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Is there an R function for finding the avg index of change (x[n]/x[1])^(1/(n-1))?
How to make a great R reproducible exampleFunction to Calculate Median in SQL ServerIs there a built-in function for finding the mode?Grouping functions (tapply, by, aggregate) and the *apply familyIs there an R function for finding the index of an element in a vector?Changing column names of a data frameFind running median from a stream of integersFunction to clear the console in R and RStudioHow can I view the source code for a function?Finding median of list in PythonRolling / moving avg by group
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
this is my first post on this website so - sorry for any mistakes.
Given a data frame
xyz <- data.frame(GEO=c("Belgium", "Belgium", "Belgium", "Germany", "Germany", "Germany", "Italy", "Italy", "Italy"),
year=c(2008, 2009, 2010, 2008, 2009, 2010, 2008, 2009, 2010),
median=c(10.3, 45.2, 34.2, 67.8, 68.7, 69.9, 53.2, 43.2, 33.3),
imigration=c(10.2, 45.2, 47.4, 33.3, 44.4, 55.5, 21.5, 76.5, 78.4))
All I want to do is to count avg index of change for median and imigration.
index=(x[n]/x[1])^(1/(n-1))
I tried a lot of solutions but I couldn't find the good one.
This is my code - maybe you will use this to help me.
First:
groupColumns = c("GEO","year")
dataColumns = c("median", "imigration")
index_median = ddply(mediana2, groupColumns, function(x) x[c((nrow(x)/x[1])^(1/nrow(x)-1)), ])
head(index_median)
Second:
xyz_index <- aggregate(xyz$median, xyz$imigration by=list(Category=xyz$"GEO/TIME"), FUN=median$index_median[nrow(median)]/median[1])^(1/(nrow(median)-1))
I want to get avg index of change for median and imigration (index_median, index_imigration) for every country (Belgium, Germany, Italy).
Something like that (numbers are fake):
GEO index_median index_imigration
1 Germany 2.3 1.3
2 Italy 1.6 1.6
3 Belgium 1.2 2.0
Thank you very much!
r grouping median
asked Mar 27 at 10:52
wloczykijekwloczykijek
112 bronze badges
|
show 1 more comment
this is my first post on this website so - sorry for any mistakes.
Given a data frame
xyz <- data.frame(GEO=c("Belgium", "Belgium", "Belgium", "Germany", "Germany", "Germany", "Italy", "Italy", "Italy"),
year=c(2008, 2009, 2010, 2008, 2009, 2010, 2008, 2009, 2010),
median=c(10.3, 45.2, 34.2, 67.8, 68.7, 69.9, 53.2, 43.2, 33.3),
imigration=c(10.2, 45.2, 47.4, 33.3, 44.4, 55.5, 21.5, 76.5, 78.4))
All I want to do is to count avg index of change for median and imigration.
index=(x[n]/x[1])^(1/(n-1))
I tried a lot of solutions but I couldn't find the good one.
This is my code - maybe you will use this to help me.
First:
groupColumns = c("GEO","year")
dataColumns = c("median", "imigration")
index_median = ddply(mediana2, groupColumns, function(x) x[c((nrow(x)/x[1])^(1/nrow(x)-1)), ])
head(index_median)
Second:
xyz_index <- aggregate(xyz$median, xyz$imigration by=list(Category=xyz$"GEO/TIME"), FUN=median$index_median[nrow(median)]/median[1])^(1/(nrow(median)-1))
I want to get avg index of change for median and imigration (index_median, index_imigration) for every country (Belgium, Germany, Italy).
Something like that (numbers are fake):
GEO index_median index_imigration
1 Germany 2.3 1.3
2 Italy 1.6 1.6
3 Belgium 1.2 2.0
Thank you very much!
r grouping median
asked Mar 27 at 10:52
wloczykijekwloczykijek
112 bronze badges
1
Can you edit your question and include your expected output: i.e. what should your result(s) look like?
