I have a vector of numbers
a = c(1:100)
digits = c(0:9)
I want to know the frequency of digits in the vector. I want the output precisely as the below example:
Digits Frequency
0 10
1 20
2 20
3 20
4 20
5 20
6 20
7 20
8 20
9 20
How to get this output using R?
You can convert the numbers to character with as.character then split to individual characters with strsplit and count the frequency with table.
table(unlist(strsplit(as.character(a), "")))
# 0 1 2 3 4 5 6 7 8 9
#11 21 20 20 20 20 20 20 20 20
Or in case for more variations of input:
table(unlist(strsplit(gsub("[^[:digit:]]", "", format(a, scientific =FALSE)), "")))
# 0 1 2 3 4 5 6 7 8 9
#11 21 20 20 20 20 20 20 20 20
Related
How to write an R-script to initialize a vector with integers, rearrange the elements by interleaving the
first half elements with the second half elements and store in the same vector without using pre-defined function and display the updated vector.
This sounds like a homework question, and it would be nice to see some effort on your own part, but it's pretty straightforward to do this in R.
Suppose your vector looks like this:
vec <- 1:20
vec
#> [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Then you can just do:
c(t(cbind(vec[1:10], vec[11:20])))
#> [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20
This works by joining the two vectors into a 10 x 2 matrix, then transposing that matrix and turning it into a vector.
We may use matrix directly and concatenate
c(matrix(vec, nrow = 2, byrow = TRUE))
-output
[1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20
data
vec <- 1:20
Or using mapply:
vec <- 1:20
c(mapply(\(x,y) c(x,y), vec[1:10], vec[11:20]))
#> [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20
We can try this using order + %%
> vec[order((seq_along(vec) - 1) %% (length(vec) / 2))]
[1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20
Another way is to use rbind on the 2 halves of the vector, which creates a matrix with two rows. Then, we can then turn the matrix into a vector, which will go through column by column (i.e., 1, 11, 2, 12...). However, this will only work for even vectors.
vec <- 1:20
c(rbind(vec[1:10], vec[11:20]))
# [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20
So, for uneven vectors, we can use order, which will return the indices of the numbers in the two seq_along vectors.
vec2 <- 1:21
order(c(seq_along(vec2[1:10]),seq_along(vec2[11:21])))
# [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 21
Say I have a vector named all_combinations with numbers from 1 to 20.
I need to extract 2 vectors (coding_1 and coding_2) of length equal to number_of_peptide_clusters, which happens to be 20 as well in my current case.
The 2 new vectors should be randomly sampled from all_combinations, so that are not overlapping at each index position.
I do the following:
set.seed(3)
all_combinations=1:20
number_of_peptide_clusters=20
coding_1 <- sample(all_combinations, number_of_peptide_clusters, replace = FALSE)
coding_1
[1] 5 12 7 4 10 8 11 15 17 16 18 13 9 20 2 14 19 1 3 6
coding_2 <- sample(all_combinations, number_of_peptide_clusters, replace = FALSE)
coding_2
[1] 5 9 19 16 18 12 8 6 15 3 13 14 7 2 11 20 10 4 17 1
This is the example that gives me trouble, cause only one number is overlapping at the same index (5 at position 1).
What I would do in these cases is spot the overlapping numbers and resample them out of the list of all overlapping numbers...
Imagine coding_1 and coding_2 were:
coding_1
[1] 5 9 7 4 10 8 11 15 17 16 18 13 12 20 2 14 19 1 3 6
coding_2
[1] 5 9 19 16 18 12 8 6 15 3 13 14 7 2 11 20 10 4 17 1
In this case I would have 5 and 9 overlapping in the same position, so I would resample them in coding_2 out of the full list of overlapping ones [resample index 1 from c(5,9) so that isn't equal to 5, and index 2 so it isn't equal to 9]. So coding_2 would be:
coding_2
[1] 9 5 19 16 18 12 8 6 15 3 13 14 7 2 11 20 10 4 17 1
However, in the particular case above, I cannot use such approach... So what would be the best way to obtain 2 samples of length 20 from a vector of length 20 as well, so that the samples aren't overlapping at the same index positions?
It would be great that I could obtain the second sample coding_2 already knowing coding_1... Otherwise obtaining the 2 at the same time would also be acceptable if it makes things easier. Thanks!
