sum.
As suggested in one of the comments in the link referenced by #SeñorO, with a little bit of work you can use zoo::rollsum:
library(zoo)
set.seed(42)
df <- data.frame(d=seq.POSIXt(as.POSIXct('2015-01-01'), as.POSIXct('2015-02-14'), by='days'),
x=sample(20, size=45, replace=T))
k <- 3
df$sum3 <- c(0, cumsum(df$x[1:(k-1)]),
head(zoo::rollsum(df$x, k=k), n=-1))
df
## d x sum3
## 1 2015-01-01 16 0
## 2 2015-01-02 12 16
## 3 2015-01-03 15 28
## 4 2015-01-04 15 43
## 5 2015-01-05 17 42
## 6 2015-01-06 10 47
## 7 2015-01-07 11 42
The 0, cumsum(...) is to pre-populate the first two rows that are ignored (rollsum(x, k) returns a vector of length length(x)-k+1). The head(..., n=-1) discards the last element, because you said that the nth entry should sum the previous 3 and not its own row.
Hello I have a dataset with multiple patients, each with multiple observations.
I want to select the earliest observation for each patient.
Example:
Patient ID Tender Swollen pt_visit
101 1 10 6
101 6 12 12
101 4 3 18
102 9 5 18
102 3 6 24
103 5 2 12
103 2 1 18
103 8 0 24
The pt_visit variable is the number of months the patient was in the study at the time of the observation. What I need is the first observation from each patient based on the lowest number of months in the pt_visit column. However I need the earliest observation for each patient ID.
My desired results:
Patient ID Tender Swollen pt_visit
101 1 10 6
102 9 5 18
103 5 2 12
Assuming your data frame is called df, use the ddply function in the plyr package:
require(plyr)
firstObs <- ddply(df, "PatientID", function(x) x[x$pt_visit == min(x$pt_visit), ])
I would use the data.table package:
Data <- data.table(Data)
setkey(Data, Patient_ID, pt_visit)
Data[,.SD[1], by=Patient_ID]
Assuming that the Patient ID column is actually named Patient_ID, here are a few approaches. DF is assumed to be the name of the input data frame:
sqldf
library(sqldf)
sqldf("select Patient_ID, Tender, Swollen, min(pt_visit) pt_visit
from DF
group by Patient_ID")
or
sqldf("select *, min(pt_visit) pt_visit from DF group by Patient_ID")[-ncol(DF)]
Note: The above two alternatives use an extension to SQL only found in SQLite so be sure you are using the SQLite backend. (SQLite is the default backend for sqldf unless RH2, RProgreSQL or RMYSQL is loaded.)
subset/ave
subset(DF, ave(pt_visit, Patient_ID, FUN = rank) == 1)
Note: This makes use of the fact that there are no duplicate pt_visit values within the same Patient_ID. If there were we would need to specify the ties= argument to rank.
I almost think they should be a subset parameter named "by" that would do the same as it does in data.table. This is a base-solution:
do.call(rbind, lapply( split(dfr, dfr$PatientID),
function(x) x[which.min(x$pt_visit),] ) )
PatientID Tender Swollen pt_visit
101 101 1 10 6
102 102 9 5 18
103 103 5 2 12
I guess you can see why #hadley built 'plyr'.