– Phil
Mar 27 at 11:06
Thank you for your advice, done!
– wloczykijek
Mar 27 at 11:10
@wloczykijek Welcome to SO I think you should read this. Meanwhile, do you want something likelibrary(dplyr);xyz %>% group_by(GEO) %>% mutate(Ind=((median/first(median))^(1/(row_number()-1))))
– A. Suliman
Mar 27 at 11:15
1
@A.Suliman I edited my post - you can see what exactly I want to get
– wloczykijek
Mar 27 at 11:29
1
Now is better, one last question whatnstand for inindex=(x[n]/x[1])^(1/(n-1)), I think it's the last obs e.g. for Belgium's median it's 34.2
– A. Suliman
Mar 27 at 11:32
|
show 1 more comment
this is my first post on this website so - sorry for any mistakes.
Given a data frame
xyz <- data.frame(GEO=c("Belgium", "Belgium", "Belgium", "Germany", "Germany", "Germany", "Italy", "Italy", "Italy"),
year=c(2008, 2009, 2010, 2008, 2009, 2010, 2008, 2009, 2010),
median=c(10.3, 45.2, 34.2, 67.8, 68.7, 69.9, 53.2, 43.2, 33.3),
imigration=c(10.2, 45.2, 47.4, 33.3, 44.4, 55.5, 21.5, 76.5, 78.4))
All I want to do is to count avg index of change for median and imigration.
index=(x[n]/x[1])^(1/(n-1))
I tried a lot of solutions but I couldn't find the good one.
This is my code - maybe you will use this to help me.
First:
groupColumns = c("GEO","year")
dataColumns = c("median", "imigration")
index_median = ddply(mediana2, groupColumns, function(x) x[c((nrow(x)/x[1])^(1/nrow(x)-1)), ])
head(index_median)
Second:
xyz_index <- aggregate(xyz$median, xyz$imigration by=list(Category=xyz$"GEO/TIME"), FUN=median$index_median[nrow(median)]/median[1])^(1/(nrow(median)-1))
I want to get avg index of change for median and imigration (index_median, index_imigration) for every country (Belgium, Germany, Italy).
Something like that (numbers are fake):
GEO index_median index_imigration
1 Germany 2.3 1.3
2 Italy 1.6 1.6
3 Belgium 1.2 2.0
Thank you very much!
r grouping median
asked Mar 27 at 10:52
wloczykijekwloczykijek
112 bronze badges
this is my first post on this website so - sorry for any mistakes.
Given a data frame
xyz <- data.frame(GEO=c("Belgium", "Belgium", "Belgium", "Germany", "Germany", "Germany", "Italy", "Italy", "Italy"),
year=c(2008, 2009, 2010, 2008, 2009, 2010, 2008, 2009, 2010),
median=c(10.3, 45.2, 34.2, 67.8, 68.7, 69.9, 53.2, 43.2, 33.3),
imigration=c(10.2, 45.2, 47.4, 33.3, 44.4, 55.5, 21.5, 76.5, 78.4))
All I want to do is to count avg index of change for median and imigration.
index=(x[n]/x[1])^(1/(n-1))
I tried a lot of solutions but I couldn't find the good one.
This is my code - maybe you will use this to help me.
First:
groupColumns = c("GEO","year")
dataColumns = c("median", "imigration")
index_median = ddply(mediana2, groupColumns, function(x) x[c((nrow(x)/x[1])^(1/nrow(x)-1)), ])
head(index_median)
Second:
xyz_index <- aggregate(xyz$median, xyz$imigration by=list(Category=xyz$"GEO/TIME"), FUN=median$index_median[nrow(median)]/median[1])^(1/(nrow(median)-1))
I want to get avg index of change for median and imigration (index_median, index_imigration) for every country (Belgium, Germany, Italy).
