I am trying to sum up the values in certain columns based on the player's birth state.
Using the Lahman package in R, I have the following code:
library(Lahman)
#filter data frames by year
#collegeInfo <- CollegePlaying %>% filter(yearID >= 1999) #to do later
battingInfo <- Batting %>% filter(yearID >= 1999)
total <- merge(battingInfo,People,by="playerID")
totalN <- total[,-c(24,25,28:47)]
filterByState <- totalN %>% group_by(birthState) %>% summarise(players = length(birthState))
filterByGame <- totalN %>% group_by(birthState) %>% summarise(gamesPlayed = length(G))
In these above two, I am trying to see how many games (G) and number
of DIFFERENT players that played in each state. However, they both return the same values for games played and number of players i.e. birthState 'AB' has a value of 11 games played and also 11 players which should not happen. Both of these values are wrong. There were 11 seasons that a player from birthstate 'AB' played, but of those 11 seasons, only 4 are from different playerIDs. So # of players from birthstate 'AB' should be 4, and adding their games played (G) it should equal 232 G. (4 players and 232 G is correct based off data from totalN)
newMerge <- merge(totalN, filterByState, by="birthState")
newTest <- newMerge %>% group_by(birthState) %>% summarise_at(vars(G, AB, R, H, X2B, X3B, HR, RBI, SB, CS, BB,
SO, IBB, HBP, SH, SF, GIDP), sum, na.rm = TRUE)
This now merges everything, and when you look at birthstate 'AB' it now has 232 games played which is correct, but doesn't show the number of players.
If possible I'd like to see the number of games and DIFFERENT players for each state in the function newTest, with the correct numbers (birthState 'AB' should have 4 players and the updated number that comes from newTest for games played is 232.
For example, the table looks something like this:
playerID birthState Hits Season GamesPlayed
player 1 NJ 17 2009 10
player 1 NJ 10 2010 20
player 2 NJ 20 2009 30
player 3 CA 45 2009 40
player 4 TX 87 2009 50
player 5 CA 50 2009 60
player 6 Outside USA 30 2009 70
And I'd like it to look like this (adding up all the hits for each state):
birthState hits Players GamesPlayed
NJ 47 (17+20+10) 2 60 (10+20+30)
CA 95 (45+50) 2 100 (40+60)
TX 87 1 50
Outside 30 1 70
We can do a group_by sum
library(dplyr)
out <- filterbyState1 %>%
group_by(birthState) %>%
summarise(hits = sum(H))
For multiple columns sum use summarise_at
filterbyState1 %>%
group_by(birthState) %>%
summarise_at(vars(H, players, AB, G), sum, na.rm = TRUE)
Related
I know this is a classic question and there are also similar ones in the archive, but I feel like the answers did not really apply to this case. Basically I want to take one dataframe (covid cases in Berlin per district), calculate the sum of the columns and create a new dataframe with a column representing the name of the district and another one representing the total number. So I wrote
covid_bln <- read.csv('https://www.berlin.de/lageso/gesundheit/infektionsepidemiologie-infektionsschutz/corona/tabelle-bezirke-gesamtuebersicht/index.php/index/all.csv?q=', sep=';')
c_tot<-data.frame('district'=c(), 'number'=c())
for (n in colnames(covid_bln[3:14])){
x<-data.frame('district'=c(n), 'number'=c(sum(covid_bln$n)))
c_tot<-rbind(c_tot, x)
next
}
print(c_tot)
Which works properly with the names but returns only the number of cases for the 8th district, but for all the districts. If you have any suggestion, even involving the use of other functions, it would be great. Thank you
Here's a base R solution:
number <- colSums(covid_bln[3:14])
district <- names(covid_bln[3:14])
c_tot <- cbind.data.frame(district, number)
rownames(c_tot) <- NULL
# If you don't want rownames:
rownames(c_tot) <- NULL
This gives us:
district number
1 mitte 16030
2 friedrichshain_kreuzberg 10679
3 pankow 10849
4 charlottenburg_wilmersdorf 10664
5 spandau 9450
6 steglitz_zehlendorf 9218
7 tempelhof_schoeneberg 12624
8 neukoelln 14922
9 treptow_koepenick 6760
10 marzahn_hellersdorf 6960
11 lichtenberg 7601
12 reinickendorf 9752
I want to provide a solution using tidyverse.
