How can I do this : { [5] [6] [7] [8:23] [24:39] [40:55] } --- these number are just an example

2015 12 22
2 个回答
更新时间：2015 12 22
1 次查看（30 天）

0 个投票

Hello,
I have two arrays a = [1 2 3 4 5 5] and b = [6 7 5 8 9]. I need to write them in CAN bit packing block with these Format
c = { [5] [6] [7] [8:23] [24:39] [40:55] } ( Number are just an example).
c is the combination between a and b. I've tried with this code :
c = unique([a b]);
strc = num2str(c);
splitc = strsplit(strc);
joinc = strjoin(splitc, ':');
strjoin ={joinc};
and it was wrong . At least how can I add a delimiter such as '[' and ']' in one element in my array? e.g. a = [1 2 3]
a1 =[1]--( it has to be string). Thanks for the help! :)

4 个评论
显示 2更早的评论 隐藏 2更早的评论

Image Analyst
Image Analyst 2015-12-22
What is "CAN bit packing block"? And how did you get those numbers like 24, 39, 40, and 55? And do you want c to be a cell a cell array?
jgg
jgg 2015-12-22
I think the best thing to do would be to show us what the output should look like for your example. If I understand your problem correctly it should be {[1:9]}, correct?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（2 个）

Renato Agurto
Renato Agurto 2015-12-22

0 个投票

Do you want something like this?
['[' strc(1) ':' strc(4) ']']

2 个评论
显示 无 隐藏 无

yeah, can you tell me how to implements that stuff in this case : a1 = [0 32]; c1 = [31 63]; Result that I attempt : { [0:31] [32:63]} --- This one is string. Thanks in advance :D
one thing, It should be automatic cause the values of a1 and c1 could varies ,e.g. a1 = [0 8 16]; c1 =[7 8 16]; then the result should be : {[0:7] [8] [16]}

请先登录，再进行评论。

Guillaume
Guillaume 2015-12-22
编辑：Guillaume 2015-12-22

0 个投票

First, learn to read the documentation. The documentation that you linked clearly says that: "The data type [...] must be a MATLAB® cell array vector. [...] The cell array elements must be of type double array". There are no strings involved in this.
Secondly, I'm not even sure you've understood what the can bit packing function does. The bit-pattern is not the same at all as the data you want to transmit. There are many bit patterns that could be valid for transmitting a = [1 2 3 4 5 5]. The following bit-pattern is a possiblity: c = {0, 1:2, 3:4, 5:7, 8:10, 11:13)}, that's the minimum pattern to transmit a but will not transmit b = [2 2 3 4 5 5] as there is not enough bits to transmit the first number. Equally valid pattern is: c = {0:9, 10:19, 20:30, 31:41, 42:52, 53:63}.
The bit-pattern is defined by:
  • how many numbers you want to pack into a double
  • the maximum value of each of these numbers
Given the maximum values you want to transmit
maxvalues = [2 2 6 4 10 8]; %for example.
You can calculate the smallest bit-pattern with:
bitsrequired = floor(log2(maxvalues)) + 1;
endbits = cumsum(bitsrequired) - 1;
assert(endbits(end) <= 63, 'more bits are required than can fit into a double');
bitpattern = arrayfun(@(s, e) (s:e), [0, endbits(1:end-1)+1], endbits, 'UniformOutput', false)

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Guillaume
Guillaume 2015-12-22
Following onto the comment you wrote to Renato's answer while I was typing mine, if given inputs
startbits = [0 8 16];
endbits = [7 8 16];
To generate the corresponding bit pattern, you'd simply do:
bitpattern = arrayfun(@(s, e) s:e, startbits, endbits, 'UniformOutput', false);

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Downloads 的更多信息

产品

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by