Create one column out of several columns Rstudio or Excel - r

I've tried to find an already existing question on this matter but I couldn't so that is why I'm asking here you:
Summary:
I want to make ONE column out of several Columns. All the values in the columns are put in the same order as they are and also, the columns should be stacked below each other.
Description and details
Below is an example of how my csv.file look like. However, note that there is >400 columns and that's why I don't want to do it manually in for example Excel. ALL columns has 24 rows each.
X1 X2 X3 X4 X5 X6 ... X470
0 1 5 10 8 0 7
0 0 0 0 0 0 0
2 3 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
I want to "stack" all the columns in one column, as I've shortly described in the summary:
Info: The sign "..." below implies the rest of the values from that column.
VALUE FROM COLUMN
0 X1
0 X1
2 X1
...
1 X2
0 X2
3 X2
...
5 X3
...
10 X4
...
8 X5
...
0 X6
...
7 X470
...
So in the end, instead of having 486 column where each of them have 24 rows. I will have 1 column with 11664 rows. It would be good if the origin column is written in a new column on the side (as showed above) but this is not "obligated".
OBS! Note that I've with this df just showed in general what I want to achieve, so clear and understandable commands are appreciated as I will apply it to my df.
It doesn't matter if the solution is done in R or Excel! As long as it is easy to do
I hope my description is clear, otherwise please let me know so I can try to describe again.
Many thanks for suggestions and help.
Kind regards, Elin

We can use stack to get the values in one column and the colnames in the next.
stack(df)
Or use unlist
data.frame(VALUE=unlist(df),
fromColumn= rep(names(df), each=nrow(df)))

Here's a VBA user defined function to do the job:
Function ConcatCols(Colrange As Variant) As Variant
Dim LongCol() As Variant, i As Long, j As Long, k As Long
Dim NumCols As Long, NumRows As Long, NumRows2 As Long
If TypeName(Colrange) = "Range" Then Colrange = Colrange.Value2
NumRows = UBound(Colrange)
NumCols = UBound(Colrange, 2)
NumRows2 = NumRows * NumCols
ReDim LongCol(1 To NumRows2, 1 To 1)
k = 1
For i = 1 To NumCols
For j = 1 To NumRows
LongCol(k, 1) = Colrange(j, i)
k = k + 1
Next j
Next i
ConcatCols = LongCol
End Function
Enter the code in a VBA module then enter =ConcatCols(A1:RL24) as an array function (Ctrl-Shift-Enter) in column RM (or wherever you want) to view the entire concatenated column, or call from a VBA sub to write the data to the spreadsheet.

The following is pretty simple but requires loading the reshape2 package which I think is included in base. As suggested above, stack() gives similar output, but reverses the columns.
library(reshape2)
df <- data.frame("A" = 1:21, "B" = 21:41, "C" = 40:60)
> df
A B C
1 1 21 40
2 2 22 41
3 3 23 42
4 4 24 43
5 5 25 44
6 6 26 45
7 7 27 46
8 8 28 47
9 9 29 48
10 10 30 49
11 11 31 50
12 12 32 51
13 13 33 52
14 14 34 53
15 15 35 54
16 16 36 55
17 17 37 56
18 18 38 57
19 19 39 58
20 20 40 59
21 21 41 60
melt(df)
> melt(df)
No id variables; using all as measure variables
variable value
1 A 1
2 A 2
3 A 3
4 A 4
5 A 5
6 A 6
7 A 7
8 A 8
9 A 9
10 A 10
11 A 11
12 A 12
13 A 13
14 A 14
15 A 15
16 A 16
17 A 17
18 A 18
19 A 19
20 A 20
21 A 21
22 B 21
23 B 22
24 B 23
25 B 24
26 B 25
27 B 26
28 B 27
29 B 28
30 B 29
31 B 30
32 B 31
33 B 32
34 B 33
35 B 34
36 B 35
37 B 36
38 B 37
39 B 38
40 B 39
41 B 40
42 B 41
43 C 40
44 C 41
45 C 42
46 C 43
47 C 44
48 C 45
49 C 46
50 C 47
51 C 48
52 C 49
53 C 50
54 C 51
55 C 52
56 C 53
57 C 54
58 C 55
59 C 56
60 C 57
61 C 58
62 C 59
63 C 60

