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Using Reshape from wide to long in R [closed]
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Closed 2 years ago.
I'm trying to calculate the total number of matches played by each team in the year 2019 and put them in a table along with the corresponding team names
teams<-c("Sunrisers Hyderabad", "Mumbai Indians", "Gujarat Lions", "Rising Pune Supergiants",
"Royal Challengers Bangalore","Kolkata Knight Riders","Delhi Daredevils",
"Kings XI Punjab", "Deccan Chargers","Rajasthan Royals", "Chennai Super Kings",
"Kochi Tuskers Kerala", "Pune Warriors", "Delhi Capitals", " Gujarat Lions")
for (j in teams) {
print(j)
ipl_table %>%
filter(season==2019 & (team1==j | team2 ==j)) %>%
summarise(match_count=n())->kl
print(kl)
match_played<-data.frame(Teams=teams,Match_count=kl)
}
The match played by last team (i.e Gujarat Lions is 0 and its filling 0's for all other teams as well.
The output match_played can be found on the link given below.
I'd be really glad if someone could help me regarding this error as I'm very new to R.
filter for the particular season, get data in long format and then count number of matches.
library(dplyr)
matches %>%
filter(season == 2019) %>%
tidyr::pivot_longer(cols = c(team1, team2), values_to = 'team_name') %>%
count(team_name) -> result
result
# team_name n
# <chr> <int>
#1 Chennai Super Kings 17
#2 Delhi Capitals 16
#3 Kings XI Punjab 14
#4 Kolkata Knight Riders 14
#5 Mumbai Indians 16
#6 Rajasthan Royals 14
#7 Royal Challengers Bangalore 14
#8 Sunrisers Hyderabad 15
Here is an example
library(tidyr)
df_2019 <- matches[matches$season == 2019, ] # get the season you need
df_long <- gather(df_2019, Team_id, Team_Name, team1:team2) # Make it long format
final_count <- data.frame(t(table(df_long$Team_Name)))[-1] # count the number of matches
names(final_count) <- c("Team", "Matches")
Team Matches
1 Chennai Super Kings 17
2 Delhi Capitals 16
3 Kings XI Punjab 14
4 Kolkata Knight Riders 14
5 Mumbai Indians 16
6 Rajasthan Royals 14
7 Royal Challengers Bangalore 14
8 Sunrisers Hyderabad 15
Or by using base R
final_count <- data.frame(t(table(c(df_2019$team1, df_2019$team2))))[-1]
names(final_count) <- c("Team", "Matches")
final_count
I want remove entire row if there are duplicates in two columns. Any quick help in doing so in R (for very large dataset) would be highly appreciated. For example:
mydf <- data.frame(p1=c('a','a','a','b','g','b','c','c','d'),
p2=c('b','c','d','c','d','e','d','e','e'),
value=c(10,20,10,11,12,13,14,15,16))
This gives:
mydf
p1 p2 value
1 a b 10
2 c c 20
3 a d 10
4 b c 11
5 d d 12
6 b b 13
7 c d 14
8 c e 15
9 e e 16
I want to get:
p1 p2 value
1 a b 10
2 a d 10
3 b c 11
4 c d 14
5 c e 15
your note in the comments suggests your actual problem is more complex. There's some preprocessing you could do to your strings before you compare p1 to p2. You will have the domain expertise to know what steps are appropriate, but here's a first start. I remove all spaced and punctuation from p1 and p2. I then convert them all to uppercase before testing for equality. You can modify the clean_str function to include more / different cleaning operations.
