I stumble upon the following thing. I read the reshape manual, but still lost.
Is there an efficient and more elegant way to reshape the matrix of even chunks?
the code to generate the matrix and reshaped matrix is below.
# current matrix
x <- matrix(sample(20*9), 20, 9)
colnames(x) <- c(paste("time",c(1:3),sep="_"),
paste("SGNL", 1, c(1:3), sep="_"),
paste("SGNL", 2, c(1:3), sep="_"))
# reshaped matrix
x.reshaped <- rbind( x[,c(1,4,7)], x[,c(2,5,8)], x[,c(3,6,9)] )
colnames(x.reshaped) <- sub("\\_1$", "", colnames(x.reshaped))
Thanks!
If you want to use an approach that is name-based and not position-based, then you should look at melt from "data.table":
library(data.table)
melt(as.data.table(x), measure.vars = patterns("time", "SGNL_1", "SGNL_2"))
Example output:
head(melt(as.data.table(x), measure.vars = patterns("time", "SGNL_1", "SGNL_2")))
# variable value1 value2 value3
# 1: 1 48 110 155
# 2: 1 67 35 140
# 3: 1 102 55 72
# 4: 1 161 39 66
# 5: 1 36 137 99
# 6: 1 158 169 85
Or, in base R:
patts <- c("time", "SGNL_1", "SGNL_2")
sapply(patts, function(y) c(x[, grep(y, colnames(x))]))
# time SGNL_1 SGNL_2
# [1,] 48 110 155
# [2,] 67 35 140
# [3,] 102 55 72
# [4,] 161 39 66
# [5,] 36 137 99
# .
# .
# .
# .
# [56,] 13 1 84
# [57,] 40 46 95
# [58,] 152 7 178
# [59,] 81 79 123
# [60,] 50 101 146
Data generated with set.seed(1).
We could create the subset of matrices (based on the index generated by the seq) in a list and then rbind it together.
do.call(rbind, lapply(1:3, function(i) x[,seq(i, length.out=3, by=3)]))
Or using a for loop
m2 <- c()
for(i in 1:3) { m2 <- rbind(m2, x[,seq(i, length.out=3, by=3)])}
x[,c(matrix(1:9, 3, byrow=TRUE))] # or shorter:
x[,matrix(1:9, 3, byrow=TRUE)]
Related
I am working on a data set which is large and having many columns. I am using data.table to speed up the calculations. However at certain points I am not sure how to go about and convert my data.table back to data.frame and do the calculation. This slows up the process. It would help a lot to have suggestions on how I can write the below in data.table. Below is a snap of my code on a dummy data -
library(data.table)
#### set the seed value
set.seed(9901)
#### create the sample variables for creating the data
p01 <- sample(1:100,1000,replace = T)
p02 <- sample(1:100,1000,replace = T)
p03 <- sample(1:100,1000,replace = T)
p04 <- sample(1:100,1000,replace = T)
p05 <- sample(1:100,1000,replace = T)
p06 <- sample(1:100,1000,replace = T)
p07 <- sample(1:100,1000,replace = T)
#### create the data.table
data <- data.table(cbind(p01,p02,p03,p04,p05,p06,p07))
###user input for last column
lcol <- 6
###calculate start column as last - 3
scol <- lcol-3
###calculate average for scol:lcol
data <- data[,avg:= apply(.SD,1,mean,na.rm=T),.SDcols=scol:lcol]
###converting to data.frame since do not know the solution in data.table
data <- as.data.frame(data)
###calculate the trend in percentage
data$t01 <- data[,lcol-00]/data[,"avg"]-1
data$t02 <- data[,lcol-01]/data[,"avg"]-1
data$t03 <- data[,lcol-02]/data[,"avg"]-1
data$t04 <- data[,lcol-03]/data[,"avg"]-1
data$t05 <- data[,lcol-04]/data[,"avg"]-1
###converting back to data.table
data <- as.data.table(data)
###calculate the min and max for the trend
data1 <- data[,`:=` (trend_min = apply(.SD,1,min,na.rm=T),
trend_max = apply(.SD,1,max,na.rm=T)),.SDcols=c(scol:lcol)]
###calculate flag if any of t04 OR t05 is an outlier for min and max values. This would be many columns in actual data
data1$flag1 <- ifelse(data1$t04 < data1$trend_min | data1$t04 > data1$trend_max,1,0)
data1$flag2 <- ifelse(data1$t05 < data1$trend_min | data1$t05 > data1$trend_max,1,0)
data1$flag <- ifelse(data1$flag1 == 1 | data1$flag2 == 1,1,0)
So basically, how can I -
calculate the percentages based on user input of column index. Note it is not simple divide but percentage
How can I create the flag variable....I think I need to use any function but not sure how....