Related

R: table frequencies of letters in string based on Alphabet

I need to compute letter frequencies of a large list of words. For each of the locations in the word (first, second,..), I need to find how many times each letter (a-z) appeared in the list and then table the data according to the word positon.
For example, if my word list is: words <- c("swims", "seems", "gills", "draws", "which", "water")
then the result table should like that:
letter
first position
second position
third position
fourth position
fifth position
a
0
1
1
0
0
b
0
0
0
0
0
c
0
0
0
1
0
d
1
0
0
0
0
e
0
1
1
1
0
f
0
0
0
0
0
...continued until z
...
...
...
...
...
All words are of same length (5).
What I have so far is:
alphabet <- letters[1:26]
words.df <- data.frame("Words" = words)
words.df <- words.df %>% mutate("First_place" = substr(words.df$words,1,1))
words.df <- words.df %>% mutate("Second_place" = substr(words.df$words,2,2))
words.df <- words.df %>% mutate("Third_place" = substr(words.df$words,3,3))
words.df <- words.df %>% mutate("Fourth_place" = substr(words.df$words,4,4))
words.df <- words.df %>% mutate("Fifth_place" = substr(words.df$words,5,5))
x1 <- words.df$First_place
x1 <- table(factor(x1,alphabet))
x2 <- words.df$Second_place
x2 <- table(factor(x2,alphabet))
x3 <- words.df$Third_place
x3 <- table(factor(x3,alphabet))
x4 <- words.df$Fourth_place
x4 <- table(factor(x4,alphabet))
x5 <- words.df$Fifth_place
x5 <- table(factor(x5,alphabet))
My code is not effective and gives tables to each letter position sepretely. All help will be appreicated.
in base R use table:
table(let = unlist(strsplit(words,'')),pos = sequence(nchar(words)))
pos
let 1 2 3 4 5
a 0 1 1 0 0
c 0 0 0 1 0
d 1 0 0 0 0
e 0 1 1 1 0
g 1 0 0 0 0
h 0 1 0 0 1
i 0 1 2 0 0
l 0 0 1 1 0
m 0 0 0 2 0
r 0 1 0 0 1
s 2 0 0 0 4
t 0 0 1 0 0
w 2 1 0 1 0
Note that if you need all the values from a-z then use
table(factor(unlist(strsplit(words,'')), letters), sequence(nchar(words)))
Also to get a dataframe you could do:
d <- table(factor(unlist(strsplit(words,'')), letters), sequence(nchar(words)))
cbind(letters = rownames(d), as.data.frame.matrix(d))
Here is a tidyverse solution using dplyr, purrr, and tidyr:
strsplit(words.df$Words, "") %>%
map_dfr(~setNames(.x, seq_along(.x))) %>%
pivot_longer(everything(),
values_drop_na = T,
names_to = "pos",
values_to = "letter") %>%
count(pos, letter) %>%
pivot_wider(names_from = pos,
names_glue = "pos{pos}",
id_cols = letter,
values_from = n,
values_fill = 0L)
Output
letter pos1 pos2 pos3 pos4 pos5 pos6 pos7 pos8 pos9 pos10 pos11
1 a 65 127 88 38 28 17 14 5 3 0 0
2 b 58 4 7 9 2 4 2 0 1 0 0
3 c 83 14 45 37 20 19 8 3 2 0 0
4 C 2 0 0 0 0 0 0 0 0 0 0
5 d 43 8 33 47 21 22 9 3 1 1 0
6 e 45 156 81 132 114 69 48 23 14 2 2
7 f 54 11 18 10 5 2 1 0 0 0 0
8 g 23 7 27 21 15 8 7 1 0 0 0
9 h 38 56 6 28 21 10 3 3 1 1 0
10 i 25 106 51 58 38 28 8 4 1 0 0
11 j 6 0 2 2 0 0 0 0 0 0 0
12 k 9 1 6 22 12 0 0 0 0 0 0
13 l 45 41 54 54 36 9 7 6 0 2 0
14 m 45 8 31 19 8 8 4 2 0 0 0
15 n 23 42 75 53 34 41 16 16 4 2 0
16 o 28 167 76 41 38 9 11 2 1 0 0
17 p 72 20 34 30 8 3 1 1 1 0 0
18 q 7 2 1 0 0 0 0 0 0 0 0
19 r 46 74 92 59 56 45 12 9 1 1 0
20 s 119 8 67 35 31 22 18 4 1 0 0
21 t 65 30 73 83 57 42 31 9 6 3 1
22 u 12 66 39 36 20 7 7 2 0 0 0
23 v 8 7 20 12 5 5 1 0 0 0 0
24 w 53 8 13 10 2 3 0 1 0 0 0
25 y 6 4 16 15 17 15 10 5 6 1 1
26 x 0 12 5 0 0 0 0 0 0 0 0
27 z 0 0 1 0 0 0 1 1 0 0 0

