Base 16 should go from 0 to F, with F being equal to 15 in base 10. But yet, when I use a base 16 converter found on google (https://www.hexator.com/) , it says that F is equal to 46.
Expected results:
0 | 0
1 | 1
2 | 2
3 | 3
4 | 4
5 | 5
6 | 6
7 | 7
8 | 8
9 | 9
a | 10
b | 11
c | 12
d | 13
e | 14
f | 15
Am I miss-interpreting something here?
That encoder is converting the ASCII value of the letter 'F' into the hexadecimal representation of it. The ASCII value of 'F' is 70, which is 46 when converted into hexadecimal. See this ascii table.
That converter is converting text into its hex representation, not Hex strings into decimal numbers.
Related
Let's say I have a table:
m:([] t: raze 3#'(2021.01.04+til 5); sym:15#`A`B`C; c: til 15)
t sym c
-----------------
2021.01.04 A 0
2021.01.04 B 1
2021.01.04 C 2
2021.01.05 A 3
2021.01.05 B 4
When I try to pivot it:
exec t!c by sym:sym from m
sym|
---| -----------------------------------------------------------------
A | 2021.01.04 2021.01.05 2021.01.06 2021.01.07 2021.01.08!0 3 6 9 12
B | 2021.01.04 2021.01.05 2021.01.06 2021.01.07 2021.01.08!1 4 7 10 13
C | 2021.01.04 2021.01.05 2021.01.06 2021.01.07 2021.01.08!2 5 8 11 14
I'd expect to get a table back, with columns sym, but I don't. What am I doing wrong?
if you're after a pivot with columns of sym you would want the following:
q)exec sym!c by t:t from m
t | A B C
----------| --------
2021.01.04| 0 1 2
2021.01.05| 3 4 5
2021.01.06| 6 7 8
2021.01.07| 9 10 11
2021.01.08| 12 13 14
It's because your column names have to be symbols:
q)exec(`$string t)!c by sym:sym from m
sym| 2021.01.04 2021.01.05 2021.01.06 2021.01.07 2021.01.08
---| ------------------------------------------------------
A | 0 3 6 9 12
B | 1 4 7 10 13
C | 2 5 8 11 14
These would be terrible column names though, so I would use .Q.id
q).Q.id exec(`$string t)!c by sym:sym from m
sym| a20210104 a20210105 a20210106 a20210107 a20210108
---| -------------------------------------------------
A | 0 3 6 9 12
B | 1 4 7 10 13
C | 2 5 8 11 14
It sounds like this isn't what you actually want though, so maybe Matthews answer is more relevant. My answer just explains why it didn't look like what you thought
I have two columns in my InfluxDB database : Values and Iterator count
I want visualise this on Grafana where my x axis is iterator count and value on y axis is basically corresponding to each iterator count.
EXAMPLE
Iterator Count(X) | Value
1 | 46
2 | 64
3 | 32
4 | 13
5 | 12
6 | 11
7 | 10
8 | 9
9 | 12
10 | 25.
Is it possible to achieve visualisation for the same, having no aspect of time
You can use plot.ly plugin
You just need to specify Iterator Count(X) as the x-axis in the trace section and Value as the y-axis.
I have some question related the RISC V manual
It has different types of instruction encoding such as R-type,I-type.
Just like the MIPS encoding.
* R-type
31 25 24 20 19 15 14 12 11 7 6 0
+------------+---------+---------+------+---------+-------------+
| funct7 | rs2 | rs1 |funct3| rd | opcode |
+------------+---------+---------+------+---------+-------------+
* I-type
31 20 19 15 14 12 11 7 6 0
+----------------------+---------+------+---------+-------------+
| imm | rs1 |funct3| rd | opcode |
+----------------------+---------+------+---------+-------------+
* S-type
31 25 24 20 19 15 14 12 11 7 6 0
+------------+---------+---------+------+---------+-------------+
| imm | rs2 | rs1 |funct3| imm | opcode |
+------------+---------+---------+------+---------+-------------+
* U-type
31 11 7 6 0
+---------------------------------------+---------+-------------+
| imm | rd | opcode |
+---------------------------------------+---------+-------------+
But it also have something called immediate format:
such as I-immediate, S-immediate and so on
* I-immediate
31 10 5 4 1 0
+-----------------------------------------+-----------+-------+--+
| <-- 31 | 30:25 | 24:21 |20|
+-----------------------------------------+-----------+-------+--+
* S-immediate
31 10 5 4 1 0
+-----------------------------------------+-----------+-------+--+
| <-- 31 | 30:25 | 11:8 |7 |
+-----------------------------------------+-----------+-------+--+
* B-immediate
31 12 11 10 5 4 1 0
+--------------------------------------+--+-----------+-------+--+
| <-- 31 |7 | 30:25 | 11:8 |z |
+--------------------------------------+--+-----------+-------+--+
* U-immediate
31 30 20 19 12 11 0
+--+-------------------+---------------+-------------------------+
|31| 30:20 | 19:12 | <-- z |
+--+-------------------+---------------+-------------------------+
* J-immediate
31 20 19 12 11 10 5 4 1 0
+----------------------+---------------+--+-----------+-------+--+
| <-- 31 | 19:12 |20| 30:25 | 24:21 |z |
+----------------------+---------------+--+-----------+-------+--+
According to the manual, it say those immediate is produced by RISC-V instruction but how are the things related?
