Difference between 2 strings - perl

I want to compare some strings like this
Previous -> Present
Something like
path 1 : 100 -> 112 --> 333 --> 500
path 2 : 100 -> 333 --> 500
path 3 : 100 -> 333 --> 500 --> 500
path 4 : 100 -> 112 --> 500
I need to compare path 1 with path 2, get the number that was in path 1 which doesn't exist in path 2 and store it in a database
Then compare path 2 with path 3 and do same thing. If it already exists then increment it. Otherwise insert the new number.
I know how to insert into a database and increment if the entry exists. What I don't know is how to loop through all those paths getting those values then deciding whether to insert into the database.
I have done some research, and I have heard of Levenshtein Edit Distance but I can't figure out how I should do it.


Your question appears to be:
Given two lists of numbers, how can I tell which ones in list A aren't in list B?
Hashes are useful for doing set arithmetic.
my #a = ( 100, 112, 333, 500 );
my #b = ( 100, 333, 500 );
my %b = map { $_ => 1 } #b;
my #missing = grep { !$b{$_} } #a;

Related

How to count the numbers of elements in parts of a text file using a loop in Perl?

I´m looking for a way to create a script in Perl to count the elements in my text file and do it in parts. For example, my text file has this form:
ID Position Potential Jury agreement NGlyc result
(PART 1)
NP_073551.1_HCoV229Egp2 23 NTSY 0.5990 (8/9) +
NP_073551.1_HCoV229Egp2 62 NTSS 0.7076 (9/9) ++
NP_073551.1_HCoV229Egp2 171 NTTI 0.5743 (5/9) +
...
(PART 2)
QJY77946.1_NA 20 NGTN 0.7514 (9/9) +++
QJY77946.1_NA 23 NTSH 0.5368 (5/9) +
QJY77946.1_NA 51 NFSF 0.7120 (9/9) ++
QJY77946.1_NA 62 NTSS 0.6947 (9/9) ++
...
(PART 3)
QJY77954.1_NA 20 NGTN 0.7694 (9/9) +++
QJY77954.1_NA 23 NTSH 0.5398 (5/9) +
QJY77954.1_NA 51 NFSF 0.7121 (9/9) ++
...
(PART N°...)
Like you can see the ID is the same in each part (one for PART 1, other to PART 2 and then...). The changes only can see in the columns Position//Potential//Jury agreement//NGlyc result Then, my main goal is to count the line with Potential 0,7 >=.
With this in mind, I´m looking for output like this:
Part 1:
1 (one value 0.7 >=)
Part 2:
2 (two values 0.7 >=)
Part 3:
2 (two values 0.7 >=)
Part N°:
X numbers of values 0.7 >=
This output tells me the number of positive values (0.7 >=) for each ID.
The pseudocode I believe would be something like this:
foreach ID in LIST
foreach LINE in FILE
if (ID is in LINE)
... count the line ...
end foreach LINE
end foreach ID
I´m looking for any suggestion (for a package or script idea) or comment to create a better script.
Thanks! Best!

To count the number of lines, for each part, that match some condition on a certain column, you can just loop over the lines, skip the header, parse the part number, and use an array to count the number of lines matching for each part.
After this you can just loop over the counts recorded in the array and print them out in your specific format.
#!/usr/bin/perl
use strict;
use warnings;
my $part = 0;
my #cnt_part;
while(my $line = <STDIN>) {
if($. == 1) {
next;
}elsif($line =~ m{^\(PART (\d+)\)}) {
$part = $1;
}else {
my #cols = split(m{\s+},$line);
if(#cols == 6) {
my $potential = $cols[3];
if(0.7 <= $potential) {
$cnt_part[$part]++;
};
};
};
};
for(my $i=1;$i<=$#cnt_part;$i++){
print "Part $i:\n";
print "$cnt_part[$i] (values 0.7 <=)\n";
};
To run it, just pipe the entire file through the Perl script:
cat in.txt | perl count.pl
and you get an output like this:
Part 1:
1 (values 0.7 <=)
Part 2:
2 (values 0.7 <=)
Part 3:
2 (values 0.7 <=)
If you want to also display the counts into words, you can use Lingua::EN::Numbers (see this program ) and you get an output very similar to the one in your post:
Part 1:
1 (one values 0.7 <=)
Part 2:
2 (two values 0.7 <=)
Part 3:
2 (two values 0.7 <=)
All the code in this post is also available here.