Additionally, you may consider approximate matching to address typos / colloquial naming conventions. Package stringdist is a good place to start.
mydf <- data.frame(p1=c('New York','New York','New York','TokYo','LosAngeles','MEMPHIS','memphis','ChIcAGo','Cleveland'),
p2=c('new York','New.York','MEMPHIS','Chicago','knoxville','tokyo','LosAngeles','Chicago','CLEVELAND'),
value=c(10,20,10,11,12,13,14,15,16),
stringsAsFactors = FALSE)
mydf[mydf$p1 != mydf$p2,]
#> p1 p2 value
#> 1 New York new York 10
#> 2 New York New.York 20
#> 3 New York MEMPHIS 10
#> 4 TokYo Chicago 11
#> 5 LosAngeles knoxville 12
#> 6 MEMPHIS tokyo 13
#> 7 memphis LosAngeles 14
#> 8 ChIcAGo Chicago 15
#> 9 Cleveland CLEVELAND 16
clean_str <- function(col){
#removes all punctuation
d <- gsub("[[:punct:][:blank:]]+", "", col)
d <- toupper(d)
return(d)
}
mydf$p1 <- clean_str(mydf$p1)
mydf$p2 <- clean_str(mydf$p2)
mydf[mydf$p1 != mydf$p2,]
#> p1 p2 value
#> 3 NEWYORK MEMPHIS 10
#> 4 TOKYO CHICAGO 11
#> 5 LOSANGELES KNOXVILLE 12
#> 6 MEMPHIS TOKYO 13
#> 7 MEMPHIS LOSANGELES 14
Created on 2020-05-03 by the reprex package (v0.3.0)
Several ways to do that. Among them :
Base R
mydf[mydf$p1 != mydf$p2, ]
dplyr
library(dplyr)
mydf %>% filter(p1 != p2)
data.table
library(data.table)
setDT(mydf)
mydf[p1 != p2]
Here's a two-step solution based on #Chase's data:
First step (as suggested by #Chase) - preprocess your data in p1and p2to make them comparable:
# set to lower-case:
mydf[,c("p1", "p2")] <- lapply(mydf[,c("p1", "p2")], tolower)
# remove anything that's not alphanumeric between words:
mydf[,c("p1", "p2")] <- lapply(mydf[,c("p1", "p2")], function(x) gsub("(\\w+)\\W(\\w+)", "\\1\\2", x))
Second step - (i) using apply, paste the rows together, (ii) use grepl and backreference \\1 to look out for immediately adjacent duplicates in these rows, and (iii) remove (-) those rows which contain these duplicates:
mydf[-which(grepl("\\b(\\w+)\\s+\\1\\b", apply(mydf, 1, paste0, collapse = " "))),]
p1 p2 value
3 newyork memphis 10
4 tokyo chicago 11
5 losangeles knoxville 12
6 memphis tokyo 13
7 memphis losangeles 14
Right now, I have a main function (let's call it performance()) that has as its arguments player1, player2, and team_of_interest.
I have a data set that looks like this:
> head(roster_van, 3)
team_name team venue num_first_last
1 VANCOUVER CANUCKS VAN Home 5 SBISA, LUCA
2 VANCOUVER CANUCKS VAN Home 8 TANEV, CHRISTOPHER
3 VANCOUVER CANUCKS VAN Home 14 BURROWS, ALEXANDRE
game_date game_id season session player_number
1 2016-10-15 2016020029 20162017 R 5
2 2016-10-15 2016020029 20162017 R 8
3 2016-10-15 2016020029 20162017 R 14
team_num first_name last_name player_name
1 VAN5 LUCA SBISA LUCA.SBISA
2 VAN8 CHRISTOPHER TANEV CHRIS.TANEV
3 VAN14 ALEXANDRE BURROWS ALEX.BURROWS
name_match player_position
1 LUCASBISA D
2 CHRISTOPHERTANEV D
3 ALEXANDREBURROWS L
This is the roster data for a hockey games played in a season.
I want to create another function (let's call it players()) that loops through every unique pair of players in a hockey team and provides their names and team to the player1, player2, and team_of_interest arguments inside the performance() function.