Summing up different elements in a matrix in R

I'm trying to perform calculations on different elements in a matrix in R. My Matrix is 18x18 and I would like to get e.g. the mean of each 6x6 array (which makes 9 arrays in total). My desired arrays would be:
A1 <- df[1:6,1:6]
A2 <- df[1:6,7:12]
A3 <- df[1:6,13:18]
B1 <- df[7:12,1:6]
B2 <- df[7:12,7:12]
B3 <- df[7:12,13:18]
C1 <- df[13:18,1:6]
C2 <- df[13:18,7:12]
C3 <- df[13:18,13:18]
The matrix looks like this:
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
5 14 17 9 10 8 4 10 12 18 9 13 14 NA NA 19 15 10 10
10 30 32 23 27 17 28 25 12 28 29 28 26 19 25 34 24 11 17
15 16 16 16 9 17 27 17 16 30 13 18 13 15 13 19 8 7 9
20 15 12 18 18 18 6 4 6 9 11 10 10 13 11 8 10 15 15
25 7 13 21 7 3 5 2 5 5 4 3 2 3 5 2 1 5 6
30 5 9 1 7 7 4 4 12 8 9 2 0 5 2 1 0 2 6
35 3 0 2 0 0 4 4 7 4 4 5 2 0 0 1 0 0 0
40 0 4 0 0 0 1 3 9 10 10 1 0 0 0 1 0 1 0
45 0 0 0 0 0 3 10 9 17 9 1 0 0 0 0 0 0 0
50 0 0 2 0 0 0 2 8 20 0 0 0 0 0 1 0 0 0
55 0 0 0 0 0 0 7 3 21 0 0 0 0 0 0 0 0 0
60 0 0 0 0 3 4 10 2 2 0 0 1 0 0 0 0 0 0
65 0 0 0 0 0 4 8 4 8 11 0 0 0 0 0 0 0 0
70 0 0 0 0 0 6 2 5 14 0 0 0 0 0 0 0 0 0
75 0 0 0 0 0 4 0 5 9 0 0 0 0 0 0 0 0 0
80 0 0 0 0 0 4 4 0 4 2 0 0 0 0 0 0 0 0
85 0 0 0 0 0 0 0 4 1 1 0 0 0 0 0 0 0 0
90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Is there a clean way to solve this issue with a loop?
Thanks a lot in advance,
Paul
Given your matrix, e.g.
x <- matrix(1:(18*18), ncol=18)
Try, for example for sub matrices of 6
step <- 6
nx <- nrow(x)
if((nx %% step) != 0) stop("nx %% step should be 0")
indI <- seq(1, nx, by=step)
nbStep <- length(indI)
for(Col in 1:nbStep){
for(Row in 1:nbStep){
name <- paste0(LETTERS[Col],Row)
theCol <- indI[Col]:(indI[Col]+step-1)
theRow <- indI[Row]:(indI[Row]+step-1)
assign(name, sum(x[theCol, theRow]))
}
}
You'll get your results in A1, A2, A3...
This is the idea. Twist the code for non square matrices, different size of sub matrices, ...
Here's one way:
# generate fake data
set.seed(47)
n = 18
m = matrix(rpois(n * n, lambda = 5), nrow = n)
# generate starting indices
n_array = 6
start_i = seq(1, n, by = n_array)
arr_starts = expand.grid(row = start_i, col = start_i)
# calculate sums
with(arr_starts, mapply(function(x, y) sum(m[(x + 1:n_array) - 1, (y + 1:n_array) - 1]), row, col))
# [1] 158 188 176 201 188 201 197 206 204