Something like that (numbers are fake):
GEO index_median index_imigration
1 Germany 2.3 1.3
2 Italy 1.6 1.6
3 Belgium 1.2 2.0
Thank you very much!
r grouping median
r grouping median
asked Mar 27 at 10:52
wloczykijekwloczykijek
112 bronze badges
asked Mar 27 at 10:52
wloczykijekwloczykijek
112 bronze badges
edited Mar 27 at 11:28
wloczykijek
asked Mar 27 at 10:52
wloczykijekwloczykijek
112 bronze badges
asked Mar 27 at 10:52
wloczykijekwloczykijek
112 bronze badges
asked Mar 27 at 10:52
wloczykijekwloczykijek
112 bronze badges
112 bronze badges
1
Can you edit your question and include your expected output: i.e. what should your result(s) look like?
– Phil
Mar 27 at 11:06
Thank you for your advice, done!
– wloczykijek
Mar 27 at 11:10
@wloczykijek Welcome to SO I think you should read this. Meanwhile, do you want something likelibrary(dplyr);xyz %>% group_by(GEO) %>% mutate(Ind=((median/first(median))^(1/(row_number()-1))))
– A. Suliman
Mar 27 at 11:15
1
@A.Suliman I edited my post - you can see what exactly I want to get
– wloczykijek
Mar 27 at 11:29
1
Now is better, one last question whatnstand for inindex=(x[n]/x[1])^(1/(n-1)), I think it's the last obs e.g. for Belgium's median it's 34.2
– A. Suliman
Mar 27 at 11:32
|
show 1 more comment
1
Can you edit your question and include your expected output: i.e. what should your result(s) look like?
– Phil
Mar 27 at 11:06
Thank you for your advice, done!
– wloczykijek
Mar 27 at 11:10
@wloczykijek Welcome to SO I think you should read this. Meanwhile, do you want something likelibrary(dplyr);xyz %>% group_by(GEO) %>% mutate(Ind=((median/first(median))^(1/(row_number()-1))))
– A. Suliman
Mar 27 at 11:15
1
@A.Suliman I edited my post - you can see what exactly I want to get
– wloczykijek
Mar 27 at 11:29
1
Now is better, one last question whatnstand for inindex=(x[n]/x[1])^(1/(n-1)), I think it's the last obs e.g. for Belgium's median it's 34.2
– A. Suliman
Mar 27 at 11:32
1
1
Can you edit your question and include your expected output: i.e. what should your result(s) look like?
– Phil
Mar 27 at 11:06
Can you edit your question and include your expected output: i.e. what should your result(s) look like?
– Phil
Mar 27 at 11:06
Thank you for your advice, done!
– wloczykijek
Mar 27 at 11:10
Thank you for your advice, done!
– wloczykijek
Mar 27 at 11:10
@wloczykijek Welcome to SO I think you should read this. Meanwhile, do you want something like
library(dplyr);xyz %>% group_by(GEO) %>% mutate(Ind=((median/first(median))^(1/(row_number()-1))))– A. Suliman
Mar 27 at 11:15
@wloczykijek Welcome to SO I think you should read this. Meanwhile, do you want something like
library(dplyr);xyz %>% group_by(GEO) %>% mutate(Ind=((median/first(median))^(1/(row_number()-1))))– A. Suliman
Mar 27 at 11:15
1
1
@A.Suliman I edited my post - you can see what exactly I want to get
– wloczykijek
Mar 27 at 11:29
@A.Suliman I edited my post - you can see what exactly I want to get
– wloczykijek
Mar 27 at 11:29
1
1
Now is better, one last question what
n stand for in index=(x[n]/x[1])^(1/(n-1)), I think it's the last obs e.g. for Belgium's median it's 34.2– A. Suliman
Mar 27 at 11:32
Now is better, one last question what
n stand for in index=(x[n]/x[1])^(1/(n-1)), I think it's the last obs e.g. for Belgium's median it's 34.2– A. Suliman
Mar 27 at 11:32
|
show 1 more comment
2 Answers
2
active
oldest
votes
We can dplyr::group_by GEO then take the first and last observation for median and immigration using dplyr::first and last in each group. To get number of observations in each group we will use dplyr::n()
