I'm trying to read the exemplary ADTF file. When reading the chunk header I see that chunk size is 96bytes, subtracting the header length (32) it leaves us with 64bytes for the actual data.
Now the data structure for the stream says we need only 43 bytes to express the data. I'm not sure how to apply padding there. The actual 64 bytes of data seems to have some padding - I cannot just read the data and push it into structures. I'm not sure how to guess the extra padding sizes. All the extracted values should be equal to 41 (decimal).
<stream description="streamid_2" name="NESTED_STRUCT" type="adtf.core.media_type">
<struct bytepos="0" name="tNestedStruct" type="tNestedStruct"/>
</stream>
<struct alignment="1" name="tNestedStruct" version="1">
<element alignment="1" arraysize="1" byteorder="LE" bytepos="0" name="sHeaderStruct" type="tHeaderStruct"/>
<element alignment="1" arraysize="1" byteorder="LE" bytepos="12" name="sSimpleStruct" type="tSimpleStruct"/>
</struct>
<struct alignment="1" name="tHeaderStruct" version="1">
<element alignment="1" arraysize="1" byteorder="LE" bytepos="0" name="ui32HeaderVal" type="tUInt32"/>
<element alignment="1" arraysize="1" byteorder="LE" bytepos="4" name="f64HeaderVal" type="tFloat64"/>
</struct>
<struct alignment="1" name="tSimpleStruct" version="1">
<element alignment="1" arraysize="1" byteorder="LE" bytepos="0" name="ui8Val" type="tUInt8"/>
<element alignment="1" arraysize="1" byteorder="LE" bytepos="1" name="ui16Val" type="tUInt16"/>
<element alignment="1" arraysize="1" byteorder="LE" bytepos="3" name="ui32Val" type="tUInt32"/>
<element alignment="1" arraysize="1" byteorder="LE" bytepos="7" name="i32Val" type="tInt32"/>
<element alignment="1" arraysize="1" byteorder="LE" bytepos="11" name="i64Val" type="tInt64"/>
<element alignment="1" arraysize="1" byteorder="LE" bytepos="19" name="f64Val" type="tFloat64"/>
<element alignment="1" arraysize="1" byteorder="LE" bytepos="27" name="f32Val" type="tFloat32"/>
</struct>
Here are the 64 data bytes:
index = value (decimal)
0 = 3
1 = 43
2 = 0
3 = 0
4 = 0
5 = -57
6 = -120
7 = 31
8 = 0
9 = 0
10 = 0
11 = 0
12 = 0
13 = 0
14 = 0
15 = 0
16 = 0
17 = 41
18 = 0
19 = 0
20 = 0
21 = 0
22 = 0
23 = 0
24 = 0
25 = 0
26 = -128
27 = 68
28 = 64
29 = 41
30 = 41
31 = 0
32 = 41
33 = 0
34 = 0
35 = 0
36 = 41
37 = 0
38 = 0
39 = 0
40 = 41
41 = 0
42 = 0
43 = 0
44 = 0
45 = 0
46 = 0
47 = 0
48 = 0
49 = 0
50 = 0
51 = 0
52 = 0
53 = -128
54 = 68
55 = 64
56 = 0
57 = 0
58 = 36
59 = 66
60 = 0
61 = 0
62 = 0
63 = 0
I don't really understand what you want to achieve... First of all, you don't need any padding in DDL, the bytepos follows the previous element size. You have to know, that the Description contains the serialized (bytepos, byteorder) and deserialized structure (alignment), please have a look at https://support.digitalwerk.net/adtf/v2/adtf_sdk_html_docs/page_a_utils_indexedfileformat.html. To access the data (read/write), just access via DDL (https://support.digitalwerk.net/adtf/v2/adtf_sdk_html_docs/page_ddl_usage_howto.html), also have a look at the example (https://support.digitalwerk.net/adtf/v2/adtf_sdk_html_docs/page_demo_media_desc_coder.html)
There is also a data offset as well as chunk headers, please have a look at https://support.digitalwerk.net/adtf_libraries/adtf-streaming-library/v2/DATFileFormatSpecification.pdf
But you don't have to care about the indexed file format to use DDL outside ADTF Framework. For that in ADTF 2.x there is the Streaming Library provided
https://support.digitalwerk.net/adtf_libraries/adtf-streaming-library/v2/api/index.html
https://support.digitalwerk.net/adtf_libraries/adtf-streaming-library/v2/StreamingLibrary.pdf
In ADTF 3.x the ADTF File Library (which comes Open Source and can also handle Files from 2.x)
https://support.digitalwerk.net/adtf_libraries/adtf-file-library/html/index.html
Both Libs support Read and Write of (ADTF)DAT Files, so I guess exactly what you need and don't need to reinvent.