I've started off with this, but don't know what next to do:
name_pairs <- function(x,y) {
x <- seq(1,19, by = 2)
y <- x+1
}
merge can make quick work of generating a cartesian join out of your dataframe.
With a shortened version of your sample dataframe and a guess at the team_of_interest column.
library(tidyverse)
roster_van <- tibble(team = "VAN",
team_num = c(5, 8, 14),
player_name = c("LUCA.SBISA", "CHRIS.TANEV", "ALEX.BURROWS"),
player_position = c("D", "D", "L"),
team_of_interest = c("SL BLUES", "BOS BRUINS", "CGY FLAMES")
)
roster_van
> roster_van
# A tibble: 3 x 5
team team_num player_name player_position team_of_interest
<chr> <dbl> <chr> <chr> <chr>
1 VAN 5 LUCA.SBISA D SL BLUES
2 VAN 8 CHRIS.TANEV D BOS BRUINS
3 VAN 14 ALEX.BURROWS L CGY FLAMES
If you only want a few of the columns repeated, then only rename the column names you wish to see joined again onto the original dataframe before you filter off the equal self joins.
roster_van_pairs <-
roster_van %>%
merge(roster_van %>%
select(team,
team_num_paired = team_num,
player_name_paired = player_name
)
) %>%
filter(player_name != player_name_paired)
roster_van_pairs
> roster_van_pairs
team team_num player_name player_position team_of_interest team_num_paired player_name_paired
1 VAN 5 LUCA.SBISA D SL BLUES 8 CHRIS.TANEV
2 VAN 5 LUCA.SBISA D SL BLUES 14 ALEX.BURROWS
3 VAN 8 CHRIS.TANEV D BOS BRUINS 5 LUCA.SBISA
4 VAN 8 CHRIS.TANEV D BOS BRUINS 14 ALEX.BURROWS
5 VAN 14 ALEX.BURROWS L CGY FLAMES 5 LUCA.SBISA
6 VAN 14 ALEX.BURROWS L CGY FLAMES 8 CHRIS.TANEV
If you want to go with a bulk approach which will join all the columns in again, you can execute a full rename of all the columns with the code below:
roster_van_copy <- roster_van
# provenience the data quickly
colnames(roster_van_copy) <- colnames(roster_van_copy) %>% paste0(., "_paired")
This makes the cross join code more concise, too:
roster_van_all_columns_paired <-
roster_van %>%
merge(roster_van_copy) %>%
filter(player_name != player_name_paired)
I imagine this will leave you with more columns than necessary, but they are very easy to remove with a select(-c(<col_x:col_y)) after all.
roster_van_all_columns_paired
> roster_van_all_columns_paired
team team_num player_name player_position team_of_interest team_paired team_num_paired player_name_paired
1 VAN 8 CHRIS.TANEV D BOS BRUINS VAN 5 LUCA.SBISA
2 VAN 14 ALEX.BURROWS L CGY FLAMES VAN 5 LUCA.SBISA
3 VAN 5 LUCA.SBISA D SL BLUES VAN 8 CHRIS.TANEV
4 VAN 14 ALEX.BURROWS L CGY FLAMES VAN 8 CHRIS.TANEV
5 VAN 5 LUCA.SBISA D SL BLUES VAN 14 ALEX.BURROWS
6 VAN 8 CHRIS.TANEV D BOS BRUINS VAN 14 ALEX.BURROWS
player_position_paired team_of_interest_paired
1 D SL BLUES
2 D SL BLUES
3 D BOS BRUINS
4 D BOS BRUINS
5 L CGY FLAMES
6 L CGY FLAMES
Base R approach could look like this:
roster.van.all.copy.baseR <- merge(roster_van, roster_van_copy)
roster.van.all.baseR <- roster.van.all.copy.baseR[ which(roster.van.all.copy.baseR$player_name != roster.van.all.copy.baseR$player_name_paired), ]
Supose I have a data frame with 3 columns (name, y, sex) where name is character, y is a numeric value and sex is a factor.
