I'm looking to recode a column, say the following:
df <- data.frame(col1 = rep(3, 100),
col2 = rep(NA, 100))
I want to recode col2 as 1 for rows 1:5, 2 for rows 6:10, 3 for 11:15, etc. So, every five rows I would add +1 to the assigned value. Any way to automate this process to avoid manually recoding 100 rows?
There are lot of ways to do that. Here are couple of them -
Using rep :
df$col2 <- rep(1:nrow(df), each = 5, length.out = nrow(df))
Using ceiling
df$col2 <- ceiling(seq(nrow(df))/5)
dplyr way
df %>% mutate(col2 = ((row_number()-1) %/% 5)+1)
OR
A simple for loop
for(i in 0:((nrow(df)/5)-1)){
df[0:nrow(df) %/% 5 == i,2] <- i+1
}
> df
col1 col2
1 3 1
2 3 1
3 3 1
4 3 1
5 3 1
6 3 2
7 3 2
8 3 2
9 3 2
10 3 2
11 3 3
12 3 3
13 3 3
14 3 3
15 3 3
16 3 4
17 3 4
18 3 4
19 3 4
20 3 4
21 3 5
22 3 5
23 3 5
24 3 5
25 3 5
26 3 6
27 3 6
28 3 6
29 3 6
30 3 6
31 3 7
32 3 7
33 3 7
34 3 7
35 3 7
36 3 8
37 3 8
38 3 8
39 3 8
40 3 8
41 3 9
42 3 9
43 3 9
44 3 9
45 3 9
46 3 10
47 3 10
48 3 10
49 3 10
50 3 10
51 3 11
52 3 11
53 3 11
54 3 11
55 3 11
56 3 12
57 3 12
58 3 12
59 3 12
60 3 12
61 3 13
62 3 13
63 3 13
64 3 13
65 3 13
66 3 14
67 3 14
68 3 14
69 3 14
70 3 14
71 3 15
72 3 15
73 3 15
74 3 15
75 3 15
76 3 16
77 3 16
78 3 16
79 3 16
80 3 16
81 3 17
82 3 17
83 3 17
84 3 17
85 3 17
86 3 18
87 3 18
88 3 18
89 3 18
90 3 18
91 3 19
92 3 19
93 3 19
94 3 19
95 3 19
96 3 20
97 3 20
98 3 20
99 3 20
100 3 20
As there is a pattern (each 5th row) you can use rep(row_number()) length.out = n() takes into account the length of column.
Learned here dplyr: Mutate a new column with sequential repeated integers of n time in a dataframe from Ronak!!!
Thanks to Ronak!
df %>% mutate(col2 = rep(row_number(), each=5, length.out = n()))
Example data.frame:
df = read.table(text = 'colA colB
2 7
2 7
2 7
2 7
1 7
1 7
1 7
89 5
89 5
89 5
88 5
88 5
70 5
70 5
70 5
69 5
69 5
44 4
44 4
44 4
43 4
42 4
42 4
41 4
41 4
120 1
100 1', header = TRUE)
I need to add an index col based on colA and colB where colB shows the exact number of rows to group but it can be duplicated. colB groups rows based on colA and colA -1.