The final result is ordered alphabetically by districts
c_tot <- covid_bln %>%
select( mitte:reinickendorf) %>%
gather(district, number, mitte:reinickendorf) %>%
group_by(district) %>%
summarise(number = sum(number))
The rusult is
# A tibble: 12 x 2
district number
* <chr> <int>
1 charlottenburg_wilmersdorf 10736
2 friedrichshain_kreuzberg 10698
3 lichtenberg 7644
4 marzahn_hellersdorf 7000
5 mitte 16064
6 neukoelln 14982
7 pankow 10885
8 reinickendorf 9784
9 spandau 9486
10 steglitz_zehlendorf 9236
11 tempelhof_schoeneberg 12656
12 treptow_koepenick 6788
This question already has answers here:
dplyr filter with condition on multiple columns
(6 answers)
Closed 2 years ago.
I have a data set like such.
df = data.frame(Business = c('HR','HR','Finance','Finance','Legal','Legal','Research'), Country = c('Iceland','Iceland','Norway','Norway','US','US','France'), Gender=c('Female','Male','Female','Male','Female','Male','Male'), Value =c(10,5,20,40,10,20,50))
I need to be filter out all rows where both male value and female value are >= 10. For example, Iceland HR should be removed as well as Research France.
I've tried df %>% group_by(Business,Country) %>% filter((Value>=10)) but this filters out any value less than 10. any ideas?
Maybe this can help:
library(reshape2)
df2 <- reshape(df,idvar = c('Business','Country'),timevar = 'Gender',direction = 'wide')
df2 %>% mutate(Index=ifelse(Value.Female>=10 & Value.Male>=10,1,0)) %>%
filter(Index==1) -> df3
df4 <- reshape2::melt(df3[,-5],idvar=c('Business','Country'))
Business Country variable value
1 Finance Norway Value.Female 20
2 Legal US Value.Female 10
3 Finance Norway Value.Male 40
4 Legal US Value.Male 20
You could just use two ave steps, one with length, one with min.
df <- df[with(df, ave(Value, Country, FUN=length)) == 2, ]
df[with(df, ave(Value, Country, FUN=min)) >= 10, ]
# Business Country Gender Value
# 3 Finance Norway Female 20
# 4 Finance Norway Male 40
# 5 Legal US Female 10
# 6 Legal US Male 20
Notice that this also works if we disturb the data frame.
set.seed(42)
df2 <- df[sample(1:nrow(df)), ]
df2 <- df2[with(df2, ave(Value, Country, FUN=length)) == 2, ]
df2[with(df2, ave(Value, Country, FUN=min)) >= 10, ]
# Business Country Gender Value
# 5 Legal US Female 10
# 6 Legal US Male 20
# 3 Finance Norway Female 20
# 4 Finance Norway Male 40
I'm using the nycflights13::flights dataframe and want to calculate the number of flights an airplane have flown before its first more than 1 hour delay. How can I do this? I've tried with a group_by and filter, but I haven't been able to. Is there a method to count the rows till a condition (e.g. till the first dep_delay >60)?