sex<-c("M","M","F","M","F","M","M","M","F")
x<-c("MARK","TOM","SUSAN","LARRY","EMMA","LEONARD","TIM","MATT","VIOLET")
name<-as.character(x)
y<-rnorm(9,8,1)
score<-data.frame(x,y,sex)
score
name y sex
1 MARK 6.767086 M
2 TOM 7.613928 M
3 SUSAN 7.447405 F
4 LARRY 8.040069 M
5 EMMA 8.306875 F
6 LEONARD 8.697268 M
7 TIM 10.385221 M
8 MATT 7.497702 M
9 VIOLET 10.177969 F
If I wanted to order it by y I would use:
score[order(score$y),]
x y sex
1 MARK 6.767086 M
3 SUSAN 7.447405 F
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
5 EMMA 8.306875 F
6 LEONARD 8.697268 M
9 VIOLET 10.177969 F
7 TIM 10.385221 M
So far, so good... The names keep the correct score BUT how could I reorder it to have M and F levels not mixed. I need to order and at the same time keep factor levels separated.
Finally I would like to take a step further to involve character, the example doesn't help, but what if there were tied y values and I would have to order again within factor (e.g. TIM and TOM got 8.4 and I have to assign alphabetical order).
I was thinking about by function but it creates a list and doesn't help really. I think there must be some function like it to apply on data frames and get data frames as return.
TO MAKE CLEAR THE POINT:
sep<-split(score,score$sex)
sep$M<-sep$M[order(sep$M[,2]),]
sep$M
x y sex
1 MARK 6.767086 M
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
6 LEONARD 8.697268 M
7 TIM 10.385221 M
sep$F<-sep$F[order(sep$F[,2]),]
sep$F
x y sex
3 SUSAN 7.447405 F
5 EMMA 8.306875 F
9 VIOLET 10.177969 F
merged<-rbind(sep$M,sep$F)
merged
x y sex
1 MARK 6.767086 M
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
6 LEONARD 8.697268 M
7 TIM 10.385221 M
3 SUSAN 7.447405 F
5 EMMA 8.306875 F
9 VIOLET 10.177969 F
I know how to do that if I have 2 or 3 factors. But what if I had serious levels of factors, say 20, should I write a for loop?
order takes multiple arguments, and it does just what you want:
with(score, score[order(sex, y, x),])
## x y sex
## 3 SUSAN 6.636370 F
## 5 EMMA 6.873445 F
## 9 VIOLET 8.539329 F
## 6 LEONARD 6.082038 M
## 2 TOM 7.812380 M
## 8 MATT 8.248374 M
## 4 LARRY 8.424665 M
## 7 TIM 8.754023 M
## 1 MARK 8.956372 M
Here is a summary of all methods mentioned in other answers/comments (to serve future searchers). I've added a data.table way of sorting.
# Base R
do.call(rbind, by(score, score$sex, function(x) x[order(x$y),]))
with(score, score[order(sex, y, x),])
score[order(score$sex,score$x),]
# Using plyr
arrange(score, sex,y)
ddply(score, c('sex', 'y'))
# Using `data.table`
library("data.table")
score_dt <- setDT(score)
# setting a key works sorts the data.table
setkey(score_dt,sex,x)
print(score_dt)
Here is Another question that deals with the same
I think there must be some function like it to apply on data frames
and get data frames as return
Yes there is:
library(plyr)
ddply(score, c('y', 'sex'))
It sounds to me like you're trying to order by score within the males and females and return a combined data frame of sorted males and sorted females.
