I have this line of code but I don't know what it means especially the note_ind part.
apply(mydat[,-c(1,2,3,note_ind:ncol(dataset))],c(1,2),as.numeric)
The notation x:y is used to create numeric vector sequences where each element is the previous element incremented by 1. It is shorthand for `seq(x, y, by = 1). It is most commonly used for integer sequences, but it works on doubles also.
1:10
[1] 1 2 3 4 5 6 7 8 9 10
1.1:10.1
[1] 1.1 2.1 3.1 4.1 5.1 6.1 7.1 8.1 9.1 10.1
1.5:10.2 # sequence stops after 9.5 because 10.2 < 9.5 + 1 - seq() behaves the same way
[1] 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
Presumably note_ind is an integer value from somewhere else in your code. ncol(data.set) is the number of columns, so note_ind:ncol(dataset) generates a seqence between those two values, incrementing by 1 for each element.
Related
I need some help with R coding here.
The data set Glass consists of 214 rows of data in which each row corresponds to a glass sample. Each row consists of 10 columns. When viewed as a classification problem, column 10
(Type) specifies the class of each observation/instance. The remaining columns are attributes that might beused to infer column 10. Here is an example of the first row
RI Na Mg Al Si K Ca Ba Fe Type
1 1.52101 13.64 4.49 1.10 71.78 0.06 8.75 0.0 0.0 1
First, I casted column 10 so that it is interpreted by R as a factor instead of an integer value.
Now I need to create a vector with indices for all observations (must have values 1-214). This needs to be done to creating training data for Naive Bayes. I know how to create a vector with 214 values, but not one that has specific indices for observations from a data frame.
If it helps this is being done to set up training data for Naive Bayes, thanks
I'm not totally sure that I get what you're trying to do... So please forgive me if my solution isn't helpful. If your df's name is 'df', just use the dplyr package for reordering your columns and write
library(dplyr)
df['index'] <- 1:214
df <- df %>% select(index,everything())
Here's an example. So that I can post full dataframes, my dataframes will only have 10 rows...
Let's say my dataframe is:
df <- data.frame(col1 = c(2.3,6.3,9.2,1.7,5.0,8.5,7.9,3.5,2.2,11.5),
col2 = c(1.5,2.8,1.7,3.5,6.0,9.0,12.0,18.0,20.0,25.0))
So it looks like
col1 col2
1 2.3 1.5
2 6.3 2.8
3 9.2 1.7
4 1.7 3.5
5 5.0 6.0
6 8.5 9.0
7 7.9 12.0
8 3.5 18.0
9 2.2 20.0
10 11.5 25.0
If I want to add another column that just is 1,2,3,4,5,6,7,8,9,10... and I'll call it 'index' ...I could do this:
library(dplyr)
df['index'] <- 1:10
df <- df %>% select(index, everything())
That will give me
index col1 col2
1 1 2.3 1.5
2 2 6.3 2.8
3 3 9.2 1.7
4 4 1.7 3.5
5 5 5.0 6.0
6 6 8.5 9.0
7 7 7.9 12.0
8 8 3.5 18.0
9 9 2.2 20.0
10 10 11.5 25.0
Hope this will help
df$ind <- seq.int(nrow(df))
I have a tibble structured as follows:
day theta
1 1 2.1
2 1 2.1
3 2 3.2
4 2 3.2
5 5 9.5
6 5 9.5
7 5 9.5
Note that the tibble contains multiple rows for each day, and for each day the same value for theta is repeated an arbitrary number of times. (The tibble contains other arbitrary columns necessitating this repeating structure.)
I'd like to use dplyr to cumulatively sum values for theta across days such that, in the example above, 2.1 is added only a single time to 3.2, etc. The tibble would be mutated so as to append the new cumulative sum (c.theta) as follows:
day theta c.theta
1 1 2.1 2.1
2 1 2.1 2.1
3 2 3.2 5.3
4 2 3.2 5.3
5 5 9.5 14.8
6 5 9.5 14.8
7 5 9.5 14.8
...
My initial efforts to group_by day and then cumsum over theta resulted only in cumulative summing over the full set of data (e.g., 2.1 + 2.1 + 3.2 ...) which is undesirable. In my Stack Overflow searches, I can find many examples of cumulative summing within groups, but never between groups, as I describe above. Nudges in the right direction would be much appreciated.