Expected output:
colA colB index_col
2 7 1
2 7 1
2 7 1
2 7 1
1 7 1
1 7 1
1 7 1
89 5 2
89 5 2
89 5 2
88 5 2
88 5 2
70 5 3
70 5 3
70 5 3
69 5 3
69 5 3
44 4 4
44 4 4
44 4 4
43 4 4
42 4 5
42 4 5
41 4 5
41 4 5
120 1 6
100 1 7
UPDATE
How can I adapt the code that works for the above df for the same purpose but by looking at colB values grouped based on colA, colA -1 and colA -2? i.e. (instead of 2 days considering 3 days)
new_df = read.table(text = 'colA colB
3 10
3 10
3 10
2 10
2 10
2 10
2 10
1 10
1 10
1 10
90 7
90 7
89 7
89 7
89 7
88 7
88 7
71 7
71 7
70 7
70 7
70 7
69 7
69 7
44 5
44 5
44 5
43 5
42 5
41 5
41 5
41 5
40 5
40 5
120 1
100 1', header = TRUE)
Expected output:
colA colB index_col
3 10 1
3 10 1
3 10 1
2 10 1
2 10 1
2 10 1
2 10 1
1 10 1
1 10 1
1 10 1
90 7 2
90 7 2
89 7 2
89 7 2
89 7 2
88 7 2
88 7 2
71 7 3
71 7 3
70 7 3
70 7 3
70 7 3
69 7 3
69 7 3
44 5 4
44 5 4
44 5 4
43 5 4
42 5 4
41 5 5
41 5 5
41 5 5
40 5 5
40 5 5
120 1 6
100 1 7
Thanks
We can use rleid
library(data.table)
index_col <-setDT(df)[, if(colB[1L] < .N) ((seq_len(.N)-1) %/% colB[1L])+1
else as.numeric(colB), rleid(colB)][, rleid(V1)]
df[, index_col := index_col]
df
# colA colB index_col
# 1: 2 7 1
# 2: 2 7 1
# 3: 2 7 1
# 4: 2 7 1
# 5: 1 7 1
# 6: 1 7 1
# 7: 1 7 1
# 8: 70 5 2
# 9: 70 5 2
#10: 70 5 2
#11: 69 5 2
#12: 69 5 2
#13: 89 5 3
#14: 89 5 3
#15: 89 5 3
#16: 88 5 3
#17: 88 5 3
#18: 120 1 4
#19: 100 1 5
Or a one-liner would be
setDT(df)[, index_col := df[, ((seq_len(.N)-1) %/% colB[1L])+1, rleid(colB)][, as.integer(interaction(.SD, drop = TRUE, lex.order = TRUE))]]
Update
Based on the new update in the OP's post
setDT(new_df)[, index_col := cumsum(c(TRUE, abs(diff(colA))> 1))
][, colB := .N , index_col]
new_df
# colA colB index_col
# 1: 3 10 1
# 2: 3 10 1
# 3: 3 10 1
# 4: 2 10 1
# 5: 2 10 1
# 6: 2 10 1
# 7: 2 10 1
# 8: 1 10 1
# 9: 1 10 1
#10: 1 10 1
#11: 71 7 2
#12: 71 7 2
#13: 70 7 2
#14: 70 7 2
#15: 70 7 2
#16: 69 7 2
#17: 69 7 2
#18: 90 7 3
#19: 90 7 3
#20: 89 7 3
#21: 89 7 3
#22: 89 7 3
#23: 88 7 3
#24: 88 7 3
#25: 44 2 4
#26: 43 2 4
#27: 120 1 5
#28: 100 1 6
An approach in base R:
df$idxcol <- cumsum(c(1,abs(diff(df$colA)) > 1) + c(0,diff(df$colB) != 0) > 0)
which gives:
> df
colA colB idxcol
1 2 7 1
2 2 7 1
3 2 7 1
4 2 7 1
5 1 7 1
6 1 7 1
7 1 7 1
8 70 5 2
9 70 5 2
10 70 5 2
11 69 5 2
12 69 5 2
13 89 5 3
14 89 5 3
15 89 5 3
16 88 5 3
17 88 5 3
18 120 1 4
19 100 1 5
On the updated example data, you need to adapt the approach to:
n <- 1
idx1 <- cumsum(c(1, diff(df$colA) < -n) + c(0, diff(df$colB) != 0) > 0)
idx2 <- ave(df$colA, cumsum(c(1, diff(df$colA) < -n)), FUN = function(x) c(0, cumsum(diff(x)) < -n ))
idx2[idx2==1 & c(0,diff(idx2))==0] <- 0
df$idxcol <- idx1 + cumsum(idx2)
which gives:
> df
colA colB idxcol
1 2 7 1
2 2 7 1
3 2 7 1
4 2 7 1
5 1 7 1
6 1 7 1
7 1 7 1
8 89 5 2
9 89 5 2
10 89 5 2
11 88 5 2
12 88 5 2
13 70 5 3
14 70 5 3
15 70 5 3
16 69 5 3
17 69 5 3
18 44 4 4
19 44 4 4
20 44 4 4
21 43 4 4
22 42 4 5
23 42 4 5
24 41 4 5
25 41 4 5
26 120 1 6
27 100 1 7
For new_df just change n tot 2 and you will get the desired output for that as well.
