I want to make something like LEAD(mes) OVER(PARTITION BY CODIGO_CLIENTE ORDER BY mes) mes_2 in R, but I dont know a similar function.
I have no clue how to work it out.
Since you shared no data and desired output, here is an example with lead() from the dplyr package. The example is from the Help page of lead(). This can give you a good idea of what you can do with this function.
df <- data.frame(year = 2000:2005, value = (0:5) ^ 2)
scrambled <- df[sample(nrow(df)), ]
year value
1 2000 0
5 2004 16
3 2002 4
4 2003 9
2 2001 1
6 2005 25
right <- mutate(scrambled, `next` = lead(value, order_by = year))
arrange(right, year)
year value next
1 2000 0 1
2 2001 1 4
3 2002 4 9
4 2003 9 16
5 2004 16 25
6 2005 25 NA
Since you're new to R I suggest you read a bit on the dplyr package. Also, to make it easier for the people trying to help you, please provide more details next time!
Related
I haven't coded for several months and now am stuck with the following issue.
I have the following dataset:
Year World_export China_exp World_import China_imp
1 1992 3445.534 27.7310 3402.505 6.2220
2 1993 1940.061 27.8800 2474.038 18.3560
3 1994 2458.337 39.6970 2978.314 3.3270
4 1995 4641.168 15.9790 5504.787 18.0130
5 1996 5680.688 74.1650 6939.291 25.1870
6 1997 7206.604 70.2440 8639.422 31.9030
7 1998 7069.725 99.6510 8530.293 41.5030
8 1999 5916.077 169.4593 6673.743 37.8139
9 2000 7331.588 136.2180 8646.253 47.3789
10 2001 7471.374 143.0542 8292.893 41.2899
11 2002 8074.975 217.4286 9092.341 46.4730
12 2003 9956.433 162.2522 11558.007 71.7753
13 2004 13751.671 282.8678 16345.452 157.0768
14 2005 15976.238 430.8655 16708.094 284.1065
15 2006 19728.935 398.6704 22344.856 553.6356
16 2007 24275.244 484.5276 28693.113 815.7914
17 2008 32570.781 613.3714 39381.251 1414.8120
18 2009 21282.228 173.9463 28563.576 1081.3720
19 2010 25283.462 475.7635 34884.450 1684.0839
20 2011 41418.670 636.5881 45759.051 2193.8573
21 2012 46027.529 432.6025 46404.382 2373.4535
22 2013 37132.301 460.7133 43022.550 2829.3705
23 2014 36046.461 640.2552 40502.268 2373.2351
24 2015 26618.982 781.0016 30264.299 2401.1907
25 2016 23537.354 472.7022 27609.884 2129.4806
What I need is simple: to compute growth rates of each variable, that is, find difference between two elements, divide it by first element and multiply by 100.
I'm trying to write a script, that ends up with error message:
trade_Ch %>%
mutate (
World_exp_grate = sapply(2:nrow(trade_Ch),function(i)((World_export[i]-World_export[i-1])/World_export[i-1]))
)
Error in mutate_impl(.data, dots) : Column World_exp_grate must
be length 25 (the number of rows) or one, not 24
although this piece of code gives me right values:
x <- sapply(2:nrow(trade_Ch),function(i)((trade_Ch$World_export[i]-trade_Ch$World_export[i-1])/trade_Ch$World_export[i-1]))
How can I correctly embedd the code into my MUTATE part from dplyr package?
OR
Is there is another elegant way to solve this issue?
library(dplyr)
df %>%
mutate_each(funs(chg = ((.-lag(.))/lag(.))*100), World_export:China_imp)
trade_Ch %>%
mutate(world_exp_grate = 100*(World_export - lag(World_export))/lag(World_export))
The problem is that you cannot calculate the World_exp_grate for your first row. Therefore you have to set it to NA.
One variant to solve this is
trade_Ch %>%
mutate (World_export_lag = lag(World_export),
World_exp_grate = (World_export - World_export_lag)/World_export_lag)) %>%
select(-World_export_lag)
lag shifts the vector by one position.
lag(1:5)
# [1] NA 1 2 3 4
This question already has answers here:
How to reshape data from long to wide format
(14 answers)
Closed 5 years ago.
I'm trying to change a dataframe in R to group multiple rows by a measurement. The table has a location (km), a size (mm) a count of things in that size bin, a site and year. I want to take the sizes, make a column from each one (2, 4 and 6 in this example), and place the corresponding count into each the row for that location, site and year.
It seems like a combination of transposing and grouping, but I can't figure out a way to accomplish this in R. I've looked at t(), dcast() and aggregate(), but those aren't really close at all.
