count positive negative values in column by group - r

I want to create two variables giving me the total number of positive and negative values by id, hopefully using dplyr.
Example data:
library(dplyr)
set.seed(42)
df <- data.frame (id=rep(1:10,each=10),
ff=rnorm(100, 0,14 ))
> head(df,20)
id ff
1 1 19.1934183
2 1 -7.9057744
3 1 5.0837978
4 1 8.8600765
5 1 5.6597565
6 1 -1.4857432
7 1 21.1613080
8 1 -1.3252265
9 1 28.2579320
10 1 -0.8779974
11 2 18.2681752
12 2 32.0130355
13 2 -19.4440498
14 2 -3.9030427
15 2 -1.8664987
16 2 8.9033056
17 2 -3.9795409
18 2 -37.1903759
19 2 -34.1665370
20 2 18.4815868
the resulting dataset should look like:
> head(df,20)
id ff pos neg
1 1 19.1934183 6 4
2 1 -7.9057744 6 4
3 1 5.0837978 6 4
4 1 8.8600765 6 4
5 1 5.6597565 6 4
6 1 -1.4857432 6 4
7 1 21.1613080 6 4
8 1 -1.3252265 6 4
9 1 28.2579320 6 4
10 1 -0.8779974 6 4
11 2 18.2681752 4 6
12 2 32.0130355 4 6
13 2 -19.4440498 4 6
14 2 -3.9030427 4 6
15 2 -1.8664987 4 6
16 2 8.9033056 4 6
17 2 -3.9795409 4 6
18 2 -37.1903759 4 6
19 2 -34.1665370 4 6
20 2 18.4815868 4 6
I have thought something similar to this will work:
df<-df%>% group_by(id) %>% mutate(pos= nrow(ff>0)) %>% ungroup()
Any help would be great, thanks.


You need sum():
df %>% group_by(id) %>%
mutate(pos = sum(ff>0),
neg = sum(ff<0))


For a fun (and a fast) solution data.table can also be used:
library(data.table)
setDT(df)
df[, ":="(pos = sum(ff > 0), neg = sum(ff < 0)), by = id]


Here's an answer that add the ifelse part of your question:
df <- df %>% group_by(id) %>%
mutate(pos = sum(ff>0), neg = sum(ff<0)) %>%
group_by(id) %>%
mutate(any_neg=ifelse(any(ff < 0), 1, 0))
Output:
> head(df, 20)
Source: local data frame [20 x 5]
Groups: id [2]
id ff pos neg any_neg
<int> <dbl> <int> <int> <dbl>
1 1 19.1934183 6 4 1
2 1 -7.9057744 6 4 1
3 1 5.0837978 6 4 1
4 1 8.8600765 6 4 1
5 1 5.6597565 6 4 1
6 1 -1.4857432 6 4 1
7 1 21.1613080 6 4 1
8 1 -1.3252265 6 4 1
9 1 28.2579320 6 4 1
10 1 -0.8779974 6 4 1
11 2 18.2681752 4 6 1
12 2 32.0130355 4 6 1
13 2 -19.4440498 4 6 1
14 2 -3.9030427 4 6 1
15 2 -1.8664987 4 6 1
16 2 8.9033056 4 6 1
17 2 -3.9795409 4 6 1
18 2 -37.1903759 4 6 1
19 2 -34.1665370 4 6 1
20 2 18.4815868 4 6 1

Related

Exclude rows where value used in another row

Imagine you have the following data set:
df = data.frame(ID = c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20), gender= c(1,2,1,2,2,2,2,1,1,2,1,2,1,2,2,2,2,1,1,2),
PID = c(1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10))
how can I write a code that removes the rows in the df whose gender and PID are the same (see picture). Please imagine that the code is over 1000 rows long (so it should be a solution that automatically searches for the right values to exclude).

base R
df[ave(rep(TRUE, nrow(df)), df[,c("gender","paar")], FUN = function(z) !any(duplicated(z))),]
# ID gender paar
# 1 1 1 1
# 2 2 2 1
# 3 3 1 2
# 4 4 2 2
# 7 7 2 4
# 8 8 1 4
# 9 9 1 5
# 10 10 2 5
# 11 11 1 6
# 12 12 2 6
# 13 13 1 7
# 14 14 2 7
# 17 17 2 9
# 18 18 1 9
# 19 19 1 10
# 20 20 2 10
dplyr
library(dplyr)
df %>%
group_by(gender, paar) %>%
filter(!any(duplicated(cbind(gender, paar)))) %>%
ungroup()

