I have a dataframe which contains information about several categories, and some associated variables. It is of the form:
ID category sales score
227 A 109 21
131 A 410 24
131 A 509 1
123 B 2 61
545 B 19 5
234 C 439 328
654 C 765 41
What I would like to do is be able to introduce two new columns, salesRank and scoreRank, where I find the item index per category, had they been ordered by sales and score, respectively. I can solve the general case like this:
dF <- dF[order(-dF$sales),]
dF$salesRank<-seq.int(nrow(dF))
but this doesn't account for the categories and so far I've only solved this by breaking up the dataframe. What I want would result in the following:
ID category sales score salesRank scoreRank
227 A 109 21 3 2
131 A 410 24 2 1
131 A 509 1 1 3
123 B 2 61 2 1
545 B 19 5 1 2
234 C 439 328 2 1
654 C 765 41 1 2
Many thanks!
Try:
library(dplyr)
df %>%
group_by(category) %>%
mutate(salesRank = row_number(desc(sales)),
scoreRank = row_number(desc(score)))
Which gives:
#Source: local data frame [7 x 6]
#Groups: category
#
# ID category sales score salesRank scoreRank
#1 227 A 109 21 3 2
#2 131 A 410 24 2 1
#3 131 A 509 1 1 3
#4 123 B 2 61 2 1
#5 545 B 19 5 1 2
#6 234 C 439 328 2 1
#7 654 C 765 41 1 2
From the help:
row_number(): equivalent to rank(ties.method = "first")
min_rank(): equivalent to rank(ties.method = "min")
desc(): transform a vector into a format that will be sorted in descending
order.
As #thelatemail pointed out, for this particular dataset you might want to use min_rank() instead of row_number() which will account for ties in sales/score more appropriately:
> row_number(c(1,2,2,4))
#[1] 1 2 3 4
> min_rank(c(1,2,2,4))
#[1] 1 2 2 4
Use ave in base R with rank (the - is to reverse the rankings from low-to-high to high-to-low):
dF$salesRank <- with(dF, ave(-sales, category, FUN=rank) )
#[1] 3 2 1 2 1 2 1
dF$scoreRank <- with(dF, ave(-score, category, FUN=rank) )
#[1] 2 1 3 1 2 1 2
I have just a base R solution with tapply.
salesRank <- tapply(dat$sales, dat$category, order, decreasing = T)
scoreRank <- tapply(dat$score, dat$category, order, decreasing = T)
cbind(dat, salesRank = unlist(salesRank), scoreRank= unlist(scoreRank))
ID category sales score salesRank scoreRank
A1 227 A 109 21 3 2
A2 131 A 410 24 2 1
A3 131 A 509 1 1 3
B1 123 B 2 61 2 1
B2 545 B 19 5 1 2
C1 234 C 439 328 2 1
C2 654 C 765 41 1 2
Related
I have an ordinal data frame which has answers in the survey format. I want to convert each factor into a possible column so as to get them by frequencies of a specific event.
I have tried lapply, dplyr to get frequencies but failed
as.data.frame(apply(mtfinal, 2, table))
and
mtfinalf<-mtfinal %>%
group_by(q28) %>%
summarise(freq=n())
Expected Results in the form of data.frame
Frequency table with respect to q28's factors
Expected Results in the form of data.frame
q28 sex1 sex2 race1 race2 race3 race4 race5 race6 race7 age1 age2
2 0
3 0
4 23
5 21
Actual Results
$age
1 2 3 4 5 6 7
6 2 184 520 507 393 170
$sex
1 2
1239 543
$grade
1 2 3 4
561 519 425 277
$race7
1 2 3 4 5 6
179 21 27 140 17 1307
7
91
$q8
1 2 3 4 5
127 259 356 501 539
$q9
1 2 3 4 5
993 224 279 86 200
$q28
2 3 4 5
1034 533 94 121
This will give you a count of number of unique combinations. What you are asking is impossible since there would be overlaps between levels of sex, race and age.
mtfinalf<-mtfinal %>%
group_by(q28,age,race,sex) %>%
tally()
This question already has answers here:
Find complement of a data frame (anti - join)
(7 answers)
Closed 4 years ago.
Dataset1:
id1 id2 abc n
1 111 yes 2
2 121 no 1
3 122 yes 2
4 224 no 2
5 441 no 3
6 665 yes 1
Dataset2:
id1 id2 age gen
1 111 45 m
1 111 46 f
2 1 52 f
121 122 41 f
121 122 44 m
4 224 54 f
4 221 56 m
5 441 44 m
5 441 45 f
5 441 58 f
6 665 54 f
I have two data sets. Both are linked by id1 and id2. How to identify those data from both data sets which fails to link???