I have data that look something like this:
id <- c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,8,8,8,9,9,9)
yr <- c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3)
gr <- c(3,4,5,3,4,5,3,4,5,4,5,6,4,5,6,4,5,6,5,6,7,5,6,7,5,6,7)
x <- c(33,48,31,41,31,36,25,38,28,17,39,53,60,60,19,39,34,47,20,28,38,15,17,49,48,45,39)
df <- data.frame(id,yr,gr,x)
id yr gr x
1 1 1 3 33
2 1 2 4 48
3 1 3 5 31
4 2 1 3 41
5 2 2 4 31
6 2 3 5 36
7 3 1 3 25
8 3 2 4 38
9 3 3 5 28
10 4 1 4 17
11 4 2 5 39
12 4 3 6 53
13 5 1 4 60
14 5 2 5 60
15 5 3 6 19
16 6 1 4 39
17 6 2 5 34
18 6 3 6 47
19 7 1 5 20
20 7 2 6 28
21 7 3 7 38
22 8 1 5 15
23 8 2 6 17
24 8 3 7 49
25 9 1 5 48
26 9 2 6 45
27 9 3 7 39
I would like to create a new variable in the data frame that contains the quantiles of "x" computed within each unique combination of "yr" and "gr". That is, rather than finding the quantiles of "x" based on all 27 rows of data in the example, I would like to compute the quantiles by two grouping variables: yr and gr. For instance, the quantiles of "x" when yr = 1 and gr = 3, yr = 1 and gr = 4, etc.
Once these values are computed, I would like them to be appended to the data frame as a single column, say "x_quant".
I am able to split the data into the separate groups I need, and I am know how to calculate quantiles, but I am having trouble combining the two steps in a way that is amenable to creating a new column in the existing data frame.
Any help y'all can provide would be greatly appretiated! Thank you much!
~kj
# turn "yr" and "gr" into sortable column
df$y <- paste(df$yr,"",df$gr)
df.ordered <- df[order(df$y),] #sort df based on group
grp <- split(df.ordered,df.ordered$y);grp
# get quantiles and turn results into string
q <- vector('list')
for (i in 1:length(grp)) {
a <- quantile(grp[[i]]$x)
q[i] <- paste(a[1],"",a[2],"",a[3],"",a[4],"",a[5])
}
x_quant <- unlist(sapply(q, `[`, 1))
x_quant <- rep(x_quant,each=3)
# append quantile back to data frame. Gave new column a more descriptive name
df.ordered$xq_0_25_50_75_100 <- x_quant
df.ordered$y <- NULL
df <- df.ordered;df </pre>
Output:
> # turn "yr" and "gr" into sortable column
> df$y <- paste(df$yr,"",df$gr)
> df.ordered <- df[order(df$y),] #sort df based on group
> grp <- split(df.ordered,df.ordered$y);grp
$`1 3`
id yr gr x y
1 1 1 3 33 1 3
4 2 1 3 41 1 3
7 3 1 3 25 1 3
$`1 4`
id yr gr x y
10 4 1 4 17 1 4
13 5 1 4 60 1 4
16 6 1 4 39 1 4
$`1 5`
id yr gr x y
19 7 1 5 20 1 5
22 8 1 5 15 1 5
25 9 1 5 48 1 5
$`2 4`
id yr gr x y
2 1 2 4 48 2 4
5 2 2 4 31 2 4
8 3 2 4 38 2 4
$`2 5`
id yr gr x y
11 4 2 5 39 2 5
14 5 2 5 60 2 5
17 6 2 5 34 2 5
$`2 6`
id yr gr x y
20 7 2 6 28 2 6
23 8 2 6 17 2 6
26 9 2 6 45 2 6
$`3 5`
id yr gr x y
3 1 3 5 31 3 5