Doing this in dplyr I came up with a very similar solution to PoGibas - use distinct to just get one row per day, find the sum and merge back in:
df = read.table(text="day theta
1 1 2.1
2 1 2.1
3 2 3.2
4 2 3.2
5 5 9.5
6 5 9.5
7 5 9.5", header = TRUE)
cumsums = df %>%
distinct(day, theta) %>%
mutate(ctheta = cumsum(theta))
df %>%
left_join(cumsums %>% select(day, ctheta), by = 'day')
Not a dplyr, but just an alternative data.table solution:
library(data.table)
# Original table is called d
setDT(d)
merge(d, unique(d)[, .(c.theta = cumsum(theta), day)])
day theta c.theta
1: 1 2.1 2.1
2: 1 2.1 2.1
3: 2 3.2 5.3
4: 2 3.2 5.3
5: 5 9.5 14.8
6: 5 9.5 14.8
7: 5 9.5 14.8
PS: If you want to preserve other columns you have to use unique(d[, .(day, theta)])
In base R you could use split<- and tapply to return the desired result.
# construct 0 vector to fill in
dat$temp <- 0
# fill in with cumulative sum for each day
split(dat$temp, dat$day) <- cumsum(tapply(dat$theta, dat$day, head, 1))
Here, tapply returns the first element of theta for each day which is is fed to cumsum. The elements of cumulative sum are fed to each day using split<-.
This returns
dat
day theta temp
1 1 2.1 2.1
2 1 2.1 2.1
3 2 3.2 5.3
4 2 3.2 5.3
5 5 9.5 14.8
6 5 9.5 14.8
7 5 9.5 14.8
I want to convert the following numbers in this way I tried to use all possible methods but i am unable to get the value which i expected
value round off value
0.0 - 4.9 0
5.0 - 5.9 6
6.0 - 6.9 7
7.0 - 7.9 8
8.0 - 8.9 9
9.0 - 10.0 10
The above table is for reference
expected output eg :- roundup(5.0) = 6 ,roundup(6.9)=7
You can try:
roundup<-function(x) c(0,6:10)[findInterval(x,c(0,5:9))]
roundup(c(5,6.9))
#[1] 6 7
I am trying to make two subvectors equal length from two vectors equal length.
Values in first vector are ordered as follows:
a<-c(9,9,9,8,8,7,6,5,5,5)
Second vector is random, but lets take
b<-c(1,2,3,4,5,6,7,8,9,10)
The first subvector is simple:it is just the vector a withouth repeats
f(a)<-c(9,8,7,6,5)
The second subvector should be made as follows:
for single value in vector a (no repeats in a)the vector g(b) has the same value as vector b on corresponding position. For repeats in a the g(b) value should be mean of values from corresponding subvector b. So:
g(b)<-c(mean(c(1,2,3)), mean(c(4,5)), 6, 7, mean(c(8,9,10)))
I have no idea where to start. Thx for help!
tapply is the function you want. See ?tapply to see how it works. Here:
res<-tapply(b,a,mean)
# 5 6 7 8 9
#9.0 7.0 6.0 4.5 2.0
If you want to preserve the order:
tapply(b,a,mean)[as.character(unique(a))]
# 9 8 7 6 5
#2.0 4.5 6.0 7.0 9.0
As you can see, it gives the unique values of a and for each of them, the desired function (in this case mean(b)) is evaluated.
We can also use ave
unique(ave(b, a))
#[1] 2.0 4.5 6.0 7.0 9.0
Or another option would be to convert the 'b' to factor with levels specified
tapply(b, factor(a, levels=unique(a)), FUN=mean)
# 9 8 7 6 5
#2.0 4.5 6.0 7.0 9.0
You can do in this way:
uniqueA <- a[!duplicated(a)] # or simply unique(a) but I'm not sure about order preservation
uniqueB <- as.numeric(by(b,match(a,uniqueA),mean))
> uniqueA
[1] 9 8 7 6 5
> uniqueB
[1] 2.0 4.5 6.0 7.0 9.0
I would like to compute a moving average (ma) over some time series data but I would like the ma to consider the order n starting from the rightmost of my series so my last ma value corresponds to the ma of the last n values of my series. The desired function rightmost_ma would produce this output:
data <- seq(1,10)
> data
[1] 1 2 3 4 5 6 7 8 9 10
rightmost_ma(data, n=2)
NA 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
I was reviewing the different ma possibilities e.g. package forecast and could not find how to cover this use case. Note that the critical requirement for me is to have valid non NA ma values for the last elements of the series or in other words I want my ma to produce valid results without "looking into the future".
Take a look at rollmean function from zoo package
> library(zoo)
> rollmean(zoo(1:10), 2, align ="right", fill=NA)
1 2 3 4 5 6 7 8 9 10
NA 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
you can also use rollapply
> rollapply(zoo(1:10), width=2, FUN=mean, align = "right", fill=NA)
1 2 3 4 5 6 7 8 9 10
NA 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
I think using stats::filter is less complicated, and might have better performance (though zoo is well written).
This:
filter(1:10, c(1,1)/2, sides=1)
gives:
Time Series:
Start = 1
End = 10
Frequency = 1
[1] NA 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
If you don't want the result to be a ts object, use as.vector on the result.