Please have a look at the Media Descritpion Example:
https://support.digitalwerk.net/adtf_libraries/adtf-streaming-library/v2/api/page_mediadescription.html
And also the Reader itself:
https://support.digitalwerk.net/adtf_libraries/adtf-streaming-library/v2/api/classadtfstreaming_1_1_i_a_d_t_f_file_reader.html
Related
%PDF-1.5
...
10737 0 obj
<</MarkInfo<</Marked true>>/Metadata 161 0 R/PageLayout/OneColumn/Pages 10732 0 R/StructTreeRoot 206 0 R/Type/Catalog>>
endobj
10738 0 obj
<</Contents[10740 0 R 10741 0 R 10747 0 R 10748 0 R 10749 0 R 10750 0 R 10751 0 R 10752 0 R]/CropBox[0.0 0.0 516.0 728.64]/MediaBox[0.0 0.0 516.0 728.64]/Parent 10733 0 R/Resources<</ColorSpace<</CS0 10771 0 R/CS1 10772 0 R>>/ExtGState<</GS0 10773 0 R>>/Font<</C2_0 10778 0 R/C2_1 10783 0 R/C2_2 10788 0 R/C2_3 10793 0 R/C2_4 10798 0 R/TT0 10800 0 R/TT1 10802 0 R/TT2 10804 0 R/TT3 10806 0 R/TT4 10808 0 R>>/XObject<</Im0 10769 0 R>>>>/Rotate 0/StructParents 0/Tabs/S/Type/Page>>
endobj
10739 0 obj
<</Filter/FlateDecode/First 410/Length 3756/N 38/Type/ObjStm>>stream
10771 0 10772 21 10773 42 10774 138 10775 190 10776 442 10777 741 10778 752 10779 869 10780 921 10781 1190 10782 2050 10783 2061 10784 2192 10785 2244 10786 2504 10787 3456 10788 3467 10789 3587 10790 3639 10791 3903 10792 6058 10793 6069 10794 6196 10795 6248 10796 6507 10797 8153 10798 8164 10799 8284 10800 8496 10801 9662 10802 9894 10803 11072 10804 11325 10805 11779 10806 11985 10807 13147 10808 13395
[/ICCBased 10753 0 R][/ICCBased 10754 0 R]
<</AIS false/BM/Normal/CA 1.0/OP false/OPM 1/SA true/SMask/None/Type/ExtGState/ca 1.0/op false>>
<</Ordering(Identity)/Registry(Adobe)/Supplement 0>><</Ascent 858/CIDSet 10757 0 R/CapHeight 719/Descent -148/Flags 4/FontBBox[-16 -148 1008 858]/FontFamily(\xfe\xff\x00H\x00Y\xc9\x11\xac\xe0\xb5\x15)/FontFile2 10758 0 R/FontName/YDRADB+H2gtrM/FontStretch/Normal/FontWeight 400/ItalicAngle 0/StemV 60/Type/FontDescriptor/XHeight 520>>
...
endstream
endobj
...
No. - Type
10732 - Pages
206 - StructTreeRoot
10771, 10772, 10773, 10778 ... - Font
Many indirect objects including 10732, 206, 10771 and 10772 do not exist in the pdf file.
But I think I found objects 10771~10808 in object 10739 stream.
Q1. Why are there no object 10732(Pages) and 206(StructTreeRoot) in the pdf file?
Q2. Why are indirect objects in stream?
I would be grateful if you would suggest any explanations or resources for reference.
Starting with version 1.5 PDF supports so called object streams, i.e. stream objects which contain other non-stream objects.
Your object 10739 is such an object stream as you can see in its Type ObjStm.
This allows those other objects to be compressed. In particular structure tree objects which otherwise can substantially increase the size of a PDF, can be compressed fairly well, reducing their impact on the document size.
For details please study the PDF specification, section 7.5.7 – Object Streams, in either the current PDF specification ISO 32000-2 or its predecessor ISO 32000-1.