We can use anti_join from the dplyr package to filter the rows with no match.
library(dplyr)
Dataset1_anti <- Dataset1 %>% anti_join(Dataset2, by = c("id1", "id2"))
Dataset1_anti
# id1 id2 abc n
# 1 2 121 no 1
# 2 3 122 yes 2
Dataset2_anti <- Dataset2 %>% anti_join(Dataset1, by = c("id1", "id2"))
Dataset2_anti
# id1 id2 age gen
# 1 2 1 52 f
# 2 121 122 41 f
# 3 121 122 44 m
# 4 4 221 56 m
DATA
Dataset1 <- read.table(text = "id1 id2 abc n
1 111 yes 2
2 121 no 1
3 122 yes 2
4 224 no 2
5 441 no 3
6 665 yes 1 ",
header = TRUE, stringsAsFactors = FALSE)
Dataset2 <- read.table(text = "id1 id2 age gen
1 111 45 m
1 111 46 f
2 1 52 f
121 122 41 f
121 122 44 m
4 224 54 f
4 221 56 m
5 441 44 m
5 441 45 f
5 441 58 f
6 665 54 f ",
header = TRUE, stringsAsFactors = FALSE)
I would like to analyze data.
My Database is composed of 1408 (704 for type 1 and 704 for type 2) observations and 49 variables. Here is part of my database.
The point is that I want to analyze gender of type 1(sellers) who overcharged.
Data
Subject ID Gender Period Matching group Group Type Overcharging
654 1 1 73 1 1 NA
654 1 2 73 1 1 NA
654 1 3 73 1 1 NA
654 1 4 73 1 1 NA
708 0 1 73 1 2 1
708 0 2 73 1 2 0
708 0 3 73 1 2 0
708 0 4 73 1 2 1
435 1 1 73 2 1 NA
435 1 2 73 2 1 NA
435 1 3 73 2 1 NA
435 1 4 73 2 1 NA
546 0 1 73 2 2 0
546 0 2 73 2 2 0
546 0 3 73 2 2 1
546 0 4 73 2 2 0
For example, if you take a look at matching group =73, there are 2 groups (1 and 2).And in each group, there are two types (1 and 2). For each type 1 (seller) we do not have information about what he did (overcharge or not). But we have informations about buyers (type 2) who were overcharged or not.
If I can identify the buyer who were over-treated, then, this means that the seller this buyer is interacting has over-treated the buyer. So all I need to look at is the gender of the seller in the same group as the buyer.
In matching group 73 we know for instance that at period 1 subject 708 was overcharged (the one in group 1). As I know that this men belongs to group 1 and matching group 73, I am able to identify the seller who has overcharged him : subject 654 with gender =1.
In group 2 (matching group 73), we know that at period 3, agent 546 was overcharged. As I know that this men belongs to group 1 and matching group 73, I am able to identify the seller who has overcharged him : subject 435 with gender =1.
....
I would do this for all the observations I have.
However I really don't know how to proceed to code and make this condition on R.
This is what I tried to do, but doesn't fit my needs !
for (matchinggroup[type==1]==matchinggroup[type==2] &
group[type==1]==group[type==2] & period[type==1]==period[type==2])
{
if ((overtreatment==1), na.rm=TRUE)
sum(gender==1[type==1], na.rm=TRUE)
}
The expected output I would like to have is :
sum(overcharging==1[gender==1&type==1])
>3
sum(overcharging==1[gender==0&type==1])
>0
sum(overcharging==0[gender==1&type==1])
>5
sum(overcharging==0[gender==0&type==1])
>0
Not exactly sure what your desired output is, but consider this:
Data <- read.table(header = T,
text = "Subject_ID Gender Period Matching_group Group Type Overcharging
654 1 1 73 1 1 NA
654 1 2 73 1 1 NA
654 1 3 73 1 1 NA
654 1 4 73 1 1 NA
708 0 1 73 1 2 1
708 0 2 73 1 2 0
708 0 3 73 1 2 0
708 0 4 73 1 2 1
435 1 1 73 2 1 NA
435 1 2 73 2 1 NA
435 1 3 73 2 1 NA
435 1 4 73 2 1 NA
546 0 1 73 2 2 0
546 0 2 73 2 2 0
546 0 3 73 2 2 1
546 0 4 73 2 2 0
")
dat1 <- subset(Data, Overcharging==1)
This will find all the Overcharging sellers. And then you could find each matching buyer using this loop:
out <- data.frame()
for(i in 1:nrow(dat1)){
dat2 <- dat1[i,]
df <- Data[Data$Period==dat2$Period & Data$Matching_group==dat2$Matching_group &
Data$Group==dat2$Group & Data$Type==1,]
out <- rbind(out, df)
}
Which will give you:
Subject_ID Gender Period Matching_group Group Type Overcharging
1 654 1 1 73 1 1 NA
4 654 1 4 73 1 1 NA
11 435 1 3 73 2 1 NA
I think "for loop" solution is not suitable in R.