Some steps can be made more efficient, i.e. instead of using the apply with MARGIN = 1, the mean, min, max can be replaced with rowMeans, pmin, pmax
library(data.table)
data[ , avg:= rowMeans(.SD, na.rm = TRUE) ,.SDcols=scol:lcol]
data[, sprintf('t%02d', 1:5) := lapply(.SD, function(x) x/avg -1),
.SDcol = patterns("^p0[1-5]")]
data[,`:=` (trend_min = do.call(pmin, c(.SD,na.rm=TRUE)),
trend_max = do.call(pmax, c(.SD,na.rm=TRUE)) ),.SDcols=c(scol:lcol)]
data
# p01 p02 p03 p04 p05 p06 p07 avg t01 t02 t03 t04 t05 trend_min trend_max
# 1: 35 53 22 82 100 59 69 65.75 -0.46768061 -0.19391635 -0.6653992 0.24714829 0.5209125 22 100
# 2: 78 75 15 65 70 69 66 54.75 0.42465753 0.36986301 -0.7260274 0.18721461 0.2785388 15 70
# 3: 15 45 27 61 63 75 99 56.50 -0.73451327 -0.20353982 -0.5221239 0.07964602 0.1150442 27 75
# 4: 41 80 13 22 63 84 17 45.50 -0.09890110 0.75824176 -0.7142857 -0.51648352 0.3846154 13 84
# 5: 53 9 75 47 25 75 66 55.50 -0.04504505 -0.83783784 0.3513514 -0.15315315 -0.5495495 25 75
# ---
# 996: 33 75 9 61 74 55 57 49.75 -0.33668342 0.50753769 -0.8190955 0.22613065 0.4874372 9 74
# 997: 24 68 74 11 43 75 37 50.75 -0.52709360 0.33990148 0.4581281 -0.78325123 -0.1527094 11 75
# 998: 62 78 82 97 56 50 74 71.25 -0.12982456 0.09473684 0.1508772 0.36140351 -0.2140351 50 97
# 999: 70 88 93 4 39 75 93 52.75 0.32701422 0.66824645 0.7630332 -0.92417062 -0.2606635 4 93
#1000: 20 50 99 94 62 66 98 80.25 -0.75077882 -0.37694704 0.2336449 0.17133956 -0.2274143 62 99
and then create the 'flag'
data[, flag := +(Reduce(`|`, lapply(.SD, function(x)
x < trend_min| x > trend_max))), .SDcols = t04:t05]
I have a matrix that consists of two columns and a number (n) of rows, while each row represents a point with the coordinates x and y (the two columns).
This is what it looks (LINK):
V1 V2
146 17
151 19
153 24
156 30
158 36
163 39
168 42
173 44
...
now, I would like to use a subset of three consecutive points starting from 1 to do some fitting, save the values from this fit in another list, an den go on to the next 3 points, and the next three, ... till the list is finished. Something like this:
Data_Fit_Kasa_1 <- CircleFitByKasa(Data[1:3,])
Data_Fit_Kasa_2 <- CircleFitByKasa(Data[3:6,])
....
Data_Fit_Kasa_n <- CircleFitByKasa(Data[i:i+2,])
I have tried to construct a loop, but I can't make it work. R either tells me that there's an "unexpected '}' in "}" " or that the "subscript is out of bonds". This is what I've tried:
minimal runnable code
install.packages("conicfit")
library(conicfit)
CFKasa <- NULL
Data.Fit <- NULL
for (i in 1:length(Data)) {
row <- Data[i:(i+2),]
CFKasa <- CircleFitByKasa(row)
Data.Fit[i] <- CFKasa[3]
}
RStudio Version 0.99.902 – © 2009-2016 RStudio, Inc.; Win10 Edu.