6 2 3 5 36 3 5
9 3 3 5 28 3 5
$`3 6`
id yr gr x y
12 4 3 6 53 3 6
15 5 3 6 19 3 6
18 6 3 6 47 3 6
$`3 7`
id yr gr x y
21 7 3 7 38 3 7
24 8 3 7 49 3 7
27 9 3 7 39 3 7
> # get quantiles and turn results into string
> q <- vector('list')
> for (i in 1:length(grp)) {
+ a <- quantile(grp[[i]]$x)
+ q[i] <- paste(a[1],"",a[2],"",a[3],"",a[4],"",a[5])
+ }
> x_quant <- unlist(sapply(q, `[`, 1))
> x_quant <- rep(x_quant,each=3)
> # append quantile back to data frame
> df.ordered$xq_0_25_50_75_100 <- x_quant
> df.ordered$y <- NULL
> df <- df.ordered
> df
id yr gr x xq_0_25_50_75_100
1 1 1 3 33 25 29 33 37 41
4 2 1 3 41 25 29 33 37 41
7 3 1 3 25 25 29 33 37 41
10 4 1 4 17 17 28 39 49.5 60
13 5 1 4 60 17 28 39 49.5 60
16 6 1 4 39 17 28 39 49.5 60
19 7 1 5 20 15 17.5 20 34 48
22 8 1 5 15 15 17.5 20 34 48
25 9 1 5 48 15 17.5 20 34 48
2 1 2 4 48 31 34.5 38 43 48
5 2 2 4 31 31 34.5 38 43 48
8 3 2 4 38 31 34.5 38 43 48
11 4 2 5 39 34 36.5 39 49.5 60
14 5 2 5 60 34 36.5 39 49.5 60
17 6 2 5 34 34 36.5 39 49.5 60
20 7 2 6 28 17 22.5 28 36.5 45
23 8 2 6 17 17 22.5 28 36.5 45
26 9 2 6 45 17 22.5 28 36.5 45
3 1 3 5 31 28 29.5 31 33.5 36
6 2 3 5 36 28 29.5 31 33.5 36
9 3 3 5 28 28 29.5 31 33.5 36
12 4 3 6 53 19 33 47 50 53
15 5 3 6 19 19 33 47 50 53
18 6 3 6 47 19 33 47 50 53
21 7 3 7 38 38 38.5 39 44 49
24 8 3 7 49 38 38.5 39 44 49
27 9 3 7 39 38 38.5 39 44 49
>
I was wondering if there is a way to produce a decision tree that solves a permutation of selecting n objects of k classes. We have the set A={1,2,...,10}, and the subsets B={1,2,..,5}, C={6,7} and D={8,9,10}. The total number of permutations can be calculated by
x <- factorial(10)/(factorial(5)*factorial(2)*factorial(3))
I would like to produce a decision tree similar to an edge list, as the following:
1 2 5 B
1 3 2 C
1 4 3 D
2 5 4 B
2 6 2 C
2 7 3 D
3 8 5 B
3 9 1 C
3 10 3 D
4 11 5 B
4 12 2 C
4 13 2 D
5 14 3 B
5 15 2 C
5 16 3 D
6 17 4 B
6 18 1 C
6 19 3 D
7 20 4 B
7 21 2 C
7 22 2 D
8 23 4 B
8 24 1 C
8 25 3 D
9 26 5 B
9 27 3 D
10 28 5 B
10 29 1 C
10 30 2 D
11 31 4 B
11 32 2 C
11 33 2 D
12 34 5 B
12 35 1 C
12 36 2 D
13 37 5 B
13 38 2 C
13 39 1 D
14 40 2 B
14 41 2 C
14 42 3 D
15 43 3 B
15 44 1 C
15 45 3 D
16 46 3 B
16 47 2 C
16 48 2 D
17 49 3 B
17 50 1 C
17 51 3 D
18 52 4 B
18 53 3 D
19 54 4 B
19 55 1 C
19 56 2 D
20 57 3 B
20 58 2 C
20 59 2 D
21 60 4 B
21 61 1 C
21 62 2 D
22 63 4 B
22 64 2 C
22 65 1 D
23 66 3 B
23 67 1 C
23 68 3 D
24 69 4 B
24 70 3 D
25 71 4 B
25 72 1 C
25 73 2 D
26 74 4 B
26 75 3 D
27 76 5 B
27 77 2 D
28 78 4 B
28 79 1 C
28 80 2 D
29 81 5 B
29 82 2 D
30 83 5 B
30 84 1 C
30 85 1 D
31 86 3 B
31 87 2 C
31 88 2 D
32 89 4 B
32 90 1 C
32 91 2 D
33 92 4 B
33 93 2 C
33 94 1 D
34 95 4 B
34 96 1 C
34 97 2 D
. . . .
