R- data frame, separating by value of col [duplicate] - r

Question

I have a dataframe and I would like to count the number of rows within each group. I reguarly use the aggregate function to sum data as follows:
df2 <- aggregate(x ~ Year + Month, data = df1, sum)
Now, I would like to count observations but can't seem to find the proper argument for FUN. Intuitively, I thought it would be as follows:
df2 <- aggregate(x ~ Year + Month, data = df1, count)
But, no such luck.
Any ideas?
Some toy data:
set.seed(2)
df1 <- data.frame(x = 1:20,
Year = sample(2012:2014, 20, replace = TRUE),
Month = sample(month.abb[1:3], 20, replace = TRUE))

Answer 1

Current best practice (tidyverse) is:
require(dplyr)
df1 %>% count(Year, Month)

Answer 2

Following #Joshua's suggestion, here's one way you might count the number of observations in your df dataframe where Year = 2007 and Month = Nov (assuming they are columns):
nrow(df[,df$YEAR == 2007 & df$Month == "Nov"])
and with aggregate, following #GregSnow:
aggregate(x ~ Year + Month, data = df, FUN = length)

Answer 3

dplyr package does this with count/tally commands, or the n() function:
First, some data:
df <- data.frame(x = rep(1:6, rep(c(1, 2, 3), 2)), year = 1993:2004, month = c(1, 1:11))
Now the count:
library(dplyr)
count(df, year, month)
#piping
df %>% count(year, month)
We can also use a slightly longer version with piping and the n() function:
df %>%
group_by(year, month) %>%
summarise(number = n())
or the tally function:
df %>%
group_by(year, month) %>%
tally()

Answer 4

An old question without a data.table solution. So here goes...
Using .N
library(data.table)
DT <- data.table(df)
DT[, .N, by = list(year, month)]

Answer 5

The simple option to use with aggregate is the length function which will give you the length of the vector in the subset. Sometimes a little more robust is to use function(x) sum( !is.na(x) ).

Answer 6

Create a new variable Count with a value of 1 for each row:
df1["Count"] <-1
Then aggregate dataframe, summing by the Count column:
df2 <- aggregate(df1[c("Count")], by=list(Year=df1$Year, Month=df1$Month), FUN=sum, na.rm=TRUE)

Answer 7

An alternative to the aggregate() function in this case would be table() with as.data.frame(), which would also indicate which combinations of Year and Month are associated with zero occurrences
df<-data.frame(x=rep(1:6,rep(c(1,2,3),2)),year=1993:2004,month=c(1,1:11))
myAns<-as.data.frame(table(df[,c("year","month")]))
And without the zero-occurring combinations
myAns[which(myAns$Freq>0),]

Answer 8

If you want to include 0 counts for month-years that are missing in the data, you can use a little table magic.
data.frame(with(df1, table(Year, Month)))
For example, the toy data.frame in the question, df1, contains no observations of January 2014.
df1
x Year Month
1 1 2012 Feb
2 2 2014 Feb
3 3 2013 Mar
4 4 2012 Jan
5 5 2014 Feb
6 6 2014 Feb
7 7 2012 Jan
8 8 2014 Feb
9 9 2013 Mar
10 10 2013 Jan
11 11 2013 Jan
12 12 2012 Jan
13 13 2014 Mar
14 14 2012 Mar
15 15 2013 Feb
16 16 2014 Feb
17 17 2014 Mar
18 18 2012 Jan
19 19 2013 Mar
20 20 2012 Jan
The base R aggregate function does not return an observation for January 2014.
aggregate(x ~ Year + Month, data = df1, FUN = length)
Year Month x
1 2012 Feb 1
2 2013 Feb 1
3 2014 Feb 5
4 2012 Jan 5
5 2013 Jan 2
6 2012 Mar 1
7 2013 Mar 3
8 2014 Mar 2
If you would like an observation of this month-year with 0 as the count, then the above code will return a data.frame with counts for all month-year combinations:
data.frame(with(df1, table(Year, Month)))
Year Month Freq
1 2012 Feb 1
2 2013 Feb 1
3 2014 Feb 5
4 2012 Jan 5
5 2013 Jan 2
6 2014 Jan 0
7 2012 Mar 1
8 2013 Mar 3
9 2014 Mar 2

Answer 9

For my aggregations I usually end up wanting to see mean and "how big is this group" (a.k.a. length).
So this is my handy snippet for those occasions;
agg.mean <- aggregate(columnToMean ~ columnToAggregateOn1*columnToAggregateOn2, yourDataFrame, FUN="mean")
agg.count <- aggregate(columnToMean ~ columnToAggregateOn1*columnToAggregateOn2, yourDataFrame, FUN="length")
aggcount <- agg.count$columnToMean
agg <- cbind(aggcount, agg.mean)

Answer 10

A sql solution using sqldf package:
library(sqldf)
sqldf("SELECT Year, Month, COUNT(*) as Freq
FROM df1
GROUP BY Year, Month")

Answer 11

Using collapse package in R
library(collapse)
library(magrittr)
df %>%
fgroup_by(year, month) %>%
fsummarise(number = fNobs(x))

Answer 12

library(tidyverse)
df_1 %>%
group_by(Year, Month) %>%
summarise(count= n())

Answer 13

Considering #Ben answer, R would throw an error if df1 does not contain x column. But it can be solved elegantly with paste:
aggregate(paste(Year, Month) ~ Year + Month, data = df1, FUN = NROW)
Similarly, it can be generalized if more than two variables are used in grouping:
aggregate(paste(Year, Month, Day) ~ Year + Month + Day, data = df1, FUN = NROW)

Answer 14

You can use by functions as by(df1$Year, df1$Month, count) that will produce a list of needed aggregation.
The output will look like,
df1$Month: Feb
x freq
1 2012 1
2 2013 1
3 2014 5
---------------------------------------------------------------
df1$Month: Jan
x freq
1 2012 5
2 2013 2
---------------------------------------------------------------
df1$Month: Mar
x freq
1 2012 1
2 2013 3
3 2014 2
>

Answer 15

There are plenty of wonderful answers here already, but I wanted to throw in 1 more option for those wanting to add a new column to the original dataset that contains the number of times that row is repeated.
df1$counts <- sapply(X = paste(df1$Year, df1$Month),
FUN = function(x) { sum(paste(df1$Year, df1$Month) == x) })
The same could be accomplished by combining any of the above answers with the merge() function.

Answer 16

If your trying the aggregate solutions above and you get the error:
invalid type (list) for variable
Because you're using date or datetime stamps, try using as.character on the variables:
aggregate(x ~ as.character(Year) + Month, data = df, FUN = length)
On one or both of the variables.

Answer 17

I usually use table function
df <- data.frame(a=rep(1:8,rep(c(1,2,3, 4),2)),year=2011:2021,month=c(1,3:10))
new_data <- as.data.frame(table(df[,c("year","month")]))