You are right that by(score, score$sex, function(x) x[order(x$y),]) returns a list of sorted data frames, one for male and one for female. You can use do.call with the rbind function to combine these data frames into a single final data frame:
do.call(rbind, by(score, score$sex, function(x) x[order(x$y),]))
# x y sex
# F.5 EMMA 7.526866 F
# F.9 VIOLET 8.182407 F
# F.3 SUSAN 9.677511 F
# M.4 LARRY 6.929395 M
# M.8 MATT 7.970015 M
# M.7 TIM 8.297137 M
# M.6 LEONARD 8.845588 M
# M.2 TOM 9.035948 M
# M.1 MARK 10.082314 M
I believe that the person asked how to sort it by the orders in the case of say 20.
I know how to do that if I have 2 or 3 factors. But what if I had
serious levels of factors, say 20, should I write a for loop?
I have one where there are 9 orders with various counts.
stage_name count
<ord> <int>
1 Closed Lost 957
2 Closed Won 1413
3 Evaluation 1773
4 Meeting Scheduled 4104
5 Nurture 1222
6 Opportunity Disqualified 805
7 Order Submitted 1673
8 Qualifying 5138
9 Quoted 4976
In this case you can see that it is displayed using alphabetical order of stage_name, but stage_name is actually an ordered factor that has a very different order.
This code orders the factor is a much different order:
# Make categoricals ----
check_stage$stage_name = ordered(check_stage$stage_name, levels=c(
'Opportunity Disqualified',
'Qualifying',
'Evaluation',
'Meeting Scheduled',
'Quoted',
'Order Submitted',
'Closed Won',
'Closed Lost',
'Nurture'))
Now we can just apply the factor as the method of ordering this is a dplyr function, but you might need forcats too. I have both libraries installed:
check_stage <- check_stage %>%
arrange(factor(stage_name))
This now gives the output in the factor order as desired:
check_stage
# A tibble: 9 × 2
stage_name count
<ord> <int>
1 Opportunity Disqualified 805
2 Qualifying 5138
3 Evaluation 1773
4 Meeting Scheduled 4104
5 Quoted 4976
6 Order Submitted 1673
7 Closed Won 1413
8 Closed Lost 957
9 Nurture 1222
Supose I have a data frame with 3 columns (name, y, sex) where name is character, y is a numeric value and sex is a factor.
sex<-c("M","M","F","M","F","M","M","M","F")
x<-c("MARK","TOM","SUSAN","LARRY","EMMA","LEONARD","TIM","MATT","VIOLET")
name<-as.character(x)
y<-rnorm(9,8,1)
score<-data.frame(x,y,sex)
score
name y sex
1 MARK 6.767086 M
2 TOM 7.613928 M
3 SUSAN 7.447405 F
4 LARRY 8.040069 M
5 EMMA 8.306875 F
6 LEONARD 8.697268 M
7 TIM 10.385221 M
8 MATT 7.497702 M
9 VIOLET 10.177969 F
If I wanted to order it by y I would use:
score[order(score$y),]
x y sex
1 MARK 6.767086 M
3 SUSAN 7.447405 F
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
5 EMMA 8.306875 F
6 LEONARD 8.697268 M
9 VIOLET 10.177969 F
7 TIM 10.385221 M
So far, so good... The names keep the correct score BUT how could I reorder it to have M and F levels not mixed. I need to order and at the same time keep factor levels separated.
Finally I would like to take a step further to involve character, the example doesn't help, but what if there were tied y values and I would have to order again within factor (e.g. TIM and TOM got 8.4 and I have to assign alphabetical order).
I was thinking about by function but it creates a list and doesn't help really. I think there must be some function like it to apply on data frames and get data frames as return.