Adobe has shared a copy of ISO 32000-1 on their web site which merely has its ISO page headers replaced. Simply google for "PDF32000_2008"; currently it is located at https://www.adobe.com/content/dam/acom/en/devnet/pdf/pdfs/PDF32000_2008.pdf but as far as I know this isn't a permalink.
I'm relatively new to lua and programming in general (self taught), so please be gentle!
Anyway, I wrote a lua script to read a UDP message from a game. The structure of the message is:
DATAxXXXXaaaaBBBBccccDDDDeeeeFFFFggggHHHH
DATAx = 4 letter ID and x = control character
XXXX = integer shows the group of the data (groups are known)
aaaa...HHHHH = 8 single-precision floating point numbers
The last ones is those numbers I need to decode.
If I print the message as received, it's something like:
DATA*{V???A?A?...etc.
Using string.byte(), I'm getting a stream of bytes like this (I have "formatted" the bytes to reflect the structure above.
68 65 84 65/42/20 0 0 0/237 222 28 66/189 59 182 65/107 42 41 65/33 173 79 63/0 0 128 63/146 41 41 65/0 0 30 66/0 0 184 65
The first 5 bytes are of course the DATA*. The next 4 are the 20th group of data. The next bytes, the ones I need to decode, and are equal to those values:
237 222 28 66 = 39.218
189 59 182 65 = 22.779
107 42 41 65 = 10.573
33 173 79 63 = 0.8114
0 0 128 63 = 1.0000
146 41 41 65 = 10.573
0 0 30 66 = 39.500
0 0 184 65 = 23.000
I've found C# code that does the decode with BitConverter.ToSingle(), but I haven't found any like this for Lua.
Any idea?
What Lua version do you have?
This code works in Lua 5.3
local str = "DATA*\20\0\0\0\237\222\28\66\189\59\182\65..."
-- Read two float values starting from position 10 in the string
print(string.unpack("<ff", str, 10)) --> 39.217700958252 22.779169082642 18
-- 18 (third returned value) is the next position in the string
For Lua 5.1 you have to write special function (or steal it from François Perrad's git repo )
local function binary_to_float(str, pos)
local b1, b2, b3, b4 = str:byte(pos, pos+3)
local sign = b4 > 0x7F and -1 or 1
local expo = (b4 % 0x80) * 2 + math.floor(b3 / 0x80)
local mant = ((b3 % 0x80) * 0x100 + b2) * 0x100 + b1
local n
if mant + expo == 0 then
n = sign * 0.0
elseif expo == 0xFF then
n = (mant == 0 and sign or 0) / 0
else
n = sign * (1 + mant / 0x800000) * 2.0^(expo - 0x7F)
end
return n
end
local str = "DATA*\20\0\0\0\237\222\28\66\189\59\182\65..."
print(binary_to_float(str, 10)) --> 39.217700958252
print(binary_to_float(str, 14)) --> 22.779169082642
It’s little-endian byte-order of IEEE-754 single-precision binary:
E.g., 0 0 128 63 is:
00111111 10000000 00000000 00000000
(63) (128) (0) (0)
Why that equals 1 requires that you understand the very basics of IEEE-754 representation, namely its use of an exponent and mantissa. See here to start.
See #Egor‘s answer above for how to use string.unpack() in Lua 5.3 and one possible implementation you could use in earlier versions.
system ubuntu 16.04
On master node where icinga2 is installed
#ls /etc/icinga2/repository.d/hosts/WIN-U52321E0BAK/
disk C%3A.conf disk.conf icinga.conf load.conf ping4.conf
ping6.conf procs.conf swap.conf users.conf
All conf files have save "dummy" check_command on them for example
#cat load.conf
object Service "load" {
import "satellite-service"
check_command = "dummy"
host_name = "WIN-U52321E0BAK"
zone = "WIN-U52321E0BAK"
}
I cant understand from where dummy command is called and how to customize the checks for warning and critical threshold
The dummy command is defined in /usr/share/icinga2/include/command-plugins.conf, like so:
144 object CheckCommand "dummy" {
145 import "plugin-check-command"
146
147 command = [
148 PluginDir + "/check_dummy",
149 "$dummy_state$",
150 "$dummy_text$"
151 ]
152
153 vars.dummy_state = 0
154 vars.dummy_text = "Check was successful."