validating model in new dataset

I have a dataset (d) in which I am looking at the prediction of hospital mortality (0 or 1) using tn1 and 3 other variables (rms package). I have built the model and now I would like to validate it in a second dataset using the same model coefficients. The variables have the same names etc, but I don’t know how to keep the coefficients from f1, rather than letting the model generate new coefficients for the second dataset.
I would be grateful for your expertise, many thanks, Annemarie
f1 <- lrm(outcomehosp ~ I(log2((tn1+0.001))) + apscore_ad + emsurg +
corrapiidiag, data = d)
record_id| corrapiidiag| tn1 |emsurg| apscore_ad |outcomehosp
7 3 0.27 1 24 1
8 9 0 1 21 0
9 7 0.11 0 22 0
11 9 0 0 13 0
12 9 13.9 0 17 0
13 22 5.02 0 37 0
21 9 9.6 0 34 0
25 9 0 0 10 0
27 9 0 0 33 1
28 25 0 0 18 1
30 9 0 0 19 0
31 9 0.16 0 26 1
32 9 0 0 13 1
34 7 0 0 18 0
35 9 0 0 20 0
36 9 3.03 0 41 1
37 9 0 0 18 0
38 9 0 0 18 0
39 9 0 0 17 0
40 9 0.14 0 23 0
41 9 0 0 10 0
42 9 0 0 8 0
43 9 2.45 0 11 0
45 9 0 1 12 0
46 9 0.16 1 17 0
49 9 0 1 22 0
50 9 0 0 15 0
51 9 0.05 1 16 0

How to combine rows into one row in TermDocumentMatrix?

