I have been trying to assign numbers with sequence. I would like to further add to repeat the sequence until the certain row numbers reached. For example repeat the sequence for every 44th row.
我一直试图用序列分配数字。我想进一步添加重复序列,直到达到某些行号。例如,每隔44行重复一次序列。
Here is what I mean
这就是我的意思
test_table <- data.frame(col=rep(0:10,each=11), row=c(rev(0:10)))
and assigning cumulative numbers in this way
并以这种方式分配累积数字
> library(dplyr)
test_table%>%
mutate(No=(row_number() - 1) %/% 11)
test_table
col row No
1 0 10 0
2 0 9 0
3 0 8 0
4 0 7 0
5 0 6 0
6 0 5 0
7 0 4 0
8 0 3 0
9 0 2 0
10 0 1 0
11 0 0 0
12 1 10 1
13 1 9 1
14 1 8 1
15 1 7 1
16 1 6 1
17 1 5 1
18 1 4 1
19 1 3 1
20 1 2 1
21 1 1 1
22 1 0 1
23 2 10 2
24 2 9 2
25 2 8 2
26 2 7 2
27 2 6 2
28 2 5 2
29 2 4 2
30 2 3 2
31 2 2 2
32 2 1 2
33 2 0 2
34 3 10 3
35 3 9 3
36 3 8 3
37 3 7 3
38 3 6 3
39 3 5 3
40 3 4 3
41 3 3 3
42 3 2 3
43 3 1 3
44 3 0 3
45 4 10 4
46 4 9 4
47 4 8 4
48 4 7 4
49 4 6 4
50 4 5 4
51 4 4 4
52 4 3 4
53 4 2 4
54 4 1 4
55 4 0 4
56 5 10 5
57 5 9 5
58 5 8 5
59 5 7 5
60 5 6 5
61 5 5 5
62 5 4 5
63 5 3 5
64 5 2 5
65 5 1 5
66 5 0 5
67 6 10 6
68 6 9 6
69 6 8 6
70 6 7 6
71 6 6 6
72 6 5 6
73 6 4 6
74 6 3 6
75 6 2 6
76 6 1 6
77 6 0 6
78 7 10 7
79 7 9 7
80 7 8 7
81 7 7 7
82 7 6 7
83 7 5 7
84 7 4 7
85 7 3 7
86 7 2 7
87 7 1 7
88 7 0 7
89 8 10 8
90 8 9 8
91 8 8 8
92 8 7 8
93 8 6 8
94 8 5 8
95 8 4 8
96 8 3 8
97 8 2 8
98 8 1 8
99 8 0 8
100 9 10 9
101 9 9 9
102 9 8 9
103 9 7 9
104 9 6 9
105 9 5 9
106 9 4 9
107 9 3 9
108 9 2 9
109 9 1 9
110 9 0 9
111 10 10 10
112 10 9 10
113 10 8 10
114 10 7 10
115 10 6 10
116 10 5 10
117 10 4 10
118 10 3 10
119 10 2 10
120 10 1 10
121 10 0 10
Ok. Good! But I would like to keep the sequence for example 0 and 1 until the 44th row reached. After that, start to the new sequence from 2 and go 88th row like this.
好。好!但是我希望将序列保持为0和1,直到达到第44行。之后,从2开始到新序列并像这样进入第88行。
So the expected output will be
所以预期的产出将是
test_table
col row No
1 0 10 0
2 0 9 0
3 0 8 0
4 0 7 0
5 0 6 0
6 0 5 0
7 0 4 0
8 0 3 0
9 0 2 0
10 0 1 0
11 0 0 0
12 1 10 1
13 1 9 1
14 1 8 1
15 1 7 1
16 1 6 1
17 1 5 1
18 1 4 1
19 1 3 1
20 1 2 1
21 1 1 1
22 1 0 1
23 2 10 0
24 2 9 0
25 2 8 0
26 2 7 0
27 2 6 0
28 2 5 0
29 2 4 0
30 2 3 0
31 2 2 0
32 2 1 0
33 2 0 0
34 3 10 1
35 3 9 1
36 3 8 1
37 3 7 1
38 3 6 1
39 3 5 1
40 3 4 1
41 3 3 1
42 3 2 1
43 3 1 1
44 3 0 1
45 4 10 2
46 4 9 2
47 4 8 2
48 4 7 2
49 4 6 2
50 4 5 2
51 4 4 2
52 4 3 2
53 4 2 2
54 4 1 2
55 4 0 2
56 5 10 3
57 5 9 3
58 5 8 3
59 5 7 3
60 5 6 3
61 5 5 3
62 5 4 3
63 5 3 3
64 5 2 3
65 5 1 3
66 5 0 3
67 6 10 2
68 6 9 2
69 6 8 2
70 6 7 2
71 6 6 2
72 6 5 2
73 6 4 2
74 6 3 2
75 6 2 2
76 6 1 2
77 6 0 2
78 7 10 3
79 7 9 3
80 7 8 3
81 7 7 3
82 7 6 3
83 7 5 3
84 7 4 3
85 7 3 3
86 7 2 3
87 7 1 3
88 7 0 3
89 8 10 4
90 8 9 4
91 8 8 4
92 8 7 4
93 8 6 4
94 8 5 4
95 8 4 4
96 8 3 4
97 8 2 4
98 8 1 4
99 8 0 4
100 9 10 5
101 9 9 5
102 9 8 5
103 9 7 5
104 9 6 5
105 9 5 5
106 9 4 5
107 9 3 5
108 9 2 5
109 9 1 5
110 9 0 5
111 10 10 4
112 10 9 4
113 10 8 4
114 10 7 4
115 10 6 4
116 10 5 4
117 10 4 4
118 10 3 4
119 10 2 4
120 10 1 4
121 10 0 4
How can we do that ? Thanks in advance!
