R plot numbers of factor levels having n, n+1, .... counts - r

I have a very large dataset (> 200000 lines) with 6 variables (only the first two shown)
>head(gt7)
ChromKey POS
1 2447 25
2 2447 183
3 26341 75
4 26341 2213
5 26341 2617
6 54011 1868
I have converted the Chromkey variable to a factor variable made up of > 55000 levels.
> gt7[1] <- lapply(gt7[1], factor)
> is.factor(gt7$ChromKey)
[1] TRUE
I can further make a table with counts of ChromKey levels
> table(gt7$ChromKey)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
88 88 44 33 11 11 33 22 121 11 22 11 11 11 22 11 33
18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
22 22 44 55 22 11 22 66 11 11 11 22 11 11 11 187 77
35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
77 11 44 11 11 11 11 11 11 22 66 11 22 11 44 22 22
... outut cropped
Which I can save in table format
> table <- table(gt7$ChromKey)
> head(table)
1 2 3 4 5 6
88 88 44 33 11 11
I would like to know whether is it possible to have a table (and histogram) of the number of levels with specific count numbers. From the example above, I would expect
88 44 33 11
2 1 1 2
I would very much appreciate any hint.

We can apply table again on the output to get the frequency count of the frequency
table(table(gt7$ChromKey))

Related

Distance Matrix from table in R

Good evening,
I need to solve a location problem in R and I'm stuck in one of the first steps.
From a .txt file I need to create a distance matrix using the euclidean method.
datos <- file.choose()
servidores <- read.table(datos)
servidores
From which I obtain the following information:
X50 shows the total number of servers.
x5 the number of hubs required.
x120 the total capacity.
The first column shows the distance of x.
The second column shows the distance of y.
The third column shows the requirements of the node.
X50 X5 X120
1 2 62 3
2 80 25 14
3 36 88 1
4 57 23 14
5 33 17 19
6 76 43 2
7 77 85 14
8 94 6 6
9 89 11 7
10 59 72 6
11 39 82 10
12 87 24 18
13 44 76 3
14 2 83 6
15 19 43 20
16 5 27 4
17 58 72 14
18 14 50 11
19 43 18 19
20 87 7 15
21 11 56 15
22 31 16 4
23 51 94 13
24 55 13 13
25 84 57 5
26 12 2 16
27 53 33 3
28 53 10 7
29 33 32 14
30 69 67 17
31 43 5 3
32 10 75 3
33 8 26 12
34 3 1 14
35 96 22 20
36 6 48 13
37 59 22 10
38 66 69 9
39 22 50 6
40 75 21 18
41 4 81 7
42 41 97 20
43 92 34 9
44 12 64 1
45 60 84 8
46 35 100 5
47 38 2 1
48 9 9 7
49 54 59 9
50 1 58 2
I tried to use the dist() function:
distance_matrix <-dist(servidores,method = "euclidean",diag = TRUE,upper = TRUE)
but since x and y are on different columns I am not sure what to do to get a 50x50 matrix with all the distances.
Anybody knows how could I create such matrix?.
Many thanks in advance.

What is the name and reason for the [1] at the output prompt?

What's the name for the [1] below.
What is its significance?
Is it always only [1]? If not, then under what conditions is it something else? (example please)
> bb <- c(5,6,7)
> bb
[1] 5 6 7
It shows the count of the variables. In your case, it shows
bb <- c(5,6,7)
> bb
# [1] 5 6 7
Try,
c(1:50)
#[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
#[35] 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
You can also avoid that being displayed by using cat
cat(c(1:50))
#1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

Splitting and iterative simple regression in r

I am pretty much new to r and I have a dummy example of a bigger table underneath. I want to split the table based on id (a,b,c,d) and do iterative simple linear regression for every subset:
x is my x variable, and columns 1:6 are y variables, to have an output of each id and each column from 1:6. Also, it would be great if I could output the model p values of the slopes into a new data frame
id x 1 2 3 4 5 6
1 a 74 18 19 NA 23 29 1
2 a 77 16 19 17 22 29 2
3 a 79 16 NA 19 23 29 3
4 a 81 17 20 18 23 29 4
5 b 74 19 20 19 23 28 11
6 b 76 15 19 18 26 28 12
7 b 79 19 21 20 24 28 NA
8 b 81 19 21 20 23 28 14
9 c 68 19 20 20 23 29 8
10 c 70 17 22 22 27 29 9
11 c 73 18 22 21 23 29 10
12 c 75 19 20 19 23 29 11
13 d 65 18 18 19 22 28 5
14 d 68 18 NA 18 20 29 6
15 d 70 18 19 18 23 28 7
16 d 72 19 17 19 22 28 8
I tried to do use plyr package but it didn't work out
regression = NULL
for ( i in 3:ncol(dumm)){
regression[i] <- dlply(dumm, .(id), function(z) lm(dumm[,i]~dumm$x, z))
}
coefs <- ldply(regression, coef)
Thanks in advance!