TO MAKE CLEAR THE POINT:
sep<-split(score,score$sex)
sep$M<-sep$M[order(sep$M[,2]),]
sep$M
x y sex
1 MARK 6.767086 M
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
6 LEONARD 8.697268 M
7 TIM 10.385221 M
sep$F<-sep$F[order(sep$F[,2]),]
sep$F
x y sex
3 SUSAN 7.447405 F
5 EMMA 8.306875 F
9 VIOLET 10.177969 F
merged<-rbind(sep$M,sep$F)
merged
x y sex
1 MARK 6.767086 M
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
6 LEONARD 8.697268 M
7 TIM 10.385221 M
3 SUSAN 7.447405 F
5 EMMA 8.306875 F
9 VIOLET 10.177969 F
I know how to do that if I have 2 or 3 factors. But what if I had serious levels of factors, say 20, should I write a for loop?
order takes multiple arguments, and it does just what you want:
with(score, score[order(sex, y, x),])
## x y sex
## 3 SUSAN 6.636370 F
## 5 EMMA 6.873445 F
## 9 VIOLET 8.539329 F
## 6 LEONARD 6.082038 M
## 2 TOM 7.812380 M
## 8 MATT 8.248374 M
## 4 LARRY 8.424665 M
## 7 TIM 8.754023 M
## 1 MARK 8.956372 M
Here is a summary of all methods mentioned in other answers/comments (to serve future searchers). I've added a data.table way of sorting.
# Base R
do.call(rbind, by(score, score$sex, function(x) x[order(x$y),]))
with(score, score[order(sex, y, x),])
score[order(score$sex,score$x),]
# Using plyr
arrange(score, sex,y)
ddply(score, c('sex', 'y'))
# Using `data.table`
library("data.table")
score_dt <- setDT(score)
# setting a key works sorts the data.table
setkey(score_dt,sex,x)
print(score_dt)
Here is Another question that deals with the same
I think there must be some function like it to apply on data frames
and get data frames as return
Yes there is:
library(plyr)
ddply(score, c('y', 'sex'))
It sounds to me like you're trying to order by score within the males and females and return a combined data frame of sorted males and sorted females.
You are right that by(score, score$sex, function(x) x[order(x$y),]) returns a list of sorted data frames, one for male and one for female. You can use do.call with the rbind function to combine these data frames into a single final data frame:
do.call(rbind, by(score, score$sex, function(x) x[order(x$y),]))
# x y sex
# F.5 EMMA 7.526866 F
# F.9 VIOLET 8.182407 F
# F.3 SUSAN 9.677511 F
# M.4 LARRY 6.929395 M
# M.8 MATT 7.970015 M
# M.7 TIM 8.297137 M
# M.6 LEONARD 8.845588 M
# M.2 TOM 9.035948 M
# M.1 MARK 10.082314 M
I believe that the person asked how to sort it by the orders in the case of say 20.
I know how to do that if I have 2 or 3 factors. But what if I had
serious levels of factors, say 20, should I write a for loop?
I have one where there are 9 orders with various counts.
stage_name count
<ord> <int>
1 Closed Lost 957
2 Closed Won 1413
3 Evaluation 1773
4 Meeting Scheduled 4104
5 Nurture 1222
6 Opportunity Disqualified 805
7 Order Submitted 1673
8 Qualifying 5138
9 Quoted 4976
In this case you can see that it is displayed using alphabetical order of stage_name, but stage_name is actually an ordered factor that has a very different order.
This code orders the factor is a much different order:
# Make categoricals ----
check_stage$stage_name = ordered(check_stage$stage_name, levels=c(
'Opportunity Disqualified',
'Qualifying',
'Evaluation',
'Meeting Scheduled',
'Quoted',
'Order Submitted',
'Closed Won',
'Closed Lost',
'Nurture'))
Now we can just apply the factor as the method of ordering this is a dplyr function, but you might need forcats too. I have both libraries installed:
check_stage <- check_stage %>%
arrange(factor(stage_name))
This now gives the output in the factor order as desired:
check_stage
# A tibble: 9 × 2
stage_name count
<ord> <int>
1 Opportunity Disqualified 805
2 Qualifying 5138
3 Evaluation 1773
4 Meeting Scheduled 4104
5 Quoted 4976
6 Order Submitted 1673
7 Closed Won 1413
8 Closed Lost 957
9 Nurture 1222