Apply function to each column of dataframe - r

I have the following (very large) dataframe:
id epoch
1 0 1.141194e+12
2 1 1.142163e+12
3 2 1.142627e+12
4 2 1.142627e+12
5 3 1.142665e+12
6 3 1.142665e+12
7 4 1.142823e+12
8 5 1.143230e+12
9 6 1.143235e+12
10 6 1.143235e+12
For every unique ID, I now want to get the difference between its maximum and minimum time (epoch timestamp). There are IDs with many more occurences than in the example above, in case it is relevant. I haven't worked much with R yet and tried the following:
unique = data.frame(as.numeric(unique(df$id)))
differences = apply(unique, 1, get_duration)
get_duration = function(id) {
maxTime = max(df$epoch[which(df$id == id)])
minTime = min(df$epoch[which(df$id == id)])
return ((maxTime - minTime) / 1000)
}
It works, but is incredibly slow. What would be a faster approach?

A couple of approaches. In base R:
tapply(df$epoch,df$id,function(x) (max(x)-min(x))/1000)
With data.table:
require(data.table)
setDT(df)
df[,list(d=(max(epoch)-min(epoch))/1000),by=id]

This can be done easily in dplyr
require(dplyr)
df %>% group_by(id) %>% summarize(diff=(max(epoch)-min(epoch))/1000)

Use the filter by id just once
subset = df$epoch[which(df$id == id)]
maxTime = max(subset)
minTime = min(subset)

Related

How to do aggregate sum by time range in R?

I have a dataframe as below:
**df**
Cust_name time freq
Andrew 0 4
Dillain 1 2
Alma 2 3
Andrew 1 4
Kiko 2 1
Sarah 2 8
Sarah 0 3
I want to calculate the sum of frequency by the time range provided for each cust_name. Example: If I select time range 0 to 2 for Andrew, it will give me sum of freq: 4+4= 8. And for Sarah, it will give me 8+3=11. I have tried it in the following ways just to get the time range, but do not know how to do the rest, as I am very new to R:
df[(df$time>=0 & df$time<=2),]
You can do this with dplyr.
To make your code reproducible, you should add the creation of your dataframe in your post. Copy and pasting everything is time consuming.
library(dplyr)
df <- data.frame(
cust_name = c('Andrew', 'Dillain', 'Alma', 'Andrew', 'Kiko', 'Sarah', 'Sarah'),
time = c(0,1,2,1,2,2,0),
freq = c(4,2,3,4,1,8,3)
)
df %>%
filter(time >=0, time <=2) %>%
group_by(cust_name) %>%
summarise(sum_freq = sum(freq))

R count and substract events from a data frame

I am trying to calculate the families sizes from a data frame, which also contains two types of events : family members who died, and those who left the family. I would like to take into account these two parameters in order to compute the actual family size.
Here is a reproductive example of my problem, with 3 families only :
family <- factor(rep(c("001","002","003"), c(10,8,15)), levels=c("001","002","003"), labels=c("001","002","003"), ordered=TRUE)
dead <- c(0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0)
left <- c(0,0,0,0,0,1,0,0,0,1,1,0,0,0,1,1,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0)
DF <- data.frame(family, dead, left) ; DF
I could count N = total family members (in each family) in a second dataframe DF2, by simply using table()
DF2 <- with(DF, data.frame(table(family)))
colnames(DF2)[2] <- "N" ; DF2
family N
1 001 10
2 002 8
3 003 15
But i can not find a proper way to get the actual number of people (for example, creating a new variable N2 into DF2) , calculated by substracting to N the number of members who died or left the family. I suppose i have to relate the two dataframes DF and DF2 in a way. i have looked for other related questions in this site but could not find the right answer...
If anyone has a good idea, it would be great !
Thank you in advance..
Deni
Logic : First we want to group_by(family) and then calculate 2 numbers : i) total #obs in each group ii) subtract the sum(dead) + sum(left) from this total .
In dplyr package : n() helps us get the total #observations in each group
In data.table : .N does the same above job
library(dplyr)
DF %>% group_by(family) %>% summarise( total = n(), current = n()-sum(dead,left, na.rm = TRUE))
# family total current
# (fctr) (int) (dbl)
#1 001 10 6
#2 002 8 4
#3 003 15 7
library(data.table)
# setDT() is preferred if incase your data was a data.frame. else just DF.
setDT(DF)[, .(total = .N, current = .N - sum(dead, left, na.rm = TRUE)), by = family]
# family total current
#1: 001 10 6
#2: 002 8 4
#3: 003 15 7
Here is a base R option
do.call(data.frame, aggregate(dl~family, transform(DF, dl = dead + left),
FUN = function(x) c(total=length(x), current=length(x) - sum(x))))
Or a modified version is
transform(aggregate(. ~ family, transform(DF, total = 1,
current = dead + left)[c(1,4:5)], FUN = sum), current = total - current)
# family total current
#1 001 10 6
#2 002 8 4
#3 003 15 7
I finally found another which works fine (from another post), allowing to compute everything from the original DF table. This uses the ddply function :
DF <- ddply(DF,.(family),transform,total=length(family))
DF <- ddply(DF,.(family),transform,actual=length(family)-sum(dead=="1")-sum(left=="1"))
DF
Thanks a lot to everyone who helped ! Deni