155 }
In order to modify the warn and crit levels, you set the custom variable at the host or service level. Using the example of ping, we see the default configuration in that same file:
36 template CheckCommand "ping-common" {
37 import "plugin-check-command"
38
39 command = [ PluginDir + "/check_ping" ]
40
41 arguments = {
42 "-H" = "$ping_address$"
43 "-w" = "$ping_wrta$,$ping_wpl$%"
44 "-c" = "$ping_crta$,$ping_cpl$%"
45 "-p" = "$ping_packets$"
46 "-t" = "$ping_timeout$"
47 }
48
49 vars.ping_wrta = 100
50 vars.ping_wpl = 5
51 vars.ping_crta = 200
52 vars.ping_cpl = 15
53 }
Here's the important bit:
49 vars.ping_wrta = 100
50 vars.ping_wpl = 5
51 vars.ping_crta = 200
52 vars.ping_cpl = 15
So: we go to our host or service definition, thusly (using /etc/icinga2/conf.d/host.conf and the NodeName/localhost definition which everybody has; comments removed):
18 object Host NodeName {
20 import "generic-host"
21
23 address = "127.0.0.1"
24 address6 = "::1"
25
27 vars.os = "Linux"
30 vars.http_vhosts["http"] = {
31 http_uri = "/"
32 }
37
39 vars.disks["disk"] = {
41 }
42 vars.disks["disk /"] = {
43 disk_partitions = "/"
44 }
45 }
And we insert before line 45 above to produce:
18 object Host NodeName {
20 import "generic-host"
21
23 address = "127.0.0.1"
24 address6 = "::1"
25
27 vars.os = "Linux"
30 vars.http_vhosts["http"] = {
31 http_uri = "/"
32 }
37
39 vars.disks["disk"] = {
41 }
42 vars.disks["disk /"] = {
43 disk_partitions = "/"
44 }
45 vars.ping_wrta = 50
46 vars.ping_wpl = 3
47 vars.ping_crta = 10
48 vars.ping_cpl = 2
49 }
...and you have successfully customized the check threshold. You can add those variables to a template or even a hostgroup (I think; better test that, I may be wrong).
Using the Images package, I can open up a color image, convert it to Gray scale and then :
using Images
img_gld = imread("...path to some color jpg...")
img_gld_gs = convert(Image{Gray},img_gld)
#change from floats to Array of values between 0 and 255:
img_gld_gs = reinterpret(Uint8,data(img_gld_gs))
Now I've got a 1920X1080 array of Uint8's:
julia> img_gld_gs
1920x1080 Array{Uint8,2}
Now I want to get a histogram of the 2D array of Uint8 values:
julia> hist(img_gld_gs)
(0.0:50.0:300.0,
6x1080 Array{Int64,2}:
1302 1288 1293 1302 1297 1300 1257 1234 … 12 13 13 12 13 15 14
618 632 627 618 623 620 663 686 189 187 187 188 185 183 183
0 0 0 0 0 0 0 0 9 9 8 7 8 7 7
0 0 0 0 0 0 0 0 10 12 9 7 13 7 9
0 0 0 0 0 0 0 0 1238 1230 1236 1235 1230 1240 1234
0 0 0 0 0 0 0 0 … 462 469 467 471 471 468 473)
But, instead of 6x1080, I'd like 256 slots in the histogram to show total number of times each value has appeared. I tried:
julia> hist(img_gld_gs,256)
But that gives:
(2.0:1.0:252.0,
250x1080 Array{Int64,2}:
So instead of a 256x1080 Array, it's 250x1080. Is there any way to force it to have 256 bins (without resorting to writing my own hist function)? I want to be able to compare different images and I want the histogram for each image to have the same number of bins.
Assuming you want a histogram for the entire image (rather than one per row), you might want
hist(vec(img_gld_gs), -1:255)
which first converts the image to a 1-dimensional vector. (You can also use img_gld_gs[:], but that copies the data.)
Also note the range here: the hist function uses a left-open interval, so it will omit counting zeros unless you use something smaller than 0.
hist also accepts a vector (or range) as an optional argument that specifies the edge boundaries, so
hist(img_gld_gs, 0:256)
should work.
I tried to create a neural network to estimate y = x ^ 2. So I created a fitting neural network and gave it some samples for input and output. I tried to build this network in C++. But the result is different than I expected.