I developed another solution for you with data.table by seperating sellers and buyers, and then joining them.
library(data.table)
Data <- data.table(read.table(header = T,
text = "Subject_ID Gender Period Matching_group Group Type Overcharging
654 1 1 73 1 1 NA
654 1 2 73 1 1 NA
654 1 3 73 1 1 NA
654 1 4 73 1 1 NA
708 0 1 73 1 2 1
708 0 2 73 1 2 0
708 0 3 73 1 2 0
708 0 4 73 1 2 1
435 1 1 73 2 1 NA
435 1 2 73 2 1 NA
435 1 3 73 2 1 NA
435 1 4 73 2 1 NA
546 0 1 73 2 2 0
546 0 2 73 2 2 0
546 0 3 73 2 2 1
546 0 4 73 2 2 0
")
)
Data[, SubjectType := ifelse(Type==1, "Seller", "Buyer")]
Subjects <- unique(Data[, .(Subject_ID, Gender)])
Matches <- dcast(Data, Matching_group+Group~SubjectType, value.var="Subject_ID", fun.aggregate = mean)
Buys <- Data[!is.na(Overcharging), .(Buyer = Subject_ID, BuyerGender = Gender, Period, Matching_group, Group, Overcharging)]
Buys <- merge(Buys, Matches, by=c("Buyer", "Matching_group", "Group"), all.x = T)
Buys <- merge(Buys, Subjects[, .(Seller = Subject_ID, SellerGender = Gender)], by="Seller", all.x = T)
Buys[Overcharging==0, .N, .(BuyerGender, SellerGender)]
Buys[Overcharging==1, .N, .(BuyerGender, SellerGender)]
I am trying to rank multiple numeric variables ( around 700+ variables) in the data and am not sure exactly how to do this as I am still pretty new to using R.
I do not want to overwrite the ranked values in the same variable and hence need to create a new rank variable for each of these numeric variables.
From reading the posts, I believe assign and transform function along with rank maybe able to solve this. I tried implementing as below ( sample data and code) and am struggling to get it to work.
The output dataset in addition to variables xcount, xvisit, ysales need to be populated
With variables xcount_rank, xvisit_rank, ysales_rank containing the ranked values.
input <- read.table(header=F, text="101 2 5 6
102 3 4 7
103 9 12 15")
colnames(input) <- c("id","xcount","xvisit","ysales")
input1 <- input[,2:4] #need to rank the numeric variables besides id
for (i in 1:3)
{
transform(input1,
assign(paste(input1[,i],"rank",sep="_")) =
FUN = rank(-input1[,i], ties.method = "first"))
}
input[paste(names(input)[2:4], "rank", sep = "_")] <-
lapply(input[2:4], cut, breaks = 10)
The problem with this approach is that it's creating the rank values as (101, 230] , (230, 450] etc whereas I would like to see the values in the rank variable to be populated as 1, 2 etc up to 10 categories as per the splits I did. Is there any way to achieve this? input[5:7] <- lapply(input[5:7], rank, ties.method = "first")
The approach I tried from the solutions provided below is:
input <- read.table(header=F, text="101 20 5 6
102 2 4 7
103 9 12 15
104 100 8 7
105 450 12 65
109 25 28 145
112 854 56 93")
colnames(input) <- c("id","xcount","xvisit","ysales")
input[paste(names(input)[2:4], "rank", sep = "_")] <-
lapply(input[2:4], cut, breaks = 3)
Current output I get is:
id xcount xvisit ysales xcount_rank xvisit_rank ysales_rank
1 101 20 5 6 (1.15,286] (3.95,21.3] (5.86,52.3]
2 102 2 4 7 (1.15,286] (3.95,21.3] (5.86,52.3]
3 103 9 12 15 (1.15,286] (3.95,21.3] (5.86,52.3]
4 104 100 8 7 (1.15,286] (3.95,21.3] (5.86,52.3]
5 105 450 12 65 (286,570] (3.95,21.3] (52.3,98.7]
6 109 25 28 145 (1.15,286] (21.3,38.7] (98.7,145]
7 112 854 56 93 (570,855] (38.7,56.1] (52.3,98.7]
Desired output:
id xcount xvisit ysales xcount_rank xvisit_rank ysales_rank
1 101 20 5 6 1 1 1
2 102 2 4 7 1 1 1
3 103 9 12 15 1 1 1
4 104 100 8 7 1 1 1
5 105 450 12 65 2 1 2
6 109 25 28 145 1 2 3
Would like to see the records in the group they would fall under if I try to rank the interval values.