The third element of the fitted circle (CFKasa[3]) represents the radius, which is what I am really interested in. I am really stuck here, please help.
Many thanks in advance!
Best, David
Turn your data into a 3D array and use apply:
DF <- read.table(text = "V1 V2
146 17
151 19
153 24
156 30
158 36
163 39", header = TRUE)
a <- t(DF)
dim(a) <-c(nrow(a), 3, ncol(a) / 3)
a <- aperm(a, c(2, 1, 3))
# , , 1
#
# [,1] [,2]
# [1,] 146 17
# [2,] 151 19
# [3,] 153 24
#
# , , 2
#
# [,1] [,2]
# [1,] 156 30
# [2,] 158 36
# [3,] 163 39
center <- function(m) c(mean(m[,1]), mean(m[,2]))
t(apply(a, 3, center))
# [,1] [,2]
#[1,] 150 20
#[2,] 159 35
center(DF[1:3,])
#[1] 150 20
I have generate a random matrix d, then make some matrix operation.
Finally, I need to store the result in vector B. Code is below
set.seed(42)
n <- 3
m <- 4
d <- matrix(sample(0:255, n*m, replace=T), nrow = n, ncol = m)
# some matrix operation
B <-c(d[1,], d[2,], d[3,])
> d
[,1] [,2] [,3] [,4]
[1,] 234 212 188 180
[2,] 239 164 34 117
[3,] 73 132 168 184
> B
[1] 234 212 188 180 239 164 34 117 73 132 168 184
>
Could some one please explain me how to rewrite last
line via a function in order to combine the n arguments in one vector?
I have tried
B <- sapply(1:n, FUN=function(i) B<-c(d[i,]))
Thank!
This function should do it (overkill, since c(t(d)) as suggested by #joran works fine):
vectorizeByRow <- function(IN) {
OUT <- rep(NA_real_, length(IN))
nc <- ncol(IN)
nr <- nrow(IN)
a <- seq(1, length(IN), nc)
b <- a + nc - 1
for (n in 1:length(a)) {
OUT[a[n]:b[n]] <- IN[n,]
}
OUT
}
Use:
vectorizeByRow(d)
Produces:
[1] 234 212 188 180 239 164 34 117 73 132
[11] 168 184
This is from the HandyStuff package. Disclaimer: I am the author.
I am new to R, and want to sort a data frame called "weights". Here are the details:
>str(weights)
'data.frame': 57 obs. of 1 variable:
$ attr_importance: num 0.04963 0.09069 0.09819 0.00712 0.12543 ...
> names(weights)
[1] "attr_importance"
> dim(weights)
[1] 57 1
> head(weights)
attr_importance
make 0.049630556
address 0.090686474
all 0.098185517
num3d 0.007122618
our 0.125433292
over 0.075182467
I want to sort by decreasing order of attr_importance BUT I want to preserve the corresponding row names also.
I tried:
> weights[order(-weights$attr_importance),]
but it gives me a "numeric" back.
I want a data frame back - which is sorted by attr_importance and has CORRESPONDING row names intact. How can I do this?
Thanks in advance.
Since your data.frame only has one column, you need to set drop=FALSE to prevent the dimensions from being dropped:
weights[order(-weights$attr_importance),,drop=FALSE]
# attr_importance
# our 0.125433292
# all 0.098185517
# address 0.090686474
# over 0.075182467
# make 0.049630556
# num3d 0.007122618
Here is the big comparison on data.frame sorting:
How to sort a dataframe by column(s)?