With the following inputs:
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 25 26
27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 -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 -48 -49 -50 -51 -52 -53 -54 -55 -56 -57 -58 -59 -60 -61 -62 -63 -64 -65 -66 -67 -68 -69 -70 -71
and the following outputs:
0 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400
441 484 529 576 625 676 729 784 841 900 961 1024 1089 1156 1225 1296
1369 1444 1521 1600 1681 1764 1849 1936 2025 2116 2209 2304 2401 2500
2601 2704 2809 2916 3025 3136 3249 3364 3481 3600 3721 3844 3969 4096
4225 4356 4489 4624 4761 4900 5041 1 4 9 16 25 36 49 64 81 100 121 144
169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841
900 961 1024 1089 1156 1225 1296 1369 1444 1521 1600 1681 1764 1849
1936 2025 2116 2209 2304 2401 2500 2601 2704 2809 2916 3025 3136 3249
3364 3481 3600 3721 3844 3969 4096 4225 4356 4489 4624 4761 4900 5041
I used fitting tool network. with matrix rows. Training is 70%, validation is 15% and testing is 15% as well. The number of hidden neurons is two. Then in command lines I wrote this:
purelin(net.LW{2}*tansig(net.IW{1}*inputTest+net.b{1})+net.b{2})
Other information :
My net.b[1] is: -1.16610230053776 1.16667147712026
My net.b[2] is: 51.3266249426358
And net.IW(1) is: 0.344272596370387 0.344111217766824
net.LW(2) is: 31.7635369693519 -31.8082184881063
When my inputTest is 3, the result of this command is 16, while it should be about 9. Have I made an error somewhere?
I found the Stack Overflow post Neural network in MATLAB that contains a problem like my problem, but there is a little difference, and the differences is in that problem the ranges of input and output are same, but in my problem is no. That solution says I need to scale out the results, but how can I scale out my result?
You are right about scaling. As was mentioned in the linked answer, the neural network by default scales the input and output to the range [-1,1]. This can be seen in the network processing functions configuration:
>> net = fitnet(2);
>> net.inputs{1}.processFcns
ans =
'removeconstantrows' 'mapminmax'
>> net.outputs{2}.processFcns
ans =
'removeconstantrows' 'mapminmax'
The second preprocessing function applied to both input/output is mapminmax with the following parameters:
>> net.inputs{1}.processParams{2}
ans =
ymin: -1
ymax: 1
>> net.outputs{2}.processParams{2}
ans =
ymin: -1
ymax: 1
to map both into the range [-1,1] (prior to training).
This means that the trained network expects input values in this range, and outputs values also in the same range. If you want to manually feed input to the network, and compute the output yourself, you have to scale the data at input, and reverse the mapping at the output.
One last thing to remember is that each time you train the ANN, you will get different weights. If you want reproducible results, you need to fix the state of the random number generator (initialize it with the same seed each time). Read the documentation on functions like rng and RandStream.
You also have to pay attention that if you are dividing the data into training/testing/validation sets, you must use the same split each time (probably also affected by the randomness aspect I mentioned).
Here is an example to illustrate the idea (adapted from another post of mine):
%%# data
x = linspace(-71,71,200); %# 1D input
y_model = x.^2; %# model
y = y_model + 10*randn(size(x)).*x; %# add some noise
%%# create ANN, train, simulate
net = fitnet(2); %# one hidden layer with 2 nodes
net.divideFcn = 'dividerand';
net.trainParam.epochs = 50;
net = train(net,x,y);
y_hat = net(x);
%%# plot
plot(x, y, 'b.'), hold on
plot(x, x.^2, 'Color','g', 'LineWidth',2)
plot(x, y_hat, 'Color','r', 'LineWidth',2)
legend({'data (noisy)','model (x^2)','fitted'})
hold off, grid on
%%# manually simulate network
%# map input to [-1,1] range
[~,inMap] = mapminmax(x, -1, 1);
in = mapminmax('apply', x, inMap);
%# propagate values to get output (scaled to [-1,1])
hid = tansig( bsxfun(#plus, net.IW{1}*in, net.b{1}) ); %# hidden layer
outLayerOut = purelin( net.LW{2}*hid + net.b{2} ); %# output layer
%# reverse mapping from [-1,1] to original data scale
[~,outMap] = mapminmax(y, -1, 1);
out = mapminmax('reverse', outLayerOut, outMap);
%# compare against MATLAB output
max( abs(out - y_hat) ) %# this should be zero (or in the order of `eps`)
I opted to use the mapminmax function, but you could have done that manually as well. The formula is a pretty simply linear mapping:
y = (ymax-ymin)*(x-xmin)/(xmax-xmin) + ymin;