So I would go from something like this:
df <- data.frame(km=c(rep(32,3),rep(50,3)), mm=rep(c(2,4,6),2), count=sample(1:25,6), site=rep("A", 6), year=rep(2013, 6))
km mm count site year
1 32 2 18 A 2013
2 32 4 2 A 2013
3 32 6 12 A 2013
4 50 2 3 A 2013
5 50 4 17 A 2013
6 50 6 21 A 2013
To this:
km site year mm_2 mm_4 mm_6
1 32 A 2013 18 2 12
2 50 A 2013 3 17 21
Edit: I tried the solution in a suggested duplicate, but I did not work for me, not really sure why. The answer below worked better.
As suggested in the comment above, we can use the sep argument in spread:
library(tidyr)
spread(df, mm, count, sep = "_")
km site year mm_2 mm_4 mm_6
1 32 A 2013 4 20 1
2 50 A 2013 15 14 22
As you mentioned dcast(), here is a method using it.
set.seed(1)
df <- data.frame(km=c(rep(32,3),rep(50,3)),
mm=rep(c(2,4,6),2),
count=sample(1:25,6),
site=rep("A", 6),
year=rep(2013, 6))
library(reshape2)
dcast(df, ... ~ mm, value.var="count")
# km site year 2 4 6
# 1 32 A 2013 13 10 20
# 2 50 A 2013 3 17 1
And if you want a bit of a challenge you can try the base function reshape().
df2 <- reshape(df, v.names="count", idvar="km", timevar="mm", ids="mm", direction="wide")
colnames(df2) <- sub("count.", "mm_", colnames(df2))
df2
# km site year mm_2 mm_4 mm_6
# 1 32 A 2013 13 10 20
# 4 50 A 2013 3 17 1
So what I have is data of cod weights at different ages. This data is taken at several locations over time.
What I would like to create is "weight at age", basically a mean value of weights at a certain age. I want do this for each location at each year.
However, the ages are not sampled the same way (all old fish caught are measured, while younger fish are sub sampled), so I can't just create a normal average, I would like to bootstrap samples.
The bootstrap should take out 5 random values of weight at an age, create a mean value and repeat this a 1000 times, and then create an average of the means. The values should be able to be used again (replace). This should be done for each age at every AreaCode for every year. Dependent factors: Year-location-Age.
So here's an example of what my data could look like.
df <- data.frame( Year= rep(c(2000:2008),2), AreaCode = c("39G4", "38G5","40G5"), Age = c(0:8), IndWgt = c(rnorm(18, mean=5, sd=3)))
> df
Year AreaCode Age IndWgt
1 2000 39G4 0 7.317489899
2 2001 38G5 1 7.846606144
3 2002 40G5 2 0.009212455
4 2003 39G4 3 6.498688035
5 2004 38G5 4 3.121134937
6 2005 40G5 5 11.283096043
7 2006 39G4 6 0.258404136
8 2007 38G5 7 6.689780137
9 2008 40G5 8 10.180511929
10 2000 39G4 0 5.972879108
11 2001 38G5 1 1.872273650
12 2002 40G5 2 5.552962065
13 2003 39G4 3 4.897882549
14 2004 38G5 4 5.649438631
15 2005 40G5 5 4.525012587
16 2006 39G4 6 2.985615831
17 2007 38G5 7 8.042884181
18 2008 40G5 8 5.847629941
AreaCode contains the different locations, in reality I have 85 different levels. The time series stretches 1991-2013, the ages 0-15. IndWgt contain the weight. My whole data frame has a row length of 185726.
Also, every age does not exist for every location and every year. Don't know if this would be a problem, just so the scripts isn't based on references to certain row number. There are some NA values in the weight column, but I could just remove them before hand.
I was thinking that I maybe should use replicate, and apply or another plyr function. I've tried to understand the boot function but I don't really know if I would write my arguments under statistics, and in that case how. So yeah, basically I have no idea.
I would be thankful for any help I can get!
How about this with plyr. I think from the question you wanted to bootstrap only the "young" fish weights and use actual means for the older ones. If not, just replace the ifelse() statement with its last argument.
require(plyr)
#cod<-read.csv("cod.csv",header=T) #I loaded your data from csv
bootstrap<-function(Age,IndWgt){
ifelse(Age>2, # treat differently for old/young fish
res<-mean(IndWgt), # old fish mean
res<-mean(replicate(1000,sample(IndWgt,5,replace = TRUE))) # young fish bootstrap
)
return(res)
}
ddply(cod,.(Year,AreaCode,Age),summarize,boot_mean=bootstrap(Age,IndWgt))
Year AreaCode Age boot_mean
1 2000 39G4 0 6.650294
2 2001 38G5 1 4.863024
3 2002 40G5 2 2.724541
4 2003 39G4 3 5.698285
5 2004 38G5 4 4.385287
6 2005 40G5 5 7.904054
7 2006 39G4 6 1.622010
8 2007 38G5 7 7.366332
9 2008 40G5 8 8.014071
PS: If you want to sample all ages in the same way, no need for the function, just:
ddply(cod,.(Year,AreaCode,Age),
summarize,
boot_mean=mean(replicate(1000,mean(sample(IndWgt,5,replace = TRUE)))))
Since you don't provide enough code, it's too hard (lazy) for me to test it properly. You should get your first step using the following code. If you wrap this into replicate, you should get your end result that you can average.