Using my now-preferred solution arrange:
dd <- data.frame(b = factor(c("Hi", "Med", "Hi", "Low"),
levels = c("Low", "Med", "Hi"), ordered = TRUE),
x = c("A", "D", "A", "C"), y = c(8, 3, 9, 9),
z = c(1, 1, 1, 2))
library(plyr)
arrange(dd,desc(z),b)
b x y z
1 Low C 9 2
2 Med D 3 1
3 Hi A 8 1
4 Hi A 9 1
rankdata.txt
regno name total maths science social cat
1 SUKUMARAN 400 78 89 73 S
2 SHYAMALA 432 65 79 87 S
3 MANOJ 500 90 129 78 C
4 MILYPAULOSE 383 59 88 65 G
5 ANSAL 278 39 77 60 O
6 HAZEENA 273 45 55 56 O
7 MANJUSHA 374 50 99 52 C
8 BILBU 408 81 97 72 S
9 JOSEPHROBIN 374 57 85 68 G
10 SHINY 381 70 79 70 S
z <- data.frame(rankdata)
z[with(z, order(-total+ maths)),] #order function maths group selection
z
z[with(z, order(name)),] # sort on name
z
I want to add many new columns simultaneously to a data.table based on by-group computations. A working example of my data would look something like this:
Time Stock x1 x2 x3
1: 2014-08-22 A 15 27 34
2: 2014-08-23 A 39 44 29
3: 2014-08-24 A 20 50 5
4: 2014-08-22 B 42 22 43
5: 2014-08-23 B 44 45 12
6: 2014-08-24 B 3 21 2
Now I want to scale and sum many of the variables to get an output like:
Time Stock x1 x2 x3 x2_scale x3_scale x2_sum x3_sum
1: 2014-08-22 A 15 27 34 -1.1175975 0.7310560 121 68
2: 2014-08-23 A 39 44 29 0.3073393 0.4085313 121 68
3: 2014-08-24 A 20 50 5 0.8102582 -1.1395873 121 68
4: 2014-08-22 B 42 22 43 -0.5401315 1.1226726 88 57
5: 2014-08-23 B 44 45 12 1.1539172 -0.3274462 88 57
6: 2014-08-24 B 3 21 2 -0.6137858 -0.7952265 88 57
A brute force implementation of my problem would be:
library(data.table)
set.seed(123)
d <- data.table(Time = rep(seq.Date( Sys.Date(), length=3, by="day" )),
Stock = rep(LETTERS[1:2], each=3 ),
x1 = sample(1:50, 6),
x2 = sample(1:50, 6),
x3 = sample(1:50, 6))
d[,x2_scale:=scale(x2),by=Stock]
d[,x3_scale:=scale(x3),by=Stock]
d[,x2_sum:=sum(x2),by=Stock]
d[,x3_sum:=sum(x3),by=Stock]
Other posts describing a similar issue (Add multiple columns to R data.table in one function call? and Assign multiple columns using := in data.table, by group) suggest the following solution:
d[, c("x2_scale","x3_scale"):=list(scale(x2),scale(x3)), by=Stock]
d[, c("x2_sum","x3_sum"):=list(sum(x2),sum(x3)), by=Stock]
But again, this would get very messy with a lot of variables and also this brings up an error message with scale (but not with sum since this isn't returning a vector).
Is there a more efficient way to achieve the required result (keeping in mind that my actual data set is quite large)?