I think the best solution is simply to use a rejection strategy:
set.seed(3)
all_combinations <- 1:20
number_of_peptide_clusters <- 20
count <- 0
repeat {
count <- count + 1
message("Try number ", count)
coding_1 <- sample(all_combinations, number_of_peptide_clusters, replace = FALSE)
coding_2 <- sample(all_combinations, number_of_peptide_clusters, replace = FALSE)
if (!any(coding_1 == coding_2))
break
}
#> Try number 1
#> Try number 2
#> Try number 3
#> Try number 4
#> Try number 5
#> Try number 6
#> Try number 7
#> Try number 8
#> Try number 9
coding_1
#> [1] 18 16 17 12 13 8 6 15 3 5 20 9 11 4 19 2 14 7 1 10
coding_2
#> [1] 5 20 14 2 11 6 7 10 19 8 4 1 15 9 13 17 18 16 12 3
Created on 2020-11-04 by the reprex package (v0.3.0)
I have a data frame d and I'd like to add a VALUE_GROUP column that looks at the value field and returns the upper limit of the bucket the value falls into
Value Value_group
0<=value<5 5
5<=value<10 10
10<=value<15 15
15<=value<20 20
You can see the Value_group is the max possible value in the bucket i.e. for value between 0 and 5 Value_group = 5
d =data.frame(group = rep("A",20),value = seq(1,20,1))
d
d$Value_Group = ??
Value_group can be added using multiple ifelse() statements but is there a better way?
The result would be:
group value Value_Group
1 A 1 5
2 A 2 5
3 A 3 5
4 A 4 5
5 A 5 5
6 A 6 10
7 A 7 10
8 A 8 10
9 A 9 10
10 A 10 10
11 A 11 15
12 A 12 15
13 A 13 15
14 A 14 15
15 A 15 15
16 A 16 20
17 A 17 20
18 A 18 20
19 A 19 20
20 A 20 20
Thank you.
I have a data frame of GPS locations with a column of seconds. How can I split create a new column based on time-gaps? i.e. for this data.frame:
df <- data.frame(secs=c(1,2,3,4,5,6,7,10,11,12,13,14,20,21,22,23,24,28,29,31))
I would like to cut the data frame when there is a time gap between locations of 3 or more seconds seconds and create a new column entitled 'bouts' which gives a running tally of the number of sections to give a data frame looking like this:
id secs bouts
1 1 1
2 2 1
3 3 1
4 4 1
5 5 1
6 6 1
7 7 1
8 10 2
9 11 2
10 12 2
11 13 2
12 14 2
13 20 3
14 21 3
15 22 3
16 23 3
17 24 3
18 28 4
19 29 4
20 31 4
Use cumsum and diff:
df$bouts <- cumsum(c(1, diff(df$secs) >= 3))
Remember that logical values get coerced to numeric values 0/1 automatically and that diff output is always one element shorter than its input.
I have a dataframe df
Reads Counts
aaaa 10
bbbb 20
cccc 25
and so on.
I want to calculate the number of reads which exceed a certain value of counts and plot that. Example I want a data frame that looks like
Counts>= #reads with Counts>=
1 3
2 3
3 3
11 2
20 2
21 1
and so on. Can you suggest how I can get such a dataframe and plot it.
Given the levels you want to plot at...
cutoffs <- 1:30
... you could do something like:
data.frame(cutoff=cutoffs, num.above=Reduce("+", lapply(dat$Counts, ">=", cutoffs)))
# cutoff num.above
# 1 1 3
# 2 2 3
# 3 3 3
# 4 4 3
# 5 5 3
# 6 6 3
# 7 7 3
# 8 8 3
# 9 9 3
# 10 10 3
# 11 11 2
# 12 12 2
# 13 13 2
# 14 14 2
# 15 15 2
# 16 16 2
# 17 17 2
# 18 18 2
# 19 19 2
# 20 20 2
# 21 21 1
# 22 22 1
# 23 23 1
# 24 24 1
# 25 25 1
# 26 26 0
# 27 27 0
# 28 28 0
# 29 29 0
# 30 30 0
Basically for each value in the original data frame you compute a vector of whether it's greater than or equal to each cutoff (using lapply with >=). Then you add them up (using Reduce with +), getting the total number greater than or equal to each cutoff.
Another option would be using outer/colSums
cutoff <- 1:30
data.frame(cutoff=cutoffs, num.above=colSums(outer(df$Counts, cutoffs, ">=")))