In base R, we may use subset after removing the observations where the group count for 'gender' and 'paar' are not 1
subset(df, ave(seq_along(gender), gender, paar, FUN = length) == 1)
Or with duplicated
df[!(duplicated(df[-1])|duplicated(df[-1], fromLast = TRUE)),]
-output
ID gender paar
1 1 1 1
2 2 2 1
3 3 1 2
4 4 2 2
7 7 2 4
8 8 1 4
9 9 1 5
10 10 2 5
11 11 1 6
12 12 2 6
13 13 1 7
14 14 2 7
17 17 2 9
18 18 1 9
19 19 1 10
20 20 2 10

Here is one more: :-)
library(dplyr)
df %>%
group_by(gender, PID) %>%
filter(is.na(ifelse(n()>1, 1, NA)))
ID gender PID
<dbl> <dbl> <dbl>
1 1 1 1
2 2 2 1
3 3 1 2
4 4 2 2
5 7 2 4
6 8 1 4
7 9 1 5
8 10 2 5
9 11 1 6
10 12 2 6
11 13 1 7
12 14 2 7
13 17 2 9
14 18 1 9
15 19 1 10
16 20 2 10

Another dplyr option could be:
df %>%
filter(with(rle(paste0(gender, PID)), rep(lengths == 1, lengths)))
ID gender PID
1 1 1 1
2 2 2 1
3 3 1 2
4 4 2 2
5 7 2 4
6 8 1 4
7 9 1 5
8 10 2 5
9 11 1 6
10 12 2 6
11 13 1 7
12 14 2 7
13 17 2 9
14 18 1 9
15 19 1 10
16 20 2 10
If the duplicated values can occur also between non-consecutive rows:
df %>%
arrange(gender, PID) %>%
filter(with(rle(paste0(gender, PID)), rep(lengths == 1, lengths)))

Using aggregate
na.omit(aggregate(. ~ gender + PID, df, function(x)
ifelse(length(x) == 1, x, NA)))
gender PID ID
1 1 1 1
2 2 1 2
3 1 2 3
4 2 2 4
6 1 4 8
7 2 4 7
8 1 5 9
9 2 5 10
10 1 6 11
11 2 6 12
12 1 7 13
13 2 7 14
15 1 9 18
16 2 9 17
17 1 10 19
18 2 10 20
With dplyr
library(dplyr)
df %>%
group_by(gender, PID) %>%
filter(n() == 1) %>%
ungroup()
# A tibble: 16 × 3
ID gender PID
<dbl> <dbl> <dbl>
1 1 1 1
2 2 2 1
3 3 1 2
4 4 2 2
5 7 2 4
6 8 1 4
7 9 1 5
8 10 2 5
9 11 1 6
10 12 2 6
11 13 1 7
12 14 2 7
13 17 2 9
14 18 1 9
15 19 1 10
16 20 2 10

Replace row value in a data frame group by the smallest value in that group

I have the following data set:
time <- c(0,1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5)
value <- c(10,8,6,5,3,2,12,10,6,5,4,2,20,15,16,9,2,2)
group <- c(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3)
data <- data.frame(time, value, group)
I want to create a new column called data$diff that is equal to data$value minus the value of data$value when data$time == 0 within each group.
I am beginning with the following code
for(i in 1:nrow(data)){
for(n in 1:max(data$group)){
if(data$group[i] == n) {
data$diff[i] <- ???????
}
}
}
But cannot figure out what to put in place of the question marks. The desired output would be this table: https://i.stack.imgur.com/1bAKj.png
Any thoughts are appreciated.