Iam trying to combine rows into on row in TermDocumentMatrix
(I know every row represents each words)
ex) cabin, staff -> crews
Because 'cabin, staff and crew' mean samething,
Iam trying to combine rows which represent 'cabin, staff'
into one row which represent 'crew.
but, it doesn't work at all.
R said argument "weighting" is missing, with no default
The codes I typed is below
r=GET('http://www.airlinequality.com/airline-reviews/cathay-pacific-airways/')
base_url=('http://www.airlinequality.com/airline-reviews/cathay-pacific-airways/')
h<-read_html(base_url)
all.reviews = c()
for (i in 1:10){
print(i)
url = paste(base_url, 'page/', i, '/', sep="")
r = GET(url)
h = read_html(r)
comment_area = html_nodes(h, '.tc_mobile')
comments= html_nodes(comment_area, '.text_content')
reviews = html_text(comments)
all.reviews=c(all.reviews, reviews)}
cps <- Corpus(VectorSource(all.reviews))
cps <- tm_map(cps, content_transformer(tolower))
cps <- tm_map(cps, content_transformer(stripWhitespace))
cps <- tm_map(cps, content_transformer(removePunctuation))
cps <- tm_map(cps, content_transformer(removeNumbers))
cps <- tm_map(cps, removeWords, stopwords("english"))
tdm <- TermDocumentMatrix(cps, control=list(
wordLengths=c(3, 20),
weighting=weightTf))
rows.cabin = grep('cabin|staff', row.names(tdm))
rows.cabin
# [1] 235 1594
count.cabin = as.array(rollup(tdm[rows.cabin,], 1))
count.cabin
#Docs
#Terms 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
#1 0 1 1 0 0 2 2 0 0 1 1 0 4 0 1 0 1 0 2 1 0 0 1 3 1 4 2 0 3 0 1 1 4 0 0 2 1 0 0 2 1 0 2 1 3 3 1
#Docs
#Terms 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91
#1 0 1 0 1 2 3 2 2 1 1 0 2 0 0 0 0 0 2 0 1 0 0 4 0 2 2 1 3 1 1 1 1 0 0 0 5 3 0 2 1 0 1 0 0
#Docs
#Terms 92 93 94 95 96 97 98 99 100
#1 1 5 2 1 0 0 0 1 0
row.crews = grep('crews', row.names(tdm))
row.crews
#[1] 408
tdm[row.crews,] = count.cabin
rows.cabin = setdiff(rows.cabin, row.crews) # ok
tdm = tdm[-rows.cabin,] # ok
dtm = as.DocumentTermMatrix(tdm)
# Error in .TermDocumentMatrix(t(x), weighting) :
# argument "weighting" is missing, with no default
maybe it is not right approach to combine rows in TermDocumentMatrix
Please fix this codes or suggest better approach to solve this problem.
Thanks in advance.
Hmm I wonder why you stick to your approach, which obviously does not work, instead of just copying+pasting+adjusting* my suggestion from here?
library(tm)
library(httr)
library(rvest)
library(slam)
# [...] # your code
inspect(tdm[grep("cabin|staff|crew", Terms(tdm), ignore.case=TRUE), 1:15])
# Docs
# Terms 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
# cabin 0 0 0 0 0 1 1 0 0 1 0 0 3 0 0
# crew 0 0 0 1 1 1 1 0 2 1 0 1 0 2 0
# crews 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
# staff 0 1 1 0 0 1 1 0 0 0 1 0 1 0 1
dict <- list(
"CREW" = grep("cabin|staff|crew", Terms(tdm), ignore.case=TRUE, value = TRUE)
)
terms <- Terms(tdm)
for (x in seq_along(dict))
terms[terms %in% dict[[x]] ] <- names(dict)[x]
tdm <- slam::rollup(tdm, 1, terms, sum)
inspect(tdm[grep("cabin|staff|crew", Terms(tdm), ignore.case=TRUE), 1:15])
# Docs
# Terms 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
# CREW 0 1 1 1 1 3 3 0 2 2 1 1 4 2 1
*I only adjusted the line inside the dict definition...

How to replace values from a matrix with another matrix based on column/row names?

I have a small matrix:
SMALL<-matrix(c(1:9),3, 3)
colnames(SMALL)<-c("25","36","48")
rownames(SMALL)<-c("18","25","48")
looks like:
25 36 48
18 1 4 7
25 2 5 8
48 3 6 9
And a large matrix:
LARGE<-matrix(0,4, 4)
colnames(LARGE)<-c("12","25","36","48")
rownames(LARGE)<-c("18","25","38","48")
looks like:
12 25 36 48
18 0 0 0 0
25 0 0 0 0
38 0 0 0 0
48 0 0 0 0
I would like to replace values from the large matrix by those from the small one based on the column/row names.
Looking for this result:
12 25 36 48
18 0 1 4 7
25 0 2 5 8
38 0 0 0 0
48 0 3 6 9
Any ideas ?
Assuming there is a match for each col and row name of SMALL in LARGE:
i <- match(rownames(SMALL), rownames(LARGE))
j <- match(colnames(SMALL), colnames(LARGE))
LARGE[i,j] <- SMALL
# 12 25 36 48
#18 0 1 4 7
#25 0 2 5 8
#38 0 0 0 0
#48 0 3 6 9

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