I am trying to do the following logic to create 'subtract' column.
I have years from 1986-2014 and around 100 firms.
year firm count sum_of_year subtract
1986 A 1 2 2
1986 B 1 2 4
1987 A 2 4 5
1987 C 1 4 2
1987 D 1 4 5
1988 C 3 5
1988 E 2 5
That is, if a firm i at t appears in t+1, then subtract its count at t+1 from the sum_of_year at t+1,
if a firm i does not appear in t+1, then just put sum_of_year at t+1 as shown in the sample.
I am having difficulties in creating this conditional code.
How can I do this in a generalized version?
Thank you for your help.
One way using dplyr with the help of tidyr::complete. We complete the missing combinations of rows for year and firm and fill count with 0. For each year, we subtract the count by sum of count for that entire year and finally for each firm, we take the value from the next year using lead.
library(dplyr)
df %>%
tidyr::complete(year, firm, fill = list(count = 0)) %>%
group_by(year) %>%
mutate(n = sum(count) - count) %>%
group_by(firm) %>%
mutate(subtract = lead(n)) %>%
filter(count != 0) %>%
select(-n)
# year firm count sum_of_year subtract
# <int> <fct> <dbl> <int> <dbl>
#1 1986 A 1 2 2
#2 1986 B 1 2 4
#3 1987 A 2 4 5
#4 1987 C 1 4 2
#5 1987 D 1 4 5
#6 1988 C 3 5 NA
#7 1988 E 2 5 NA
Related
My dataset is:
CLASS YEAR VALUE
A 1990 4
A 1991 3
A 1992 7
B 1989 5
B 1990 23
B 1991 3
C 1990 7
C 1991 4
C 1992 6
I want to apply the CAGR formula for each class, I was trying with this code:
df <- df %>%
arrange(CLASS, YEAR) %>%
group_by(CLASS) %>%
mutate(cagr = ((VALUE / lag(VALUE, n)) ^ (1 / n)) - 1)
The dataset that I am using is quite huge, the issue is that I get the first n values of the first class as N/A but it does not happen for the other classes. Therefore I think that in this way the formula takes into account the values of the class above for the first n cases which is wrong.
See if this works for your CAGR:
library(dplyr)
library(xts)
df %>% group_by(CLASS) %>% mutate(cagr = (last(VALUE)/first(VALUE))^(1/(n()-1)) - 1)
# A tibble: 9 x 4
# Groups: CLASS [3]
CLASS YEAR VALUE cagr
<chr> <dbl> <dbl> <dbl>
1 A 1990 4 0.323
2 A 1991 3 0.323
3 A 1992 7 0.323
4 B 1989 5 -0.225
5 B 1990 23 -0.225
6 B 1991 3 -0.225
7 C 1990 7 -0.0742
8 C 1991 4 -0.0742
9 C 1992 6 -0.0742
I have a dataframe like the following one:
day year value
1 2014 5
1 2015 16
1 2016 0
2 2014 3
2 2015 1
2 2016 4
and I want to calculate the average value by day for the three year period (2014, 2015, 2016). The following code works for this purpose:
data %>%
group_by(day) %>%
mutate(MEAN = mean(value))
and produces this output:
day year value MEAN
1 2014 5 7
1 2015 16 7
1 2016 0 7
2 2014 3 3
2 2015 1 3
2 2016 4 3
but I want to add the average values as new rows in the same dataframe as follows:
day year value
1 2014 5
1 2015 16
1 2016 0
2 2014 3
2 2015 1
2 2016 4
1 avg 7 <--
2 avg 3 <--
Any suggestions about how can I possibly do this? Thanks!
We can use summarise (instead of mutate - which adds a new column in the original dataset) to calculate the mean and then with bind_rows can bind with original data. The tidyverse functions are very particular about type, so make sure the class are the same before we do the binding
library(dplyr)
data %>%
group_by(day) %>%
summarise(year = 'avg', value = mean(value)) %>%
bind_rows(data %>%
mutate(year = as.character(year)), .)