library(dplyr)
xyz %>% group_by(GEO) %>%
#medain[n()] similar to last(median)
summarise(Ind_median=((last(median)/first(median))^(1/(n()-1))),
Ind_imigration=((last(imigration)/first(imigration))^(1/(n()-1))))
answered Mar 27 at 11:45
A. SulimanA. Suliman
7,4914 gold badges13 silver badges26 bronze badges
add a comment |
You can also use base R only like this:
avgchange <- function(z, fld) x <- z[,fld]; n<- length(x);(x[n]/x[1])^(1/(n-1))
Argument fld is the field name you want to summarise
Use base R's by
z <- by(xyz, list(Category=xyz$GEO), fld="median",FUN=avgchange, simplify=FALSE)
z
And this to get a named vector only
unlist(z)
And similarly for the imigration column
z <- by(xyz, list(Category=xyz$GEO), fld="imigration",FUN=avgchange, simplify=FALSE)
z
unlist(z)
answered Mar 27 at 14:00
BhasBhas
1,5211 gold badge9 silver badges8 bronze badges
add a comment |
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2 Answers
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active
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2 Answers
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active
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oldest
votes
We can dplyr::group_by GEO then take the first and last observation for median and immigration using dplyr::first and last in each group. To get number of observations in each group we will use dplyr::n()
library(dplyr)
xyz %>% group_by(GEO) %>%
#medain[n()] similar to last(median)
summarise(Ind_median=((last(median)/first(median))^(1/(n()-1))),
Ind_imigration=((last(imigration)/first(imigration))^(1/(n()-1))))
answered Mar 27 at 11:45
A. SulimanA. Suliman
7,4914 gold badges13 silver badges26 bronze badges
add a comment |
We can dplyr::group_by GEO then take the first and last observation for median and immigration using dplyr::first and last in each group. To get number of observations in each group we will use dplyr::n()
library(dplyr)
xyz %>% group_by(GEO) %>%
#medain[n()] similar to last(median)
summarise(Ind_median=((last(median)/first(median))^(1/(n()-1))),
Ind_imigration=((last(imigration)/first(imigration))^(1/(n()-1))))
answered Mar 27 at 11:45
A. SulimanA. Suliman
7,4914 gold badges13 silver badges26 bronze badges
add a comment |
We can dplyr::group_by GEO then take the first and last observation for median and immigration using dplyr::first and last in each group. To get number of observations in each group we will use dplyr::n()
library(dplyr)
xyz %>% group_by(GEO) %>%
#medain[n()] similar to last(median)
summarise(Ind_median=((last(median)/first(median))^(1/(n()-1))),
Ind_imigration=((last(imigration)/first(imigration))^(1/(n()-1))))
answered Mar 27 at 11:45
A. SulimanA. Suliman
7,4914 gold badges13 silver badges26 bronze badges
We can dplyr::group_by GEO then take the first and last observation for median and immigration using dplyr::first and last in each group. To get number of observations in each group we will use dplyr::n()
library(dplyr)
xyz %>% group_by(GEO) %>%
#medain[n()] similar to last(median)
summarise(Ind_median=((last(median)/first(median))^(1/(n()-1))),
Ind_imigration=((last(imigration)/first(imigration))^(1/(n()-1))))
answered Mar 27 at 11:45
A. SulimanA. Suliman
7,4914 gold badges13 silver badges26 bronze badges
answered Mar 27 at 11:45
A. SulimanA. Suliman
7,4914 gold badges13 silver badges26 bronze badges
answered Mar 27 at 11:45