Since in your example data$time == 0 is always the first element of the group, you can use this data.table approach.
library(data.table)
setDT(data)
data[, diff := value[1] - value, by = group]
In case that data$time == 0 is not the first element in each group you can use this:
data[, diff := value[time==0] - value, by = group]
Output:
> data
time value group diff
1: 0 10 1 0
2: 1 8 1 2
3: 2 6 1 4
4: 3 5 1 5
5: 4 3 1 7
6: 5 2 1 8
7: 0 12 2 0
8: 1 10 2 2
9: 2 6 2 6
10: 3 5 2 7
11: 4 4 2 8
12: 5 2 2 10
13: 0 20 3 0
14: 1 15 3 5
15: 2 16 3 4
16: 3 9 3 11
17: 4 2 3 18
18: 5 2 3 18

Here is a base R approach.
within(data, diff <- ave(
seq_along(value), group,
FUN = \(i) value[i][time[i] == 0] - value[i]
))
Output
time value group diff
1 0 10 1 0
2 1 8 1 2
3 2 6 1 4
4 3 5 1 5
5 4 3 1 7
6 5 2 1 8
7 0 12 2 0
8 1 10 2 2
9 2 6 2 6
10 3 5 2 7
11 4 4 2 8
12 5 2 2 10
13 0 20 3 0
14 1 15 3 5
15 2 16 3 4
16 3 9 3 11
17 4 2 3 18
18 5 2 3 18

Here is a short way to do it with dplyr.
library(dplyr)
data %>%
group_by(group) %>%
mutate(diff = value[which(time == 0)] - value)
Which gives
# Groups: group [3]
time value group diff
<dbl> <dbl> <dbl> <dbl>
1 0 10 1 0
2 1 8 1 2
3 2 6 1 4
4 3 5 1 5
5 4 3 1 7
6 5 2 1 8
7 0 12 2 0
8 1 10 2 2
9 2 6 2 6
10 3 5 2 7
11 4 4 2 8
12 5 2 2 10
13 0 20 3 0
14 1 15 3 5
15 2 16 3 4
16 3 9 3 11
17 4 2 3 18
18 5 2 3 18

library(dplyr)
vals2use <- data %>%
group_by(group) %>%
filter(time==0) %>%
select(c(2,3)) %>%
rename(value4diff=value)
dataNew <- merge(data, vals2use, all=T)
dataNew$diff <- dataNew$value4diff-dataNew$value
dataNew <- dataNew[,c(1,2,3,5)]
dataNew
group time value diff
1 1 0 10 0
2 1 1 8 2
3 1 2 6 4
4 1 3 5 5
5 1 4 3 7
6 1 5 2 8
7 2 0 12 0
8 2 1 10 2
9 2 2 6 6
10 2 3 5 7
11 2 4 4 8
12 2 5 2 10
13 3 0 20 0
14 3 1 15 5
15 3 2 16 4
16 3 3 9 11
17 3 4 2 18
18 3 5 2 18

Apply same function to several data replicates in R

Consider the following data simulation mechanism:
set.seed(1)
simulW <- function(G)
{
# Let G be the number of groups
n<-2*G #Assume 2 individuals per group
i<-rep(1:G, rep(2,G)) # Group index
j<-rep (1:n)
Y<-rbinom(n, 1, 0.5) # binary
data.frame(id=1:n, i,Y)
}
r<-5 #5 replicates
dat1 <- replicate(r, simulW(G = 10 ), simplify=FALSE)
#For example the first data replicate will be
> dat1[[1]]
id i Y
1 1 1 0
2 2 1 1
3 3 2 0
4 4 2 0
5 5 3 0
6 6 3 0
7 7 4 0
8 8 4 1
9 9 5 1
10 10 5 0
The code below can perform group wise (i is the group) sum of Y but by default considers only the first replicate i.e dat1[[1]].
Di<-aggregate( Y, by=list ( i ),FUN=sum) #Sum per group for the first dataset
e<-colSums(Di [ 2 ] ) #Total sum of Y for all groups for dataset 1
> e
x
8
di<-Di [ 2 ] # Groupwise sum for replicate 1
> di
x
1 2
2 2
3 2
4 0
5 2
How can I use the same function to perform the group wise sum for the other replicates.
Maybe something like:
for (m in 1:r )
{
Di[m]<-
e[m]<-
di[m]<-
}

You may use aggregate in lapply -
result <- lapply(dat1, function(x) aggregate(Y~i, x, sum))
result
#[[1]]
# i Y
#1 1 1
#2 2 1
#3 3 0
#4 4 0
#5 5 1
#6 6 1
#7 7 0
#8 8 2
#9 9 1
#10 10 1
#[[2]]
# i Y
#1 1 2
#2 2 2
#3 3 2
#4 4 0
#5 5 2
#6 6 1
#7 7 0
#8 8 0
#9 9 1
#10 10 1
#...
#...