Encoding Spotify URI to Spotify Codes

Spotify Codes are little barcodes that allow you to share songs, artists, users, playlists, etc.
They encode information in the different heights of the "bars". There are 8 discrete heights that the 23 bars can be, which means 8^23 different possible barcodes.
Spotify generates barcodes based on their URI schema. This URI spotify:playlist:37i9dQZF1DXcBWIGoYBM5M gets mapped to this barcode:
The URI has a lot more information (62^22) in it than the code. How would you map the URI to the barcode? It seems like you can't simply encode the URI directly. For more background, see my "answer" to this question: https://stackoverflow.com/a/62120952/10703868

The patent explains the general process, this is what I have found.
This is a more recent patent
When using the Spotify code generator the website makes a request to https://scannables.scdn.co/uri/plain/[format]/[background-color-in-hex]/[code-color-in-text]/[size]/[spotify-URI].
Using Burp Suite, when scanning a code through Spotify the app sends a request to Spotify's API: https://spclient.wg.spotify.com/scannable-id/id/[CODE]?format=json where [CODE] is the media reference that you were looking for. This request can be made through python but only with the [TOKEN] that was generated through the app as this is the only way to get the correct scope. The app token expires in about half an hour.
import requests
head={
"X-Client-Id": "58bd3c95768941ea9eb4350aaa033eb3",
"Accept-Encoding": "gzip, deflate",
"Connection": "close",
"App-Platform": "iOS",
"Accept": "*/*",
"User-Agent": "Spotify/8.5.68 iOS/13.4 (iPhone9,3)",
"Accept-Language": "en",
"Authorization": "Bearer [TOKEN]",
"Spotify-App-Version": "8.5.68"}
response = requests.get('https://spclient.wg.spotify.com:443/scannable-id/id/26560102031?format=json', headers=head)
print(response)
print(response.json())
Which returns:
<Response [200]>
{'target': 'spotify:playlist:37i9dQZF1DXcBWIGoYBM5M'}
So 26560102031 is the media reference for your playlist.
The patent states that the code is first detected and then possibly converted into 63 bits using a Gray table. For example 361354354471425226605 is encoded into 010 101 001 010 111 110 010 111 110 110 100 001 110 011 111 011 011 101 101 000 111.
However the code sent to the API is 6875667268, I'm unsure how the media reference is generated but this is the number used in the lookup table.
The reference contains the integers 0-9 compared to the gray table of 0-7 implying that an algorithm using normal binary has been used. The patent talks about using a convolutional code and then the Viterbi algorithm for error correction, so this may be the output from that. Something that is impossible to recreate whithout the states I believe. However I'd be interested if you can interpret the patent any better.
This media reference is 10 digits however others have 11 or 12.
Here are two more examples of the raw distances, the gray table binary and then the media reference:
1.
022673352171662032460
000 011 011 101 100 010 010 111 011 001 100 001 101 101 011 000 010 011 110 101 000
67775490487
2.
574146602473467556050
111 100 110 001 110 101 101 000 011 110 100 010 110 101 100 111 111 101 000 111 000
57639171874
edit:
Some extra info:
There are some posts online describing how you can encode any text such as spotify:playlist:HelloWorld into a code however this no longer works.
I also discovered through the proxy that you can use the domain to fetch the album art of a track above the code. This suggests a closer integration of Spotify's API and this scannables url than previously thought. As it not only stores the URIs and their codes but can also validate URIs and return updated album art.
https://scannables.scdn.co/uri/800/spotify%3Atrack%3A0J8oh5MAMyUPRIgflnjwmB

Your suspicion was correct - they're using a lookup table. For all of the fun technical details, the relevant patent is available here: https://data.epo.org/publication-server/rest/v1.0/publication-dates/20190220/patents/EP3444755NWA1/document.pdf