我们怎么能这样做?提前致谢!
3 个解决方案
#1
1
This would do it in more general way
这将以更一般的方式进行
num.seq = 11L # total number of sequences in the first column
num.rows = N * num.seq # total number of rows
seq.length.3 = 44 # length of the pattern in the 3rd column
# number of paterns in the 3rd column
num.seq.3 = ( num.rows - 1 ) %/% seq.length.3 +1
# starting number in the sequence of the 3rd column
nseq=0
# vector for the 3rd column (could be done right in data frame def.)
No = (rep(rep( nseq:(nseq+1), each = N, times= 2), times=num.seq.3) +
rep(0:(num.seq.3 -1)*2, each= seq.length.3)) [1:num.rows]
test_table <- data.frame(col=rep(0:10,each=11),
row=c(rev(0:10)),
No=No)
An alternative way:
另一种方式:
library(dplyr)
dt2 <- test_table%>%
mutate(No = (row_number() - 1) %/% 11)
dt2$No <- dt2$No %% 2 + (rep(0:num.seq.3, each =44, times=num.seq.3 )*2)
[1:num.rows]
#2
1
The arithmetic, which is totally dependent on row numbers, seems right this way.
算术完全依赖于行号,这种方式似乎是正确的。
test_table%>%
mutate(No=((row_number() - 1) %/% 11) %% 2) %>% # alternating 11 rows of 0's and 1's
mutate(No = No + ((row_number() - 1) %/% 44) * 2) # add 2 every after 44 rows
Here is the result, as intended.
这是预期的结果。
structure(list(col = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L,
8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L,
10L, 10L), row = c(10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L,
10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L,
6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L,
2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L,
9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L,
5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L,
1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L,
8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L,
4L, 3L, 2L, 1L, 0L), No = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4)), class = "data.frame", .Names = c("col", "row",
"No"), row.names = c(NA, -121L))
#3
0
This would deliver what I understand to be the requested vector (except I think your sequencing"skipped a beat"):
这将提供我理解为请求的向量(除了我认为您的排序“跳过一个节拍”):
c( rep( c(1,2,1,2), each=11) , rep(c(3,4,3,4), each=11), rep(c(5,6,5,6), each=11) )
[1] 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 3 3 3
[48] 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5
[95] 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6
A more general way:
更通用的方式:
c( sapply( seq(1, 6, by=2), function(start) {
rep( rep(start:(start+1) , 2), each=11) }))
The outer c() will remove the matrix character that sapply defaults to.
外部c()将删除sapply默认的矩阵字符。
#1
1
This would do it in more general way
这将以更一般的方式进行
num.seq = 11L # total number of sequences in the first column
num.rows = N * num.seq # total number of rows
seq.length.3 = 44 # length of the pattern in the 3rd column
# number of paterns in the 3rd column
num.seq.3 = ( num.rows - 1 ) %/% seq.length.3 +1
# starting number in the sequence of the 3rd column
nseq=0
# vector for the 3rd column (could be done right in data frame def.)
No = (rep(rep( nseq:(nseq+1), each = N, times= 2), times=num.seq.3) +
rep(0:(num.seq.3 -1)*2, each= seq.length.3)) [1:num.rows]
test_table <- data.frame(col=rep(0:10,each=11),
row=c(rev(0:10)),
No=No)
An alternative way:
另一种方式:
library(dplyr)
dt2 <- test_table%>%
mutate(No = (row_number() - 1) %/% 11)
dt2$No <- dt2$No %% 2 + (rep(0:num.seq.3, each =44, times=num.seq.3 )*2)
[1:num.rows]
#2
1
The arithmetic, which is totally dependent on row numbers, seems right this way.
算术完全依赖于行号,这种方式似乎是正确的。
test_table%>%
mutate(No=((row_number() - 1) %/% 11) %% 2) %>% # alternating 11 rows of 0's and 1's
mutate(No = No + ((row_number() - 1) %/% 44) * 2) # add 2 every after 44 rows
Here is the result, as intended.
这是预期的结果。
structure(list(col = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L,
8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L,
10L, 10L), row = c(10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L,
10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L,
6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L,
2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L,
9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L,
5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L,
1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L,
8L, 7L, 6L, 5L, 4L, 3L, 2L, 1L, 0L, 10L, 9L, 8L, 7L, 6L, 5L,
4L, 3L, 2L, 1L, 0L), No = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4)), class = "data.frame", .Names = c("col", "row",
"No"), row.names = c(NA, -121L))
#3
0
This would deliver what I understand to be the requested vector (except I think your sequencing"skipped a beat"):
这将提供我理解为请求的向量(除了我认为您的排序“跳过一个节拍”):
c( rep( c(1,2,1,2), each=11) , rep(c(3,4,3,4), each=11), rep(c(5,6,5,6), each=11) )
[1] 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 3 3 3
[48] 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5
[95] 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6
A more general way:
更通用的方式:
c( sapply( seq(1, 6, by=2), function(start) {
rep( rep(start:(start+1) , 2), each=11) }))
The outer c() will remove the matrix character that sapply defaults to.
外部c()将删除sapply默认的矩阵字符。