I think with a small modification to your last code you can easily do both for as many variables you want
vars <- c("x2", "x3") # <- Choose the variable you want to operate on
d[, paste0(vars, "_", "scale") := lapply(.SD, function(x) scale(x)[, 1]), .SDcols = vars, by = Stock]
d[, paste0(vars, "_", "sum") := lapply(.SD, sum), .SDcols = vars, by = Stock]
## Time Stock x1 x2 x3 x2_scale x3_scale x2_sum x3_sum
## 1: 2014-08-22 A 13 14 32 -1.1338934 1.1323092 87 44
## 2: 2014-08-23 A 25 39 9 0.7559289 -0.3701780 87 44
## 3: 2014-08-24 A 18 34 3 0.3779645 -0.7621312 87 44
## 4: 2014-08-22 B 44 8 6 -0.4730162 -0.7258662 59 32
## 5: 2014-08-23 B 49 3 18 -0.6757374 1.1406469 59 32
## 6: 2014-08-24 B 15 48 8 1.1487535 -0.4147807 59 32
For simple functions (that don't need special treatment like scale) you could easily do something like
vars <- c("x2", "x3") # <- Define the variable you want to operate on
funs <- c("min", "max", "mean", "sum") # <- define your function
for(i in funs){
d[, paste0(vars, "_", i) := lapply(.SD, eval(i)), .SDcols = vars, by = Stock]
}
Another variation using data.table
vars <- c("x2", "x3")
d[, paste0(rep(vars, each=2), "_", c("scale", "sum")) := do.call(`cbind`,
lapply(.SD, function(x) list(scale(x)[,1], sum(x)))), .SDcols=vars, by=Stock]
d
# Time Stock x1 x2 x3 x2_scale x2_sum x3_scale x3_sum
#1: 2014-08-22 A 15 27 34 -1.1175975 121 0.7310560 68
#2: 2014-08-23 A 39 44 29 0.3073393 121 0.4085313 68
#3: 2014-08-24 A 20 50 5 0.8102582 121 -1.1395873 68
#4: 2014-08-22 B 42 22 43 -0.5401315 88 1.1226726 57
#5: 2014-08-23 B 44 45 12 1.1539172 88 -0.3274462 57
#6: 2014-08-24 B 3 21 2 -0.6137858 88 -0.7952265 57
Based on comments from #Arun, you could also do:
cols <- paste0(rep(vars, each=2), "_", c("scale", "sum"))
d[,(cols):= unlist(lapply(.SD, function(x) list(scale(x)[,1L], sum(x))),
rec=F), by=Stock, .SDcols=vars]
You're probably looking for a pure data.table solution, but you could also consider using dplyr here since it works with data.tables as well (no need for conversion). Then, from dplyr you could use the function mutate_all as I do in this example here (with the first data set you showed in your question):
library(dplyr)
dt %>%
group_by(Stock) %>%
mutate_all(funs(sum, scale), x2, x3)
#Source: local data table [6 x 9]
#Groups: Stock
#
# Time Stock x1 x2 x3 x2_sum x3_sum x2_scale x3_scale
#1 2014-08-22 A 15 27 34 121 68 -1.1175975 0.7310560
#2 2014-08-23 A 39 44 29 121 68 0.3073393 0.4085313
#3 2014-08-24 A 20 50 5 121 68 0.8102582 -1.1395873
#4 2014-08-22 B 42 22 43 88 57 -0.5401315 1.1226726
#5 2014-08-23 B 44 45 12 88 57 1.1539172 -0.3274462
#6 2014-08-24 B 3 21 2 88 57 -0.6137858 -0.7952265
You can easily add more functions to be calculated which will create more columns for you. Note that mutate_all applies the function to each column except the grouping variable (Stock) by default. But you can either specify the columns you only want to apply the functions to (which I did in this example) or you can specify which columns you don't want to apply the functions to (that would be, e.g. -c(x2,x3) instead of where I wrote x2, x3).
EDIT: replaced mutate_each above with mutate_all as mutate_each will be deprecated in the near future.
EDIT: cleaner version using functional. I think this is the closest to the dplyr answer.
library(functional)
funs <- list(scale=Compose(scale, c), sum=sum) # See data.table issue #783 on github for the need for this
cols <- paste0("x", 2:3)
cols.all <- outer(cols, names(funs), paste, sep="_")
d[,
c(cols.all) := unlist(lapply(funs, Curry(lapply, X=.SD)), rec=F),
.SDcols=cols,
by=Stock
]
Produces:
Time Stock x1 x2 x3 x2_scale x3_scale x2_sum x3_sum
1: 2014-08-22 A 15 27 34 -1.1175975 0.7310560 121 68
2: 2014-08-23 A 39 44 29 0.3073393 0.4085313 121 68
3: 2014-08-24 A 20 50 5 0.8102582 -1.1395873 121 68
4: 2014-08-22 B 42 22 43 -0.5401315 1.1226726 88 57
5: 2014-08-23 B 44 45 12 1.1539172 -0.3274462 88 57
6: 2014-08-24 B 3 21 2 -0.6137858 -0.7952265 88 57