Very interesting discussion. Always been attracted to barcodes so I had to take a look. I did some analysis of the barcodes alone (didn't access the API for the media refs) and think I have the basic encoding process figured out. However, based on the two examples above, I'm not convinced I have the mapping from media ref to 37-bit vector correct (i.e. it works in case 2 but not case 1). At any rate, if you have a few more pairs, that last part should be simple to work out. Let me know.
For those who want to figure this out, don't read the spoilers below!
It turns out that the basic process outlined in the patent is correct, but lacking in details. I'll summarize below using the example above. I actually analyzed this in reverse which is why I think the code description is basically correct except for step (1), i.e. I generated 45 barcodes and all of them matched had this code.
1. Map the media reference as integer to 37 bit vector.
Something like write number in base 2, with lowest significant bit
on the left and zero-padding on right if necessary.
57639171874 -> 0100010011101111111100011101011010110
2. Calculate CRC-8-CCITT, i.e. generator x^8 + x^2 + x + 1
The following steps are needed to calculate the 8 CRC bits:
Pad with 3 bits on the right:
01000100 11101111 11110001 11010110 10110000
Reverse bytes:
00100010 11110111 10001111 01101011 00001101
Calculate CRC as normal (highest order degree on the left):
-> 11001100
Reverse CRC:
-> 00110011
Invert check:
-> 11001100
Finally append to step 1 result:
01000100 11101111 11110001 11010110 10110110 01100
3. Convolutionally encode the 45 bits using the common generator
polynomials (1011011, 1111001) in binary with puncture pattern
110110 (or 101, 110 on each stream). The result of step 2 is
encoded using tail-biting, meaning we begin the shift register
in the state of the last 6 bits of the 45 long input vector.
Prepend stream with last 6 bits of data:
001100 01000100 11101111 11110001 11010110 10110110 01100
Encode using first generator:
(a) 100011100111110100110011110100000010001001011
Encode using 2nd generator:
(b) 110011100010110110110100101101011100110011011
Interleave bits (abab...):
11010000111111000010111011110011010011110001...
1010111001110001000101011000010110000111001111
Puncture every third bit:
111000111100101111101110111001011100110000100100011100110011
4. Permute data by choosing indices 0, 7, 14, 21, 28, 35, 42, 49,
56, 3, 10..., i.e. incrementing 7 modulo 60. (Note: unpermute by
incrementing 43 mod 60).
The encoded sequence after permuting is
111100110001110101101000011110010110101100111111101000111000
5. The final step is to map back to bar lengths 0 to 7 using the
gray map (000,001,011,010,110,111,101,100). This gives the 20 bar
encoding. As noted before, add three bars: short one on each end
and a long one in the middle.
UPDATE: I've added a barcode (levels) decoder (assuming no errors) and an alternate encoder that follows the description above rather than the equivalent linear algebra method. Hopefully that is a bit more clear.
UPDATE 2: Got rid of most of the hard-coded arrays to illustrate how they are generated.
The linear algebra method defines the linear transformation (spotify_generator) and mask to map the 37 bit input into the 60 bit convolutionally encoded data. The mask is result of the 8-bit inverted CRC being convolutionally encoded. The spotify_generator is a 37x60 matrix that implements the product of generators for the CRC (a 37x45 matrix) and convolutional codes (a 45x60 matrix). You can create the generator matrix from an encoding function by applying the function to each row of an appropriate size generator matrix. For example, a CRC function that add 8 bits to each 37 bit data vector applied to each row of a 37x37 identity matrix.
import numpy as np
import crccheck
# Utils for conversion between int, array of binary
# and array of bytes (as ints)
def int_to_bin(num, length, endian):
if endian == 'l':
return [num >> i & 1 for i in range(0, length)]
elif endian == 'b':
return [num >> i & 1 for i in range(length-1, -1, -1)]
def bin_to_int(bin,length):
return int("".join([str(bin[i]) for i in range(length-1,-1,-1)]),2)
def bin_to_bytes(bin, length):
b = bin[0:length] + [0] * (-length % 8)