We may use tidyverse
library(purrr)
library(dplyr)
map(dat1, ~ .x %>%
group_by(i) %>%
summarise(Y = sum(Y)))
-output
[[1]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 0
2 2 2
3 3 1
4 4 2
5 5 1
6 6 0
7 7 1
8 8 1
9 9 2
10 10 1
[[2]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 1
2 2 1
3 3 0
4 4 0
5 5 1
6 6 1
7 7 0
8 8 2
9 9 1
10 10 1
[[3]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 2
2 2 2
3 3 2
4 4 0
5 5 2
6 6 1
7 7 0
8 8 0
9 9 1
10 10 1
[[4]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 1
2 2 0
3 3 1
4 4 1
5 5 1
6 6 1
7 7 0
8 8 1
9 9 1
10 10 2
[[5]]
# A tibble: 10 × 2
i Y
<int> <int>
1 1 1
2 2 0
3 3 1
4 4 1
5 5 0
6 6 0
7 7 2
8 8 2
9 9 0
10 10 2

Retrieve a value by another column criteria in R

i need some help:
i got this df:
df <- data.frame(month = c(1,1,1,1,1,2,2,2,2,2),
day = c(1,2,3,4,5,1,2,3,4,5),
flow = c(2,5,7,8,5,4,6,7,9,2))
month day flow
1 1 1 2
2 1 2 5
3 1 3 7
4 1 4 8
5 1 5 5
6 2 1 4
7 2 2 6
8 2 3 7
9 2 4 9
10 2 5 2
but i want to know the day of min per month:
month day flow dayminflowofthemonth
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5
this repetition is not a problem, i will use pivot fuction
tks people!

We can use which.min to return the index of 'min'imum 'flow' per group and use that to get the corresponding 'day' to create the column with mutate
library(dplyr)
df <- df %>%
group_by(month) %>%
mutate(dayminflowofthemonth = day[which.min(flow)]) %>%
ungroup
-output
df
# A tibble: 10 x 4
# month day flow dayminflowofthemonth
# <dbl> <dbl> <dbl> <dbl>
# 1 1 1 2 1
# 2 1 2 5 1
# 3 1 3 7 1
# 4 1 4 8 1
# 5 1 5 5 1
# 6 2 1 4 5
# 7 2 2 6 5
# 8 2 3 7 5
# 9 2 4 9 5
#10 2 5 2 5

Another option using indexing inside dplyr pipeline:
library(dplyr)
#Code
newdf <- df %>% group_by(month) %>% mutate(Val=day[flow==min(flow)][1])
Output:
# A tibble: 10 x 4
# Groups: month [2]
month day flow Val
<dbl> <dbl> <dbl> <dbl>
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5

Here is a base R option using ave
transform(
df,
dayminflowofthemonth = ave(day*(ave(flow,month,FUN = min)==flow),month,FUN = max)
)
which gives
month day flow dayminflowofthemonth
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5

One more base R approach:
df$dayminflowofthemonth <- by(
df,
df$month,
function(x) x$day[which.min(x$flow)]
)[df$month]

How to find difference between values in two rows in an R dataframe using dplyr

I have an R dataframe such as:
df <- data.frame(period=rep(1:4,2),
farm=c(rep('A',4),rep('B',4)),
cumVol=c(1,5,15,31,10,12,16,24),
other = 1:8);
period farm cumVol other
1 1 A 1 1
2 2 A 5 2
3 3 A 15 3
4 4 A 31 4
5 1 B 10 5
6 2 B 12 6
7 3 B 16 7
8 4 B 24 8
How do I find the change in cumVol at each farm in each period, ignoring the 'other' column? I would like a dataframe like this (optionally with the cumVol column remaining):
period farm volume other
1 1 A 0 1
2 2 A 4 2
3 3 A 10 3
4 4 A 16 4
5 1 B 0 5
6 2 B 2 6
7 3 B 4 7
8 4 B 8 8
In practice there may be many 'farm'-like columns, and many 'other'-like (ie. ignored) columns. I'd like to be able to specify all the column names using variables.
I am using the dplyr package.

