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
Related
I am having difficulty importing my data in the way I would like to from a .csv file to tidydata.
My data set is made up of descriptive data (age, country, etc.) and then 15 condition columns that I would like to have in just one column (long format). I have previously tried 'melting' the data in a few ways, but it does not turn out the way I intended it to. These are a few things I have tried, I know it is kind of messy. There are quite a few NAs in the data, which seem to be causing an issue. I am trying to create this specific column "Vignette" which will serve as the collective column for the 15 vignette columns I would like in long format.
head(dat)
ID Frequency Gender Country Continent Age
1 5129615189 At least weekly female France Europe 30-50 years
2 5128877943 At least daily female Spain Europe > 50 years
3 5126775994 At least weekly female Spain Europe 30-50 years
4 5126598863 At least daily male Albania Europe 30-50 years
5 5124909744 At least daily female Ireland Europe > 50 years
6 5122047758 At least weekly female Denmark Europe 30-50 years
Practice Specialty Seniority AMS
1 University public hospital centre Infectious diseases 6-10 years Yes
2 Other public hospital Infectious diseases > 10 years Yes
3 University public hospital centre Intensive care > 10 years Yes
4 University public hospital centre Infectious diseases > 10 years No
5 Private hospial/clinic Clinical microbiology > 10 years Yes
6 University public hospital centre Infectious diseases 0-5 years Yes
Durations V01 V02 V03 V04 V05 V06 V07 V08 V09 V10 V11 V12 V13 V14 V15
1 range 7 2 7 7 7 5 7 14 7 42 42 90 7 NA 5
2 range 7 10 10 5 14 5 7 14 10 42 21 42 14 14 14
3 range 7 5 5 7 14 5 5 13 10 42 42 42 5 0 7
4 range 10 7 7 5 7 10 7 5 7 28 14 42 10 10 7
5 range 7 5 7 7 14 7 7 14 10 42 42 90 10 0 7
6 fixed duration 7 3 3 7 10 10 7 14 7 90 90 90 10 7 7
dat_long %>%
gather(Days, Age, -Vignette)
dat$new_sp = NULL
names(dat) <- gsub("new_sp", "", names(dat))
dat_tidy<-melt(
data=dat,
id=0:180,
variable.name="Vignette",
value.name="Days",
na.rm=TRUE
)
dat_tidy<- mutate(dat_tidy,
Days= sub("^V", "", Days)
)
It keeps saying "Error: id variables not found in data: NA"
I have tried to get rid of NAs but it doesn't seem to do anything.
I am guessing you are loading the melt function from reshape2. I will recommend that you try tidyr which is basically the next generation of reshape2.
Your error is presumable that the argument id=0:180. This is basically asking it to keep columns 0-180 as "identifier" columns, and melt the rest (i.e. create a new row for each value in each column).
When you subset more column indices than columns in a data.frame, the non-existing columns are filled in with pure NA - you asked for them, so you get them!
I would recommend loading tidyr, as it is newer. There should be some new verbs in the package that are more intuitive, but I'll give you a solution with the older semantic:
library(tidyr)
dat_tidy <- dat %>% gather('Vignette', 'Days', starts_with('V'))
# or a bit more verbose
dat_tidy <- dat %>% gather('Vignette', 'Days', V01, V02, V03, V04)
And check out the comment #heck1 for asking even better questions.
This question already has answers here:
How to reshape data from long to wide format
(14 answers)
Closed 5 years ago.
I'm trying to change a dataframe in R to group multiple rows by a measurement. The table has a location (km), a size (mm) a count of things in that size bin, a site and year. I want to take the sizes, make a column from each one (2, 4 and 6 in this example), and place the corresponding count into each the row for that location, site and year.
It seems like a combination of transposing and grouping, but I can't figure out a way to accomplish this in R. I've looked at t(), dcast() and aggregate(), but those aren't really close at all.
So I would go from something like this:
df <- data.frame(km=c(rep(32,3),rep(50,3)), mm=rep(c(2,4,6),2), count=sample(1:25,6), site=rep("A", 6), year=rep(2013, 6))
km mm count site year
1 32 2 18 A 2013
2 32 4 2 A 2013
3 32 6 12 A 2013
4 50 2 3 A 2013
5 50 4 17 A 2013
6 50 6 21 A 2013
To this:
km site year mm_2 mm_4 mm_6
1 32 A 2013 18 2 12
2 50 A 2013 3 17 21
Edit: I tried the solution in a suggested duplicate, but I did not work for me, not really sure why. The answer below worked better.
As suggested in the comment above, we can use the sep argument in spread:
library(tidyr)
spread(df, mm, count, sep = "_")
km site year mm_2 mm_4 mm_6
1 32 A 2013 4 20 1
2 50 A 2013 15 14 22
As you mentioned dcast(), here is a method using it.
set.seed(1)
df <- data.frame(km=c(rep(32,3),rep(50,3)),
mm=rep(c(2,4,6),2),
count=sample(1:25,6),
site=rep("A", 6),
year=rep(2013, 6))
library(reshape2)
dcast(df, ... ~ mm, value.var="count")
# km site year 2 4 6
# 1 32 A 2013 13 10 20
# 2 50 A 2013 3 17 1
And if you want a bit of a challenge you can try the base function reshape().
df2 <- reshape(df, v.names="count", idvar="km", timevar="mm", ids="mm", direction="wide")
colnames(df2) <- sub("count.", "mm_", colnames(df2))
df2
# km site year mm_2 mm_4 mm_6
# 1 32 A 2013 13 10 20
# 4 50 A 2013 3 17 1
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.
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
I have a dataset in which there are following columns: flavor, flavorid and unitSoled.
Flavor Flavorid unitsoled
beans 350 6
creamy 460 2
.
.
.
I want to find top ten flavors and then calculate market share for each flavor. My logic is market share for each flavor = units soled for particular flavor divided by total units soled.
How do I implement this. For output I just want two col Flavorid and corresponding market share. Do I need to save top ten flavors in some table first?
One way is with the dplyr package:
An example data set:
flavor <- rep(letters[1:15],each=5)
flavorid <- rep(1:15,each=5)
unitsold <- 1:75
df <- data.frame(flavor,flavorid,unitsold)
> df
flavor flavorid unitsold
1 a 1 1
2 a 1 2
3 a 1 3
4 a 1 4
5 a 1 5
6 b 2 6
7 b 2 7
8 b 2 8
9 b 2 9
...
...
Solution:
library(dplyr)
df %>%
select(flavorid,unitsold) %>% #select the columns you want
group_by(flavorid) %>% #group by flavorid
summarise(total=sum(unitsold)) %>% #sum the total units sold per id
mutate(marketshare=total/sum(total)) %>% #calculate the market share per id
arrange( desc(marketshare)) %>% #order by marketshare descending
head(10) #pick the 10 first
#and you can add another select(flavorid,marketshare) if you only want those two
Output:
Source: local data frame [10 x 3]
flavorid total marketshare
1 15 365 0.12807018
2 14 340 0.11929825
3 13 315 0.11052632
4 12 290 0.10175439
5 11 265 0.09298246
6 10 240 0.08421053
7 9 215 0.07543860
8 8 190 0.06666667
9 7 165 0.05789474
10 6 140 0.04912281