Create new index / re-index in dplyr [duplicate]

This question already has answers here:
How to number/label data-table by group-number from group_by?
(6 answers)
Closed 6 years ago.
I am using a dplyr table in R. Typical fields would be a primary key, an id number identifying a group, a date field, and some values. There are numbersI did some manipulation that throws out a bunch of data in some preliminary steps.
In order to do the next step of my analysis (in MC Stan), It'll be easier if both the date and the group id fields are integer indices. So basically, I need to re-index them as integers between 1 and whatever the total number of distinct elements are (about 750 for group_id and about 250 for date_id, the group_id is already integer, but the date is not). This is relatively straightforward to do after exporting it to a data frame, but I was curious if it is possible in dplyr.
My attempt at creating a new date_val (called date_val_new) is below. Per the discussion in the comments I have some fake data. I purposefully made the group and date values not be 1 to whatever, but I didn't make the date an actual date. I made the data unbalanced, removing some values to illustrate the issue. The dplyr command re-starts the index at 1 for each new group, regardless of what date_val it is. So every group starts at 1, even if the date is different.
df1 <- data.frame(id = 1:40,
group_id = (10 + rep(1:10, each = 4)),
date_val = (20 + rep(rep(1:4), 10)),
val = runif(40))
for (i in c(5, 17, 33))
{
df1 <- df1[!df1$id == i, ]
}
df_new <- df1 %>%
group_by(group_id) %>%
arrange(date_val) %>%
mutate(date_val_new=row_number(group_id)) %>%
ungroup()
This is the base R method:
df1 %>% mutate(date_val_new = match(date_val, unique(date_val)))
Or with a data.table, df1[, date_val_new := .GRP, by=date_val].
Use group_indices_() to generate a unique id for each group:
df1 %>% mutate(date_val_new = group_indices_(., .dots = "date_val"))
Update
Since group_indices() does not handle class tbl_postgres, you could try dense_rank()
copy_to(my_db, df1, name = "df1")
tbl(my_db, "df1") %>%
mutate(date_val_new = dense_rank(date_val))
Or build a custom query using sql()
tbl(my_db, sql("SELECT *,
DENSE_RANK() OVER (ORDER BY date_val) AS DATE_VAL_NEW
FROM df1"))
Alternatively, I think you can try getanID() from the splitstackshape package.
library(splitstackshape)
getanID(df1, "group_id")[]
# id group_id date_val val .id
# 1: 1 11 21 0.01857242 1
# 2: 2 11 22 0.57124557 2
# 3: 3 11 23 0.54318903 3
# 4: 4 11 24 0.59555088 4
# 5: 6 12 22 0.63045007 1
# 6: 7 12 23 0.74571297 2
# 7: 8 12 24 0.88215668 3

How to subset data for a specific column with ddply?