What is the point to have immediate format?
The 2nd set of diagrams is showing you how the immediate bits are concatenated and sign-extended into a 32-bit integer (so they can work as a source operand for normal 32-bit ALU instructions like addi which need both their inputs to be the same size).
For I-type instructions it's trivial, just arithmetic right-shift the instruction word by 20 bits, because there's only one immediate field, and it's contiguous at the top of the instruction word.
For S-type immediate instructions, there are two separate fields in the instruction word: [31:25] and [11:7], and this shows you that they're in that order, not [11:7, 31:25] and not with any implicit zeros between them.
B-type immediate instructions apparently put bit 7 in front of [30:25], and the low bit is an implicit zero. (So the resulting number is always even). I assume B-type is for branches.
U-type is also interesting, padding the 20-bit immediate with trailing zeros. It's used for lui to create the upper bits of 32-bit constants (with addi supplying the rest). It's not a coincidence that U-type and I-type together have 32 total immediate bits.
To access static data, lui can create the high part of an address while lw can supply the low part directly, instead of using an addi to create the full address in a register. This is typical for RISC ISAs like MIPS and PowerPC as well (see an example on the Godbolt compiler explorer). But unlike most other RISC ISAs, RISC-V has auipc which adds the U-type immediate to the program counter, for efficient PIC without having to load addresses from a GOT (global offset table). (A recent MIPS revision also added an add-to-PC instruction, but for a long time MIPS was quite bad at PIC).
lui can encode any 4k-aligned address, i.e. a page-start address with 4k pages.
In matlab, I want to fit a piecewise regression and find where on the x-axis the first change-point occurs. For example, for the following data, the output might be changepoint=20 (I don't actually want to plot it, just want the change point).
data = [1 4 4 3 4 0 0 4 5 4 5 2 5 10 5 1 4 15 4 9 11 16 23 25 24 17 31 42 35 45 49 54 74 69 63 46 35 31 27 15 10 5 10 4 2 4 2 2 3 5 2 2];
x = 1:52;
plot(x,data,'.')
If you have the Signal Processing Toolbox, you can directly use the findchangepts function (see https://www.mathworks.com/help/signal/ref/findchangepts.html for documentation):
data = [1 4 4 3 4 0 0 4 5 4 5 2 5 10 5 1 4 15 4 9 11 16 23 25 24 17 31 42 35 45 49 54 74 69 63 46 35 31 27 15 10 5 10 4 2 4 2 2 3 5 2 2];
x = 1:52;
ipt = findchangepts(data);
x_cp = x(ipt);
data_cp = data(ipt);
plot(x,data,'.',x_cp,data_cp,'o')
The index of the change point in this case is 22.
Plot of data and its change point circled in red:
I know this is an old question but just want to provide some extra thoughts. In Maltab, an alternative implemented by me is a Bayesian changepoint detection algorithm that estimates not just the number and locations of the changepoints but also reports the occurrence probability of changepoints. In its current implementation, it deals with only time-series-like data (aka, 1D sequential data). More info about the tool is available at this FileExchange entry (https://www.mathworks.com/matlabcentral/fileexchange/72515-bayesian-changepoint-detection-time-series-decomposition).