return [(b[i]<<7) + (b[i+1]<<6) + (b[i+2]<<5) + (b[i+3]<<4) +
(b[i+4]<<3) + (b[i+5]<<2) + (b[i+6]<<1) + b[i+7] for i in range(0,len(b),8)]
# Return the circular right shift of an array by 'n' positions
def shift_right(arr, n):
return arr[-n % len(arr):len(arr):] + arr[0:-n % len(arr)]
gray_code = [0,1,3,2,7,6,4,5]
gray_code_inv = [[0,0,0],[0,0,1],[0,1,1],[0,1,0],
[1,1,0],[1,1,1],[1,0,1],[1,0,0]]
# CRC using Rocksoft model:
# NOTE: this is not quite any of their predefined CRC's
# 8: number of check bits (degree of poly)
# 0x7: representation of poly without high term (x^8+x^2+x+1)
# 0x0: initial fill of register
# True: byte reverse data
# True: byte reverse check
# 0xff: Mask check (i.e. invert)
spotify_crc = crccheck.crc.Crc(8, 0x7, 0x0, True, True, 0xff)
def calc_spotify_crc(bin37):
bytes = bin_to_bytes(bin37, 37)
return int_to_bin(spotify_crc.calc(bytes), 8, 'b')
def check_spotify_crc(bin45):
data = bin_to_bytes(bin45,37)
return spotify_crc.calc(data) == bin_to_bytes(bin45[37:], 8)[0]
# Simple convolutional encoder
def encode_cc(dat):
gen1 = [1,0,1,1,0,1,1]
gen2 = [1,1,1,1,0,0,1]
punct = [1,1,0]
dat_pad = dat[-6:] + dat # 6 bits are needed to initialize
# register for tail-biting
stream1 = np.convolve(dat_pad, gen1, mode='valid') % 2
stream2 = np.convolve(dat_pad, gen2, mode='valid') % 2
enc = [val for pair in zip(stream1, stream2) for val in pair]
return [enc[i] for i in range(len(enc)) if punct[i % 3]]
# To create a generator matrix for a code, we encode each row
# of the identity matrix. Note that the CRC is not quite linear
# because of the check mask so we apply the lamda function to
# invert it. Given a 37 bit media reference we can encode by
# ref * spotify_generator + spotify_mask (mod 2)
_i37 = np.identity(37, dtype=bool)
crc_generator = [_i37[r].tolist() +
list(map(lambda x : 1-x, calc_spotify_crc(_i37[r].tolist())))
for r in range(37)]
spotify_generator = 1*np.array([encode_cc(crc_generator[r]) for r in range(37)], dtype=bool)
del _i37
spotify_mask = 1*np.array(encode_cc(37*[0] + 8*[1]), dtype=bool)
# The following matrix is used to "invert" the convolutional code.
# In particular, we choose a 45 vector basis for the columns of the
# generator matrix (by deleting those in positions equal to 2 mod 4)
# and then inverting the matrix. By selecting the corresponding 45
# elements of the convolutionally encoded vector and multiplying
# on the right by this matrix, we get back to the unencoded data,
# assuming there are no errors.
# Note: numpy does not invert binary matrices, i.e. GF(2), so we
# hard code the following 3 row vectors to generate the matrix.
conv_gen = [[0,1,0,1,1,1,1,0,1,1,0,0,0,1]+31*[0],
[1,0,1,0,1,0,1,0,0,0,1,1,1] + 32*[0],
[0,0,1,0,1,1,1,1,1,1,0,0,1] + 32*[0] ]
conv_generator_inv = 1*np.array([shift_right(conv_gen[(s-27) % 3],s) for s in range(27,72)], dtype=bool)
# Given an integer media reference, returns list of 20 barcode levels
def spotify_bar_code(ref):
bin37 = np.array([int_to_bin(ref, 37, 'l')], dtype=bool)
enc = (np.add(1*np.dot(bin37, spotify_generator), spotify_mask) % 2).flatten()
perm = [enc[7*i % 60] for i in range(60)]
return [gray_code[4*perm[i]+2*perm[i+1]+perm[i+2]] for i in range(0,len(perm),3)]
# Equivalent function but using CRC and CC encoders.
def spotify_bar_code2(ref):
bin37 = int_to_bin(ref, 37, 'l')
enc_crc = bin37 + calc_spotify_crc(bin37)
enc_cc = encode_cc(enc_crc)
perm = [enc_cc[7*i % 60] for i in range(60)]
return [gray_code[4*perm[i]+2*perm[i+1]+perm[i+2]] for i in range(0,len(perm),3)]
# Given 20 (clean) barcode levels, returns media reference
def spotify_bar_decode(levels):
level_bits = np.array([gray_code_inv[levels[i]] for i in range(20)], dtype=bool).flatten()
conv_bits = [level_bits[43*i % 60] for i in range(60)]
cols = [i for i in range(60) if i % 4 != 2] # columns to invert
conv_bits45 = np.array([conv_bits[c] for c in cols], dtype=bool)
bin45 = (1*np.dot(conv_bits45, conv_generator_inv) % 2).tolist()
if check_spotify_crc(bin45):
return bin_to_int(bin45, 37)
else:
print('Error in levels; Use real decoder!!!')
return -1
And example:
>>> levels = [5,7,4,1,4,6,6,0,2,4,3,4,6,7,5,5,6,0,5,0]
>>> spotify_bar_decode(levels)
57639171874
>>> spotify_barcode(57639171874)
[5, 7, 4, 1, 4, 6, 6, 0, 2, 4, 3, 4, 6, 7, 5, 5, 6, 0, 5, 0]