How can I change the column names of a chunk of columns?

I have the following example table and would like to change the column names of e,f,g,h,i. In this example, let's say I just want to paste a "2" onto the end (so e2, f2, etc.) Is there a way to do this simply without a for loop?
m <- matrix(seq_len(12*5), nrow=5, ncol=12)
m <- data.frame(m)
names(m) <- letters[1:12]
m
a b c d e f g h i j k l
1 1 6 11 16 21 26 31 36 41 46 51 56
2 2 7 12 17 22 27 32 37 42 47 52 57
3 3 8 13 18 23 28 33 38 43 48 53 58
4 4 9 14 19 24 29 34 39 44 49 54 59
5 5 10 15 20 25 30 35 40 45 50 55 60
After diligent searching, and trial/error I have not found the answer.
Both sprintf and paste0 will work. If the two who posted good answers in the comments wish to post answers, I'll remove this since they should get the credit.
Here's a paste0 answer.
> names(m)[5:9] <- paste0(names(m[5:9]), 2)
> m
a b c d e2 f2 g2 h2 i2 j k l
1 1 6 11 16 21 26 31 36 41 46 51 56
2 2 7 12 17 22 27 32 37 42 47 52 57
3 3 8 13 18 23 28 33 38 43 48 53 58
4 4 9 14 19 24 29 34 39 44 49 54 59
5 5 10 15 20 25 30 35 40 45 50 55 60

Clean bad data automatically [duplicate]

This question already has answers here:
How to convert a factor to integer\numeric without loss of information?
(12 answers)
Closed 9 years ago.
I am building an App using shiny and openair to analyze wind data.
Right now the data needs to be “cleaned” before uploading by the user.
I am interested in doing this automatically.
Some of the data is empty, some of is not numeric, so it is not possible to build a wind rose.
I want to:
1. Estimate how much of the data is not numeric
2. Cut it out and leave only numeric data
here is an example of the data:
the "NO2.mg" is read as a factor and not int becuse it does not consist only numbers
OK
here is a reproducible example:
no2<-factor(c(5,4,"c1",54,"c5",seq(2:50)))
no2
[1] 5 4 c1 54 c5 1 2 3 4 5 6 7 8 9 10 11 12 13 14
[20] 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
[39] 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
52 Levels: 1 10 11 12 13 14 15 16 17 18 19 2 20 21 22 ... c5
> as.numeric(no2)
[1] 45 34 51 46 52 1 12 23 34 45 47 48 49 50 2 3 4 5 6
[20] 7 8 9 10 11 13 14 15 16 17 18 19 20 21 22 24 25 26 27
[39] 28 29 30 31 32 33 35 36 37 38 39 40 41 42 43 44
Worst R haiku ever:
Some of the data is empty,
some of is not numeric,
so it is not possible to build a wind rose.
To convert a factor to numeric, you need to convert to character first:
no2<-factor(c(5,4,"c1",54,"c5",seq(2:50)))
no2_num <- as.numeric(as.character(no2))
#Warning message:
# NAs introduced by coercion
no2_clean <- na.omit(no2_num) #remove NAs resulting from the bad data
# [1] 5 4 54 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
# [40] 37 38 39 40 41 42 43 44 45 46 47 48 49
# attr(,"na.action")
# [1] 3 5
# attr(,"class")
# [1] "omit"
length(attr(no2_clean,"na.action"))/length(no2)*100
#[1] 3.703704
OK this is how i did it i am sure someone has abetter way
i'd love it if you share with me
this is my data:
no2<-factor(c(5,4,"c1",54,"c5",seq(2:50)))
to count the "bad data:"
sum(is.na((as.numeric(as.vector(no2)))))
and to estimate the percent of bad data:
sum(is.na((as.numeric(as.vector(no2)))))/length(no2)*100

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