Here is its quick application to your sample data:
% Automatically install the Rbeast or BEAST library to local drive
eval(webread('http://b.link/beast')) %
data = [1 4 4 3 4 0 0 4 5 4 5 2 5 10 5 1 4 15 4 9 11 16 23 25 24 17 31 42 35 45 49 54 74 69 63 46 35 31 27 15 10 5 10 4 2 4 2 2 3 5 2 2];
out = beast(data, 'season','none') % season='none': there is no seasonal/periodic variation in the data
printbeast(out)
plotbeast(out)
Below is a summary of the changepoint, given by printbeast():
#####################################################################
# Trend Changepoints #
#####################################################################
.-------------------------------------------------------------------.
| Ascii plot of probability distribution for number of chgpts (ncp) |
.-------------------------------------------------------------------.
|Pr(ncp = 0 )=0.000|* |
|Pr(ncp = 1 )=0.000|* |
|Pr(ncp = 2 )=0.000|* |
|Pr(ncp = 3 )=0.859|*********************************************** |
|Pr(ncp = 4 )=0.133|******** |
|Pr(ncp = 5 )=0.008|* |
|Pr(ncp = 6 )=0.000|* |
|Pr(ncp = 7 )=0.000|* |
|Pr(ncp = 8 )=0.000|* |
|Pr(ncp = 9 )=0.000|* |
|Pr(ncp = 10)=0.000|* |
.-------------------------------------------------------------------.
| Summary for number of Trend ChangePoints (tcp) |
.-------------------------------------------------------------------.
|ncp_max = 10 | MaxTrendKnotNum: A parameter you set |
|ncp_mode = 3 | Pr(ncp= 3)=0.86: There is a 85.9% probability |
| | that the trend component has 3 changepoint(s).|
|ncp_mean = 3.15 | Sum{ncp*Pr(ncp)} for ncp = 0,...,10 |
|ncp_pct10 = 3.00 | 10% percentile for number of changepoints |
|ncp_median = 3.00 | 50% percentile: Median number of changepoints |
|ncp_pct90 = 4.00 | 90% percentile for number of changepoints |
.-------------------------------------------------------------------.
| List of probable trend changepoints ranked by probability of |
| occurrence: Please combine the ncp reported above to determine |
| which changepoints below are practically meaningful |
'-------------------------------------------------------------------'
|tcp# |time (cp) |prob(cpPr) |
|------------------|---------------------------|--------------------|
|1 |33.000000 |1.00000 |
|2 |42.000000 |0.98271 |
|3 |19.000000 |0.69183 |
|4 |26.000000 |0.03950 |
|5 |11.000000 |0.02292 |
.-------------------------------------------------------------------.
Here is the graphic output. Three major changepoints are detected:
You can use sgolayfilt function, that is a polynomial fit to the data, or reproduce OLS method: http://www.utdallas.edu/~herve/Abdi-LeastSquares06-pretty.pdf (there is a+bx notation instead of ax+b)
For linear fit of ax+b:
If you replace x with constant vector of length 2n+1: [-n, ... 0 ... n] on each step, you get the following code for sliding regression coeffs:
for i=1+n:length(y)-n
yi = y(i-n : i+n);
sum_xy = sum(yi.*x);
a(i) = sum_xy/sum_x2;
b(i) = sum(yi)/n;
end
Notice that in this code b means sliding average of your data, and a is a least-square slope estimate (first derivate).
I am creating an iphone app where I have a grid view of 25 images as:
0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19
20 21 22 23 24
Now when any 5 consecutive images are selected it should say bingo, like if 0,6, 12, 18, 24 are selected it should say Bingo.
How will i do that, please help me.
Many Thanks for your help.
Rs
iPhone Developer
-----------------------------------
| 0 | 1 | 2 | 3 | 4 | 5 |
-----------------------------------
| 6 | 7 | 8 | 9 | 10 | 11 |
-----------------------------------
| 12 | 13 | 14 | 15 | 16 | 17 |
-----------------------------------
| 18 | 19 | 20 | 21 | 22 | 23 |
-----------------------------------
| 24 |
-----------------------------------
Hope this is how your grid looks like.
Associate each column with an array. The array will contain the list of all neighbour elements of that column,
For example, the neighbor array of the column [ 6 ] will ollk like array(0, 7, 12), which are all the immediate neighbors of [ 6 ].
Set counter = 0;
Now, when someone clicks an element, increment the counter (Now counter = 1)
When he clicks the second element, check if the element is in the neighbor list of the previous element OR the 1st element.
If the element clicked is in the neighbor list, increment the counter (now counter = 2)
ELSE
If the element clicked is not in the neighbor array, reset the counter (counter = 0) and start over.
Check if the value of counter = 5. If it is, Say Bingo!
The algorithm is not fully correct, but I hope you got the idea :)