In dplyr:
require(dplyr)
df %>%
group_by(farm) %>%
mutate(volume = cumVol - lag(cumVol, default = cumVol[1]))
Source: local data frame [8 x 5]
Groups: farm
period farm cumVol other volume
1 1 A 1 1 0
2 2 A 5 2 4
3 3 A 15 3 10
4 4 A 31 4 16
5 1 B 10 5 0
6 2 B 12 6 2
7 3 B 16 7 4
8 4 B 24 8 8
Perhaps the desired output should actually be as follows?
df %>%
group_by(farm) %>%
mutate(volume = cumVol - lag(cumVol, default = 0))
period farm cumVol other volume
1 1 A 1 1 1
2 2 A 5 2 4
3 3 A 15 3 10
4 4 A 31 4 16
5 1 B 10 5 10
6 2 B 12 6 2
7 3 B 16 7 4
8 4 B 24 8 8
Edit: Following up on your comments I think you are looking for arrange(). It that is not the case it might be best to start a new question.
df1 <- data.frame(period=rep(1:4,4), farm=rep(c(rep('A',4),rep('B',4)),2), crop=(c(rep('apple',8), rep('pear',8))), cumCropVol=c(1,5,15,31,10,12,16,24,11,15,25,31,20,22,26,34), other = rep(1:8,2) );
df1 %>%
arrange(desc(period), desc(farm)) %>%
group_by(period, farm) %>%
summarise(cumVol=sum(cumCropVol))
Edit: Follow up #2
df1 <- data.frame(period=rep(1:4,4), farm=rep(c(rep('A',4),rep('B',4)),2), crop=(c(rep('apple',8), rep('pear',8))), cumCropVol=c(1,5,15,31,10,12,16,24,11,15,25,31,20,22,26,34), other = rep(1:8,2) );
df <- df1 %>%
arrange(desc(period), desc(farm)) %>%
group_by(period, farm) %>%
summarise(cumVol=sum(cumCropVol))
ungroup(df) %>%
arrange(farm) %>%
group_by(farm) %>%
mutate(volume = cumVol - lag(cumVol, default = 0))
Source: local data frame [8 x 4]
Groups: farm
period farm cumVol volume
1 1 A 12 12
2 2 A 20 8
3 3 A 40 20
4 4 A 62 22
5 1 B 30 30
6 2 B 34 4
7 3 B 42 8
8 4 B 58 16

In dplyr -- so you don't have to replace NAs
library(dplyr)
df %>%
group_by(farm)%>%
mutate(volume = c(0,diff(cumVol)))
period farm cumVol other volume
1 1 A 1 1 0
2 2 A 5 2 4
3 3 A 15 3 10
4 4 A 31 4 16
5 1 B 10 5 0
6 2 B 12 6 2
7 3 B 16 7 4
8 4 B 24 8 8

Would creating a new column in your original dataset be an option?
Here is an option using the data.table operator :=.
require("data.table")
DT <- data.table(df)
DT[, volume := c(0,diff(cumVol)), by="farm"]
or
diff_2 <- function(x) c(0,diff(x))
DT[, volume := diff_2(cumVol), by="farm"]
Output:
# > DT
# period farm cumVol other volume
# 1: 1 A 1 1 0
# 2: 2 A 5 2 4
# 3: 3 A 15 3 10
# 4: 4 A 31 4 16
# 5: 1 B 10 5 0
# 6: 2 B 12 6 2
# 7: 3 B 16 7 4
# 8: 4 B 24 8 8

tapply and transform?
> transform(df, volumen=unlist(tapply(cumVol, farm, function(x) c(0, diff(x)))))
period farm cumVol other volumen
A1 1 A 1 1 0
A2 2 A 5 2 4
A3 3 A 15 3 10
A4 4 A 31 4 16
B1 1 B 10 5 0
B2 2 B 12 6 2
B3 3 B 16 7 4
B4 4 B 24 8 8
ave is a better option, see # thelatemail's comment
with(df, ave(cumVol,farm,FUN=function(x) c(0,diff(x))) )

Resources