. . . .
. . . .
The first two columns correspond to the edge list, the third column is the number of elements in each subset decreasing by each ramification and the fourth column is just the subset name.
One computed the edge list, I'm thinking on plotting the graph with this command:
plot(g, layout = layout.reingold.tilford(g, root=1)
Basically I want an autoincremented id column based on my cohorts - in this case .(kmer, cvCut)
> myDataFrame
size kmer cvCut cumsum
1 8132 23 10 8132
10000 778 23 10 13789274
30000 324 23 10 23658740
50000 182 23 10 28534840
100000 65 23 10 33943283
200000 25 23 10 37954383
250000 584 23 12 16546507
300000 110 23 12 29435303
400000 28 23 12 34697860
600000 127 23 2 47124443
600001 127 23 2 47124570
I want a column added that has new row names based on the kmer/cvCut group
> myDataFrame
size kmer cvCut cumsum newID
1 8132 23 10 8132 1
10000 778 23 10 13789274 2
30000 324 23 10 23658740 3
50000 182 23 10 28534840 4
100000 65 23 10 33943283 5
200000 25 23 10 37954383 6
250000 584 23 12 16546507 1
300000 110 23 12 29435303 2
400000 28 23 12 34697860 3
600000 127 23 2 47124443 1
600001 127 23 2 47124570 2
I'd do it like this:
library(plyr)
ddply(df, c("kmer", "cvCut"), transform, newID = seq_along(kmer))
Just add a new column each time plyr calls you:
R> DF <- data.frame(kmer=sample(1:3, 50, replace=TRUE), \
cvCut=sample(LETTERS[1:3], 50, replace=TRUE))
R> library(plyr)
R> ddply(DF, .(kmer, cvCut), function(X) data.frame(X, newId=1:nrow(X)))
kmer cvCut newId
1 1 A 1
2 1 A 2
3 1 A 3
4 1 A 4
5 1 A 5
6 1 A 6
7 1 A 7
8 1 A 8
9 1 A 9
10 1 A 10
11 1 A 11
12 1 B 1
13 1 B 2
14 1 B 3
15 1 B 4
16 1 B 5
17 1 B 6
18 1 C 1
19 1 C 2
20 1 C 3
21 2 A 1
22 2 A 2
23 2 A 3
24 2 A 4
25 2 A 5
26 2 B 1
27 2 B 2
28 2 B 3
29 2 B 4
30 2 B 5
31 2 B 6
32 2 B 7
33 2 C 1
34 2 C 2
35 2 C 3
36 2 C 4
37 3 A 1
38 3 A 2
39 3 A 3
40 3 A 4
41 3 B 1
42 3 B 2
43 3 B 3
44 3 B 4
45 3 C 1
46 3 C 2
47 3 C 3
48 3 C 4
49 3 C 5
50 3 C 6
R>
I think that this is what you want:
Load the data:
x <- read.table(textConnection(
"id size kmer cvCut cumsum
1 8132 23 10 8132
10000 778 23 10 13789274
30000 324 23 10 23658740
50000 182 23 10 28534840
100000 65 23 10 33943283
200000 25 23 10 37954383
250000 584 23 12 16546507
300000 110 23 12 29435303
400000 28 23 12 34697860
600000 127 23 2 47124443
600001 127 23 2 47124570"), header=TRUE)
Use ddply:
library(plyr)
ddply(x, .(kmer, cvCut), function(x) cbind(x, 1:nrow(x)))