# day year value
#1 1 2014 5.00
#2 1 2015 16.00
#3 1 2016 0.00
#4 2 2014 3.00
#5 2 2015 1.00
#6 2 2016 4.00
#7 1 avg 7.00
#8 2 avg 2.67
Another option is to split by the 'day' and then with add_row (from tibble) create a new row on each of the list elements
library(tibble)
library(purrr)
data %>%
mutate(year = as.character(year)) %>%
group_split(day) %>%
map_dfr(~ .x %>% add_row(day = first(.$day),
year = 'avg', value = mean(.$value)))
Here is a base R option using aggregate
rbind(df,cbind(aggregate(value~day,df,mean),year = "avg")[c(1,3,2)])
or a variation (by #thelatemail from comments)
rbind(df, aggregate(df["value"], cbind(df["day"], year="avg"), FUN=mean))
which gives
day year value
1 1 2014 5.000000
2 1 2015 16.000000
3 1 2016 0.000000
4 2 2014 3.000000
5 2 2015 1.000000
6 2 2016 4.000000
7 1 avg 7.000000
8 2 avg 2.666667
I have a data frame from which I created a reproducible example:
country <- c('A','A','A','B','B','C','C','C','C')
year <- c(2010,2011,2015,2008,2009,2008,2009,2011,2015)
score <- c(1,2,2,1,4,1,1,3,2)
country year score
1 A 2010 1
2 A 2011 2
3 A 2015 2
4 B 2008 1
5 B 2009 4
6 C 2008 1
7 C 2009 1
8 C 2011 3
9 C 2015 2
And I am trying to calculate the average percentage increase (or decrease) in the score for each country by calculating [(final score - initial score) รท (initial score)] for each year and averaging it over the number of years.
country year score change
1 A 2010 1 NA
2 A 2011 2 1
3 A 2015 2 0
4 B 2008 1 NA
5 B 2009 4 3
6 C 2008 1 NA
7 C 2009 1 0
8 C 2011 3 2
9 C 2015 2 -0.33
The final result I am hoping to obtain:
country avg_change
1 A 0.5
2 B 3
3 C 0.55
As you can see, the trick is that countries have spans over different years, sometimes with a missing year in between. I tried different ways to do it manually but I do struggle. If someone could hint me a solution would be great. Many thanks.
With dplyr, we can group_by country and get mean of difference between scores.
library(dplyr)
df %>%
group_by(country) %>%
summarise(avg_change = mean(c(NA, diff(score)), na.rm = TRUE))
# country avg_change
# <fct> <dbl>
#1 A 0.500
#2 B 3.00
#3 C 0.333
Using base R aggregate with same logic
aggregate(score~country, df, function(x) mean(c(NA, diff(x)), na.rm = TRUE))
We can use data.table to group by 'country' and take the mean of the difference between the 'score' and the lag of 'score'
library(data.table)
setDT(df1)[, .(avg_change = mean(score -lag(score), na.rm = TRUE)), .(country)]
# country avg_change
#1: A 0.5000000
#2: B 3.0000000
#3: C 0.3333333
I'm trying, for each row, to calculate the difference with the closest previous row belonging to the same group which meets a certain criterion.
Suppose I have the following dataframe:
s <- read.table(text = "Visit_num Patient Day Admitted
1 1 2015/01/01 Yes
2 1 2015/01/10 No
3 1 2015/01/15 Yes
4 1 2015/02/10 No
5 1 2015/03/08 Yes
6 2 2015/01/01 Yes
7 2 2015/04/01 No
8 2 2015/04/10 No
9 3 2015/04/01 No
10 3 2015/04/10 No", header = T, sep = "")
For each Visit_num and for each Patient, I'd like to get the difference with the closest row for which the patient was admitted (i.e. Yes). Note column day is ordered by day, and time unit for this example is days.
Here is what I wanted my dataframe to look like:
Visit_num Patient Day Admitted Diff_days
1 1 2015/01/01 Yes NA
2 1 2015/01/10 No 9
3 1 2015/01/15 Yes 14
4 1 2015/02/10 No 26
5 1 2015/03/08 Yes 52
6 2 2015/01/01 Yes NA
7 2 2015/04/01 No 90
8 2 2015/04/10 No 99
9 3 2015/04/01 No NA
10 3 2015/04/10 No NA
Any help is appreciated.