Thanks.
library(dplyr)
library(nycflights13)
data("flights")
There may be more elegant ways, but this code counts the total number of flights made by each plane (omitting cancelled flights) and joins this with flights that were not cancelled, grouping on the unique plane identifier (tailnum), sorting on departure date/time, assigning the row_number less 1, filtering on delays>60, and taking the first row.
select(
filter(flights, !is.na(dep_time)) %>%
count(tailnum, name="flights") %>% left_join(
filter(flights, !is.na(dep_time)) %>%
group_by(tailnum) %>%
arrange(month, day, dep_time) %>%
mutate(not_delayed=row_number() -1) %>%
filter(dep_delay>60) %>% slice(1)),
tailnum, flights, not_delayed)
# A tibble: 4,037 x 3
tailnum flights not_delayed
<chr> <int> <dbl>
1 D942DN 4 0
2 N0EGMQ 354 53
3 N10156 146 9
4 N102UW 48 25
5 N103US 46 NA
6 N104UW 47 3
7 N10575 272 0
8 N105UW 45 22
9 N107US 41 20
10 N108UW 60 36
# ... with 4,027 more rows
The plane with tailnum N103US has made 46 flights, of which none have been delayed by more than 1 hour. So the number of flights it has made before its first 1 hour delay is undefined or NA.
I got the answer:
flights %>%
#Eliminate the NAs
filter(!is.na(dep_time)) %>%
#Sort by date and time
arrange(time_hour) %>%
group_by(tailnum) %>%
#cumulative number of flights delayed more than one hour
mutate(acum_delay = cumsum(dep_delay > 60)) %>%
#count the number of flights
summarise(before_1hdelay = sum(acum_delay < 1))
I'm working on the baseball data set:
data(baseball, package="plyr")
library(dplyr)
baseball[,1:4] %>% head
id year stint team
4 ansonca01 1871 1 RC1
44 forceda01 1871 1 WS3
68 mathebo01 1871 1 FW1
99 startjo01 1871 1 NY2
102 suttoez01 1871 1 CL1
106 whitede01 1871 1 CL1
First I want to group the data set by team in order to find the first year each team appears, and the number of distinct players that has ever played for each team:
baseball[,1:4] %>% group_by(team) %>%
summarise("first_year"=min(year), "num_distinct_players"=n_distinct(id))
# A tibble: 132 × 3
team first_year num_distinct_players
<chr> <int> <int>
1 ALT 1884 1
2 ANA 1997 29
3 ARI 1998 43
4 ATL 1966 133
5 BAL 1954 158
Now I want to add a column showing the maximum number of years any player (id) has played for the team in question. To do this, I need to somehow group by player within the existing group (team), and select the maximum number of rows. How do I do this?
Perhaps this helps
baseball %>%
select(1:4) %>%
group_by(id, team) %>%
dplyr::mutate(nyear = n_distinct(year)) %>%
group_by(team) %>%
dplyr::summarise(first_year = min(year),
num_distinct_players = n_distinct(id),
maxYear = max(nyear))
I tried doing this with base R and came up with this. It's fairly slow.
df = data.frame(t(sapply(split(baseball, baseball$team), function(x)
cbind( min(x$year),
length(unique(x$id)),
max(sapply(split(x,x$id), function(y)
nrow(y))),
names(which.max(sapply(split(x,x$id), function(y)
nrow(y)))) ))))
colnames(df) = c("Year", "Unique Players", "Longest played duration",
"Longest Playing Player")
First, split by team into different groups
For each group, obtain the minimum year as first year when the team appears
Get length of unique ids which is the number of players in that team
Split each group into subgroup by id and obtain the maximum number of rows that will give the maximum duration played by a player in that team
For each subgroup, get names of the id with maximum rows which gives the name of the player that played for the longest time in that team
I have searched the forum, but found nothing that could answer or provide hint on how to do what I wish to on the forum.
I have yearly measurement of exposure data from which I wish to calculate individual level annual average based on entry of each individual into the study. For each row the one year exposure assignment should include data from the preceding 12 months starting from the last month before joining the study.
As an example the first person in the sample data joined the study on Feb 7, 2002. His exposure will include a contribution of January 2002 (annual average is 18) and February to December 2001 (annual average is 19). The time weighted average for this person would be (1/12*18) + (11/12*19). The two year average exposure for the same person would extend back from January 2002 to February 2000.
Similarly, for last person who joined the study in December 2004 will include contribution on 11 months in 2004 and one month in 2003 and his annual average exposure will be (11/12*5 ) derived form 2004 and (1/12*6) which comes from the annual average of 2003.
