[R] How to avoid the three loops in R?
David L Carlson
dcarlson at tamu.edu
Fri Aug 1 16:47:27 CEST 2014
Here's another approach:
# First put the data in a format that is easier to transmit using dput():
dta <- structure(list(Country = structure(c(1L, 3L, 2L, 1L, 3L, 2L,
1L, 2L, 1L, 3L, 2L, 1L, 3L), .Label = c("AE", "CN", "DE"), class = "factor"),
Product = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 1L, 1L, 1L, 2L,
2L), Price = c(20L, 20L, 28L, 28L, 28L, 22L, 28L, 28L, 20L,
20L, 28L, 28L, 28L), Year_Month = c(201204L, 201204L, 201204L,
201204L, 201204L, 201204L, 201204L, 201204L, 201205L, 201205L,
201205L, 201205L, 201205L)), .Names = c("Country", "Product",
"Price", "Year_Month"), class = "data.frame", row.names = c(NA,
-13L))
# Then
> grp <- aggregate(Price~Product+Year_Month, dta, function(x) c(sum(x),
+ length(x)))
> dtmrg <- merge(dta, grp, by=c("Product", "Year_Month"))
> newdta <- with(dtmrg, data.frame(Country, Product, Price=Price.x,
+ Year_Month, Price_average_in_other_area=(Price.y[,1]-Price.x)/
+ (Price.y[,2]-1)))
> newdta
Country Product Price Year_Month Price_average_in_other_area
1 AE 1 20 201204 24
2 DE 1 20 201204 24
3 CN 1 28 201204 20
4 AE 1 20 201205 24
5 DE 1 20 201205 24
6 CN 1 28 201205 20
7 AE 2 28 201204 25
8 DE 2 28 201204 25
9 CN 2 22 201204 28
10 AE 2 28 201205 28
11 DE 2 28 201205 28
12 AE 3 28 201204 28
13 CN 3 28 201204 28
-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of John McKown
Sent: Friday, August 1, 2014 9:07 AM
To: Lingyi Ma
Cc: r-help
Subject: Re: [R] How to avoid the three loops in R?
On Fri, Aug 1, 2014 at 6:41 AM, Lingyi Ma <lingyi.ma at gmail.com> wrote:
> I have the following data set:
>
> Country Product Price Year_Month
> AE 1 20 201204
> DE 1 20 201204
> CN 1 28 201204
> AE 2 28 201204
> DE 2 28 201204
> CN 2 22 201204
> AE 3 28 201204
> CN 3 28 201204
> AE 1 20 201205
> DE 1 20 201205
> CN 1 28 201205
> AE 2 28 201205
> DE 2 28 201205
>
> I want to create the one more column which is "The average price of the
> product in other areas".
> in other word, for each month, for each product, I calculate the average of
> such product in the other area.
>
> I want sth like:
>
> Country Product Price Year_Month Price_average_In_Other_area
> AE 1 20 201204 14
> AE 2 28 201204 25
The output above looks wrong. The Price_average_In_Other_area for AE,
product 1 should be 24?
My possible solution:
# Initialize data.frame & call it "x".
Country <- c("AE","DE","CN","AE","DE","CN","AE","CN","AE","DE","CN","AE","DE");
Product <- c(1,1,1,2,2,2,3,3,1,1,1,2,2);
Price <- c(20,20,28,28,28,22,28,28,20,20,28,28,28);
Year_Month <- c(201204,201204,201204,201204,201204,201204,201204,201204,201205,201205,201205,201205,201205);
x <- data.frame(Country,Product,Price,Year_Month,stringsAsFactors=FALSE);
#
#
library("dplyr");
#
# Get the total Price of all Products and number of Products for each
Product & Year_Month"
y <- summarize(group_by(x, Product,
Year_Month),sumPrice=sum(Price),NoPrice=length(Price));
#
# Merge the above data back into the original data.frame, based on
# Product and Year_Month (similar to SQL inner join).
x <- merge(x=x,y=y);
#
# Now calculate the "other area" average by subtracting the cost in this area
# from the total cost in all areas and divide by the number of areas, minus one.
# Please note that if a Product and Year_Month is unique, i.e. no other areas
# for this Product & Year_Month, this will try to divide by zero.
# This gives "Inf" as an answer.
x$Prive_average_In_Other_area <- (x$sumPrice-x$Price)/(x$NoPrice-1);
# Possible alternate to handle above consideration
x$Avg_other <- ifelse(x$NoPrice>1,(x$sumPrice-x$Price)/(x$NoPrice-1),NA);
>
> Please avoid the three for loop, I have tried and it never end. I have
> 1070427 rows. Is there better way to speed up my program?
--
There is nothing more pleasant than traveling and meeting new people!
Genghis Khan
Maranatha! <><
John McKown
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