A. SulimanA. Suliman
7,4914 gold badges13 silver badges26 bronze badges
answered Mar 27 at 11:45
A. SulimanA. Suliman
7,4914 gold badges13 silver badges26 bronze badges
7,4914 gold badges13 silver badges26 bronze badges
add a comment |
add a comment |
You can also use base R only like this:
avgchange <- function(z, fld) x <- z[,fld]; n<- length(x);(x[n]/x[1])^(1/(n-1))
Argument fld is the field name you want to summarise
Use base R's by
z <- by(xyz, list(Category=xyz$GEO), fld="median",FUN=avgchange, simplify=FALSE)
z
And this to get a named vector only
unlist(z)
And similarly for the imigration column
z <- by(xyz, list(Category=xyz$GEO), fld="imigration",FUN=avgchange, simplify=FALSE)
z
unlist(z)
answered Mar 27 at 14:00
BhasBhas
1,5211 gold badge9 silver badges8 bronze badges
add a comment |
You can also use base R only like this:
avgchange <- function(z, fld) x <- z[,fld]; n<- length(x);(x[n]/x[1])^(1/(n-1))
Argument fld is the field name you want to summarise
Use base R's by
z <- by(xyz, list(Category=xyz$GEO), fld="median",FUN=avgchange, simplify=FALSE)
z
And this to get a named vector only
unlist(z)
And similarly for the imigration column
z <- by(xyz, list(Category=xyz$GEO), fld="imigration",FUN=avgchange, simplify=FALSE)
z
unlist(z)
answered Mar 27 at 14:00
BhasBhas
1,5211 gold badge9 silver badges8 bronze badges
add a comment |
You can also use base R only like this:
avgchange <- function(z, fld) x <- z[,fld]; n<- length(x);(x[n]/x[1])^(1/(n-1))
Argument fld is the field name you want to summarise
Use base R's by
z <- by(xyz, list(Category=xyz$GEO), fld="median",FUN=avgchange, simplify=FALSE)
z
And this to get a named vector only
unlist(z)
And similarly for the imigration column
z <- by(xyz, list(Category=xyz$GEO), fld="imigration",FUN=avgchange, simplify=FALSE)
z
unlist(z)
answered Mar 27 at 14:00
BhasBhas
1,5211 gold badge9 silver badges8 bronze badges
You can also use base R only like this:
avgchange <- function(z, fld) x <- z[,fld]; n<- length(x);(x[n]/x[1])^(1/(n-1))
Argument fld is the field name you want to summarise
Use base R's by
z <- by(xyz, list(Category=xyz$GEO), fld="median",FUN=avgchange, simplify=FALSE)
z
And this to get a named vector only
unlist(z)
And similarly for the imigration column
z <- by(xyz, list(Category=xyz$GEO), fld="imigration",FUN=avgchange, simplify=FALSE)
z
unlist(z)
answered Mar 27 at 14:00
BhasBhas
1,5211 gold badge9 silver badges8 bronze badges
answered Mar 27 at 14:00
BhasBhas
1,5211 gold badge9 silver badges8 bronze badges
answered Mar 27 at 14:00
BhasBhas
1,5211 gold badge9 silver badges8 bronze badges
answered Mar 27 at 14:00
BhasBhas
1,5211 gold badge9 silver badges8 bronze badges
1,5211 gold badge9 silver badges8 bronze badges
add a comment |
add a comment |
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1
Can you edit your question and include your expected output: i.e. what should your result(s) look like?
– Phil
Mar 27 at 11:06
Thank you for your advice, done!
– wloczykijek
Mar 27 at 11:10
@wloczykijek Welcome to SO I think you should read this. Meanwhile, do you want something like
library(dplyr);xyz %>% group_by(GEO) %>% mutate(Ind=((median/first(median))^(1/(row_number()-1))))– A. Suliman
Mar 27 at 11:15
1
@A.Suliman I edited my post - you can see what exactly I want to get
– wloczykijek
Mar 27 at 11:29
1
Now is better, one last question what
nstand for inindex=(x[n]/x[1])^(1/(n-1)), I think it's the last obs e.g. for Belgium's median it's 34.2– A. Suliman
Mar 27 at 11:32