How can I calculate the 1, 2 and 5 year average exposure going back from the date of entry into study? How can I use lags in the manner taht I hve described?
Sample data is accessed from this link
https://drive.google.com/file/d/0B_4NdfcEvU7La1ZCd2EtbEdaeGs/view?usp=sharing
This is not an elegant answer. But, I would like to leave what I tried. I first arranged the data frame. I wanted to identify which year will be the key year for each subject. So, I created id. variable comes from the column names (e.g., pol_2000) in your original data set. entryYear comes from entry in your data. entryMonth comes from entry as well. check was created in order to identify which year is the base year for each participant. In my next step, I extracted six rows for each participant using getMyRows in the SOfun package. In the next step, I used lapply and did math as you described in your question. For the calculation for two/five year average, I divided the total values by year (2 or 5). I was not sure how the final output would look like. So I decided to use the base year for each subject and added three columns to it.
library(stringi)
library(SOfun)
devtools::install_github("hadley/tidyr")
library(tidyr)
library(dplyr)
### Big thanks to BondedDust for this function
### http://stackoverflow.com/questions/6987478/convert-a-month-abbreviation-to-a-numeric-month-in-r
mo2Num <- function(x) match(tolower(x), tolower(month.abb))
### Arrange the data frame.
ana <- foo %>%
mutate(id = 1:n()) %>%
melt(id.vars = c("id","entry")) %>%
arrange(id) %>%
mutate(variable = as.numeric(gsub("^.*_", "", variable)),
entryYear = as.numeric(stri_extract_last(entry, regex = "\\d+")),
entryMonth = mo2Num(substr(entry, 3,5)) - 1,
check = ifelse(variable == entryYear, "Y", "N"))
### Find a base year for each subject and get some parts of data for each participant.
indx <- which(ana$check == "Y")
bob <- getMyRows(ana, pattern = indx, -5:0)
### Get one-year average
cathy <- lapply(bob, function(x){
x$one <- ((x[6,6] / 12) * x[6,4]) + (((12-x[5,6])/12) * x[5,4])
x
})
one <- unnest(lapply(cathy, `[`, i = 6, j = 8))
### Get two-year average
cathy <- lapply(bob, function(x){
x$two <- (((x[6,6] / 12) * x[6,4]) + x[5,4] + (((12-x[4,6])/12) * x[4,4])) / 2
x
})
two <- unnest(lapply(cathy, `[`, i = 6, j =8))
### Get five-year average
cathy <- lapply(bob, function(x){
x$five <- (((x[6,6] / 12) * x[6,4]) + x[5,4] + x[4,4] + x[3,4] + x[2,4] + (((12-x[2,6])/12) * x[1,4])) / 5
x
})
five <- unnest(lapply(cathy, `[`, i =6 , j =8))
### Combine the results with the key observations
final <- cbind(ana[which(ana$check == "Y"),], one, two, five)
colnames(final) <- c(names(ana), "one", "two", "five")
# id entry variable value entryYear entryMonth check one two five
#6 1 07feb2002 2002 18 2002 1 Y 18.916667 18.500000 18.766667
#14 2 06jun2002 2002 16 2002 5 Y 16.583333 16.791667 17.150000
#23 3 16apr2003 2003 14 2003 3 Y 15.500000 15.750000 16.050000
#31 4 26may2003 2003 16 2003 4 Y 16.666667 17.166667 17.400000
#39 5 11jun2003 2003 13 2003 5 Y 13.583333 14.083333 14.233333
#48 6 20feb2004 2004 3 2004 1 Y 3.000000 3.458333 3.783333
#56 7 25jul2004 2004 2 2004 6 Y 2.000000 2.250000 2.700000
#64 8 19aug2004 2004 4 2004 7 Y 4.000000 4.208333 4.683333
#72 9 19dec2004 2004 5 2004 11 Y 5.083333 5.458333 4.800000