Using dplyr
library(dplyr)
nm1 <- paste("rank", names(input)[2:4], sep="_")
input[nm1] <- mutate_each(input[2:4],funs(rank(., ties.method="first")))
input
# id xcount xvisit ysales rank_xcount rank_xvisit rank_ysales
#1 101 2 5 6 1 2 1
#2 102 3 4 7 2 1 2
#3 103 9 12 15 3 3 3
Update
Based on the new input and using cut
input[nm1] <- mutate_each(input[2:4], funs(cut(., breaks=3, labels=FALSE)))
input
# id xcount xvisit ysales rank_xcount rank_xvisit rank_ysales
#1 101 20 5 6 1 1 1
#2 102 2 4 7 1 1 1
#3 103 9 12 15 1 1 1
#4 104 100 8 7 1 1 1
#5 105 450 12 65 2 1 2
#6 109 25 28 145 1 2 3
#7 112 854 56 93 3 3 2
I have two dataframes that are structured as follows:
Dataframe A:
id sqft traf month
1 1030 16 35 1
1 1030 15 32 2
2 1027 1 31 1
2 1027 2 31 2
Dataframe B:
id price frequency month day
1 1030 8 196 1 1
2 1030 9 101 1 15
3 1030 10 156 1 30
4 1030 3 137 2 1
5 1030 7 190 2 15
6 1027 6 188 1 1
7 1027 1 198 1 15
8 1027 2 123 1 30
9 1027 4 185 2 1
10 1027 5 122 2 15
I want to output certain types of summary statistics (centered around each unique ID) from both these columns. This would be easy with ddply if say I wanted the mean price for each ID for each month (split by id and month) from Dataframe B or if I wanted the average ratio of sqft to traf for each id (split by id).
But what would be a potential solution if I wanted to make combined variables from both dataframes. For instance, how would I get the average price for each id/month (Dataframe B) divided by sqft for each id/month?
The varying frequencies at of the dataframes are measured makes combining them not easily doable. The only solution I've found so far is to ddply the first dataframe to extract average sqft/id/month and then pass that value into a second ddply call on the second dataframe.
Is there a more efficient/less convoluted way to do this? I would be splitting both dataframes on the same variables (id and month).
Thanks in advance for any suggestions!
In the case of the sample data, you could merge the two data sets like this (by specifying all.y = TRUE you can make sure that all rows of dfb are kept and, in this case, corresponding entries of dfa are repeated accordingly)
dfall <- merge(dfa, dfb, by = c("id", "month"), all.y=TRUE)
# id month sqft traf price frequency day
#1 1027 1 1 31 6 188 1
#2 1027 1 1 31 1 198 15
#3 1027 1 1 31 2 123 30
#4 1027 2 2 31 4 185 1
#5 1027 2 2 31 5 122 15
#6 1030 1 16 35 8 196 1
#7 1030 1 16 35 9 101 15
#8 1030 1 16 35 10 156 30
#9 1030 2 15 32 3 137 1
#10 1030 2 15 32 7 190 15
Then, you can use ddply as usual:
ddply(dfall, .(id, month), mutate, newcol = mean(price)/sqft)
# id month sqft traf price frequency day newcol
#1 1027 1 1 31 6 188 1 3.0000000
#2 1027 1 1 31 1 198 15 3.0000000
#3 1027 1 1 31 2 123 30 3.0000000
#4 1027 2 2 31 4 185 1 2.2500000
#5 1027 2 2 31 5 122 15 2.2500000
#6 1030 1 16 35 8 196 1 0.5625000
#7 1030 1 16 35 9 101 15 0.5625000
#8 1030 1 16 35 10 156 30 0.5625000
#9 1030 2 15 32 3 137 1 0.3333333
#10 1030 2 15 32 7 190 15 0.3333333
Edit: if you're looking for better performance, consider using dplyr instead of plyr. The equivalent dplyr code (including the merge) is:
library(dplyr)
dfall <- dfb %>%
left_join(., dfa, by = c("id", "month")) %>%
group_by(id, month) %>%
dplyr::mutate(newcol = mean(price)/sqft) # I added dplyr:: to avoid confusion with plyr::mutate
Of course, you could also check out data.table which is also very efficient.
AFAIK ddply is not designed to be used with different data frames at the same time.
dplyr does well here. This code merges the data frames, gets price and sqft means by unique id/month combination, then creates a new variable pricePerSqft.
require(dplyr)
dfa %>%
left_join(dfb, by = c("id", "month")) %>%
group_by(id, month) %>%
summarize(
avgPrice = mean(price),
avgSqft = mean(sqft)) %>%
mutate(pricePerSqft = round(avgPrice / avgSqft, 2))
Here's the result:
id month avgPrice avgSqft pricePerSqft
1 1027 1 3.0 1 3.00
2 1027 2 4.5 2 2.25
3 1030 1 9.0 16 0.56
4 1030 2 5.0 15 0.33