part.result <- aggregate(IndWgt ~ Year + AreaCode + Age, data = data, FUN = function(x) {
rws <- length(x)
get.em <- sample(x, size = 5, replace = TRUE)
out <- mean(get.em)
out
})
To handle any missing combination of year/age/location, you could probably add an if statement checking for NULL/NA and producing a warning and/or skipping the iteration.
Hello I am working with a table with these characteristics:
2000 0.051568
2000 0.04805
2002 0.029792
2002 0.056141
2008 0.047285
2008 0.038989
And I need to convert it to something like this:
2000 2002 2008
0.051568 0.029792 0.047285
0.04805 0.056141 0.038989
I would be grateful if somebody could give me a solution.
Here's a relatively simple solution:
# CREATE ORIGINAL DATA.FRAME
df <- read.table(text="2000 0.051568
2000 0.04805
2002 0.029792
2002 0.056141
2008 0.047285
2008 0.038989", header=FALSE)
names(df) <- c("year", "value")
# MODIFY ITS LAYOUT
df2 <- as.data.frame(split(df$value, df$year))
df2
# X2000 X2002 X2008
# 1 0.051568 0.029792 0.047285
# 2 0.048050 0.056141 0.038989
I'm guessing you are new to R, so I'm going to guess what you mean and give you some more correct terminology. If I guess wrong, then at least this may help you to clarify the question.
In R, a table is a special case of a matrix that arises from cross-tabulation. What I think you have (or want) to start with is a data.frame. A data.frame is a set of columns with potentially different types, but all the same length; it is "rectangular" in that sense. Generally, elements in the same positions in the columns (that is, each row) of a data.frame are related to each other. The columns of a data.frame have names, as can the rows.
long <- data.frame(year=c(2000,2000,2002,2002,2008,2008),
val=c(0.051568, 0.04805, 0.029792,
0.056141, 0.047285, 0.038989))
Which when printed looks like
> long
year val
1 2000 0.051568
2 2000 0.048050
3 2002 0.029792
4 2002 0.056141
5 2008 0.047285
6 2008 0.038989
By itself, this isn't enough, because for your desired output, you need to specify which value for, say, 2000 is in the first row and which is in the second (etc., if there were more). In your example, it is just the order they are in.
long$targetrow = 1:2
Which makes long now look like
> long
year val targetrow
1 2000 0.051568 1
2 2000 0.048050 2
3 2002 0.029792 1
4 2002 0.056141 2
5 2008 0.047285 1
6 2008 0.038989 2
Now you can use reshape on it.
reshape(long, idvar="targetrow", timevar="year", direction="wide")
which gives
> reshape(long, idvar="targetrow", timevar="year", direction="wide")
targetrow val.2000 val.2002 val.2008
1 1 0.051568 0.029792 0.047285
2 2 0.048050 0.056141 0.038989
More complicated transformations are possible using the reshape2 package, but this should get you started.
probably i am understanding this wrong but is ?reshape what you are looking for?
from the examples:
summary(Indometh)
wide <- reshape(Indometh, v.names="conc", idvar="Subject", timevar="time", direction="wide")
wide
I have the following data read into R as a data frame named "data_old":
yes year month
1 15 2004 5
2 9 2005 6
3 15 2006 3
4 12 2004 5
5 14 2005 1
6 15 2006 7
. . ... .
. . ... .
I have written a small loop which goes through the data and sums up the yes variable for each month/year combination:
year_f <- c(2004:2006)
month_f <- c(1:12)
for (i in year_f){
for (j in month_f){
x <- subset(data_old, month == j & year == i, select="yes")
if (nrow(x) > 0){
print(sum(x))
}
else{print("Nothing")}
}
}
My question is this: I can print the sum for each month/year combination in the terminal, but how do i store it in a vector? (the nested loop is giving me headaches trying to figure this out).
Thomas
Another way,
library(plyr)
ddply(data_old,.(year,month),function(x) sum(x[1]))
year month V1
1 2004 5 27
2 2005 1 14
3 2005 6 9
4 2006 3 15
5 2006 7 15
Forget the loops, you want to use an aggregation function. There's a recent discussion of them in this SO question.
with(data_old, tapply(yes, list(year, month), sum))
is one of many solutions.
Also, you don't need to use c() when you aren't concatenating anything. Plain 1:12 is fine.
Just to add a third option:
aggregate(yes ~ year + month, FUN=sum, data=data_old)