Here is an option with tidyverse. Convert the 'Day' to Date class, arrange by 'Patient', 'Day', grouped by 'Patient' get the difference of adjacent 'Day', create a group 'grp' based on the occurrence of 'Yes' in 'Admitted' and take the cumulative sum of 'Diff_days'
library(tidyverse)
s %>%
mutate(Day = ymd(Day)) %>%
arrange(Patient, Day) %>%
group_by(Patient) %>%
mutate(Diff_days = c(NA, diff(Day))) %>%
group_by(grp = cumsum(lag(Admitted == "Yes", default = TRUE)), add = TRUE) %>%
mutate(Diff_days = cumsum(replace_na(Diff_days, 0))) %>%
ungroup %>%
select(-grp) %>%
mutate(Diff_days = na_if(Diff_days, 0))
# A tibble: 8 x 5
# Visit_num Patient Day Admitted Diff_days
# <int> <int> <date> <fct> <dbl>
#1 1 1 2015-01-01 Yes NA
#2 2 1 2015-01-10 No 9
#3 3 1 2015-01-15 Yes 14
#4 4 1 2015-02-10 No 26
#5 5 1 2015-03-08 Yes 52
#6 6 2 2015-01-01 Yes NA
#7 7 2 2015-04-01 No 90
#8 8 2 2015-04-10 No 99
This question already has answers here:
Count number of rows within each group
(17 answers)
Closed 4 years ago.
I have a data frame df as follows:
df
Code Time Country Type
1 n001 2000 France 1
2 n002 2001 Japan 5
3 n003 2003 USA 2
4 n004 2004 USA 2
5 n005 2004 Canada 1
6 n006 2005 Britain 2
7 n007 2005 USA 1
8 n008 2005 USA 2
9 n010 2005 USA 1
10 n011 2005 Canada 1
11 n012 2005 USA 2
12 n013 2005 USA 5
13 n014 2005 Canada 1
14 n015 2006 USA 2
15 n017 2006 Canada 1
16 n018 2006 Britain 1
17 n019 2006 Canada 1
18 n020 2006 USA 1
...
where Type is the type of news, and Time is the year when the news was published.
My aim is to count the number of each type of news each year.
I was thinking about a result like this:
...
$2005
Type: 1 Count: 4
Type: 2 Count: 3
Type: 5 Count: 1
$2006
Type: 1 Count: 4
...
I used the following code:
gp = group_by(df, Time)
summarise(gp, table(Time)
Error in summarise_impl(.data, dots) :
Evaluation error: unique() applies only to vectors.
Then I tried split( ), thinking it may be able to separate the dataframe by year so I could count the number of each type by year
split(df, 'Time')
$Time
Code Time Country Type
1 n001 2000 France 1
2 n002 2001 Japan 5
3 n003 2003 USA 2
4 n004 2004 USA 2
...
Everything is almost the same, apart from the "$Time" sign.
I was wondering what I did wrong, and how to fix it.
We can split Type Column by Time and calculate it's frequency by table.
lapply(split(df$Type, df$Time), table)
#$`2000`
#1
#1
#$`2001`
#5
#1
#$`2003`
#2
#1
#$`2004`
#1 2
#1 1
#$`2005`
#1 2 5
#4 3 1
#$`2006`
#1 2
#4 1
How about this?
df %>%
group_by(Time, Type) %>%
count() %>%
spread(Type, n)
You could use something like this. split on Time, then group by Type and tally the result
df %>%
split(.$Time) %>%
map(~ group_by(., Type) %>% tally())
......
$`2004`
# A tibble: 2 x 2
Type n
<int> <int>
1 1 1
2 2 1
$`2005`
# A tibble: 3 x 2
Type n
<int> <int>
1 1 4
2 2 3
3 5 1
$`2006`
# A tibble: 2 x 2
......
Or use summarise instead of tally if you want a column called count instead of n
df1 %>%
split(.$Time) %>%
map(~ group_by(., Type) %>% summarise(count = n()))