I would like to know if there is a simple way to achieve what I describe below using ddply. My data frame describes an experiment with two conditions. Participants had to select between options A and B, and we recorded how long they took to decide, and whether their responses were accurate or not.
I use ddply to create averages by condition. The column nAccurate summarizes the number of accurate responses in each condition. I also want to know how much time they took to decide and express it in the column RT. However, I want to calculate average response times only when participants got the response right (i.e. Accuracy==1). Currently, the code below can only calculate average reaction times for all responses (accurate and inaccurate ones). Is there a simple way to modify it to get average response times computed only in accurate trials?
See sample code below and thanks!
library(plyr)
# Create sample data frame.
Condition = c(rep(1,6), rep(2,6)) #two conditions
Response = c("A","A","A","A","B","A","B","B","B","B","A","A") #whether option "A" or "B" was selected
Accuracy = rep(c(1,1,0),4) #whether the response was accurate or not
RT = c(110,133,121,122,145,166,178,433,300,340,250,674) #response times
df = data.frame(Condition,Response, Accuracy,RT)
head(df)
Condition Response Accuracy RT
1 1 A 1 110
2 1 A 1 133
3 1 A 0 121
4 1 A 1 122
5 1 B 1 145
6 1 A 0 166
# Calculate averages.
avg <- ddply(df, .(Condition), summarise,
N = length(Response),
nAccurate = sum(Accuracy),
RT = mean(RT))
# The problem: response times are calculated over all trials. I would like
# to calculate mean response times *for accurate responses only*.
avg
Condition N nAccurate RT
1 6 4 132.8333
2 6 4 362.5000
With plyr, you can do it as follows:
ddply(df,
.(Condition), summarise,
N = length(Response),
nAccurate = sum(Accuracy),
RT = mean(RT[Accuracy==1]))
this gives:
Condition N nAccurate RT
1: 1 6 4 127.50
2: 2 6 4 300.25
If you use data.table, then this is an alternative way:
library(data.table)
setDT(df)[, .(N = .N,
nAccurate = sum(Accuracy),
RT = mean(RT[Accuracy==1])),
by = Condition]
Using dplyr package:
library(dplyr)
df %>%
group_by(Condition) %>%
summarise(N = n(),
nAccurate = sum(Accuracy),
RT = mean(RT[Accuracy == 1]))

Count based on multiple conditions from other data.frame

I am migrating analysis from Excel to R, and would like some input on how best to perform something similar to Excel's COUNTIFS in R.
I have a two data.frames, statedf and memberdf.
statedf=data.frame(state=c('MD','MD','MD','NY','NY','NY'), week = 5:7)
memberdf=data.frame(memID = 1:15, state = c('MD','MD','NY','NY','MD'),
finalweek = c(3,3,5,3,3,5,3,5,3,5,6,5,2,3,5),
orders = c(1,2,3))
This data is for a subscription-based business. I would like to know the number of members who newly lapsed for each week/state combo in statedf, where newly lapse is defined by statedf$week - 1 = memberdf$finalweek. Further I would like to have separate counts for each order value (1,2,3).
The desired output would look like
out <- data.frame(state=c('MD','MD','MD','NY','NY','NY'), week = 5:7,
oneorder = c(0,1,0,0,0,0),
twoorder = c(0,0,1,0,1,0),
threeorder = c(0,3,0,0,1,0))
I asked (and got a great response for) a simpler version of this question yesterday - the answers revolved around creating a new data.frame based on member.df. However, I need to append the data to statedf, because statedf has member/week combos that don't exist in memberdf, and vice versa. If this was in Excel, I'd use COUNTIFS but am struggling for a solution in R.
Thanks.
Here is a solution with the dplyr and tidyr packages:
library(tidyr) ; library(dplyr)
counts <- memberdf %>%
mutate(lapsedweek = finalweek + 1) %>%
group_by(state, lapsedweek, orders) %>%
tally()
counts <- counts %>% spread(orders, n, fill = 0)
out <- left_join(statedf, counts, by = c("state", "week" = "lapsedweek"))
out[is.na(out)] <- 0 # convert rows with all NAs to 0s
names(out)[3:5] <- paste0("order", names(out)[3:5]) # rename columns
We could create a new variable ('week1') in the 'statedf' dataset, merge the 'memberdf' with 'statedf', and then reshape from 'long' to 'wide' format with dcast. I changed the 'orders' column to match the column names in the 'out'.
statedf$week1 <- statedf$week-1
df1 <- merge(memberdf[-1], statedf, by.x=c('state', 'finalweek'),
by.y=c('state', 'week1'), all.y=TRUE)
lvls <- paste0(c('one', 'two', 'three'), 'order')
df1$orders <- factor(lvls[df1$orders],levels=lvls)
library(reshape2)
out1 <- dcast(df1, state+week~orders, value.var='orders', length)[-6]
out1
# state week oneorder twoorder threeorder
#1 MD 5 0 0 0
#2 MD 6 1 0 3
#3 MD 7 0 1 0
#4 NY 5 0 0 0
#5 NY 6 0 1 1
#6 NY 7 0 0 0
all.equal(out, out1)
#[1] TRUE

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