Matlab - read unstructured file

I'm quite new with Matlab and I've been searching, unsucessfully, for the following issue: I have an unstructure txt file, with several rows I don't need, but there are a number of rows inside that file that have an structured format. I've been researching how to "load" the file to edit it, but cannot find anything.
Since i don't know if I was clear, let me show you the content in the file:
8782 PROJCS["UTM-39",GEOGC.......
1 676135.67755473056 2673731.9365976951 -15 0
2 663999.99999999302 2717629.9999999981 -14.00231124135486 3
3 709999.99999999162 2707679.2185399458 -10 2
4 679972.20003752434 2674637.5679516452 0.070000000000000007 1
5 676124.87132483651 2674327.3183533219 -18.94794942571912 0
6 682614.20527054626 2671000.0000000549 -1.6383425512446661 0
...........
8780 682247.4593014461 2676571.1515358146 0.1541080392180566 0
8781 695426.98657108378 2698111.6168302582 -8.5039945992245904 0
8782 674723.80100125563 2675133.5486935056 -19.920312922947179 0
16997 3 21
1 2147 658 590
2 1855 2529 5623
.........
I'd appreciate if someone can just tell me if there is the possibility to open the file to later load only the rows starting with 1 to the one starting with 8782. First row and all the others are not important.
I know than manually copy and paste to a new file would be a solution, but I'd like to know about the possibility to read the file and edit it for other ideas I have.
Thanks!

% Now lines{i} is the string of the i'th line.
lines = strsplit(fileread('filename'), '\n')
% Now elements{i}{j} is the j'th field of the i'th line.
elements = arrayfun(#(x){strsplit(x{1}, ' ')}, lines)
% Remove the first row:
elements(1) = []
% Take the first several rows:
n_rows = 8782
elements = elements(1:n_rows)
Or if the number of rows you need to take is not fixed, you can replace the last two statements above by:
firsts = arrayfun(#(x)str2num(x{1}{1}), elements)
n_rows = find((firsts(2:end) - firsts(1:end-1)) ~= 1, 1, 'first')
elements = elements(1:n_rows)

How to sum values in a column grouped by values in the other

I have a large file consisting data in 2 columns
100 5
100 10
100 10
101 2
101 4
102 10
102 2
I want to sum the values in 2nd column with matching values in column 1. For this example, the output I'm expecting is
100 25
101 6
102 12
I'm trying to work on this using bash script preferably. Can someone explain me how can I do this

Using awk:
awk '{a[$1]+=$2}END{for(i in a){print i, a[i]}}' inputfile
For your input, it'd produce:
100 25
101 6
102 12

In a perl oneliner
perl -lane "$s{$F[0]} += $F[1]; END { print qq{$_ $s{$_}} for keys %s}" file.txt

You can use an associative array. The first column is the index and the second becomes what you add to it.
#!/bin/bash
declare -A columns=()
while read -r -a line ; do
columns[${line[0]}]=$((${columns[${line[0]}]} + ${line[1]}))
done < "${1}"
for idx in ${!columns[#]} ; do
echo "${idx} ${columns[${idx}]}"
done

Using awk and maintain the order:
awk '!($1 in a){a[$1]=$2; b[++i]=$1;next} {a[$1]+=$2} END{for (k=1; k<=i; k++) print b[k], a[b[k]]}' file
100 25
101 6
102 12

Python is my choice:
d = {}
for line in f.readlines():
key,value = line.split()
if d[key] == None:
d[key] = 0
d[key] += value
print d
Why would you want a bash script?

Example of 'a subroutine may have several entry and exit points'

I'm reading the paper of Non structured programming, and found it says:
Unlike a procedure, a subroutine may have several entry and exit points, and a direct jump into or out of subroutine is (theoretically) allowed
I can't understand it, could anyone give me an code sample of:
a subroutine may have several entry and exit points
a direct jump into or out of subroutine
Thanks

10 A = 1
20 GOSUB 100
30 A = 2
40 GOSUB 110
50 A = 3
60 GOTO 130
70 END
100 PRINT A
110 PRINT "HELLO"
120 IF A = 1 THEN RETURN
130 PRINT "THERE"
140 IF A = 3 THEN GOTO 70
150 RETURN
The subroutine has three entry points (lines 100, 110, and 130) and three exit points (lines 120, 140, and 150). There is a direct jump into line 130 (from line 60) and a direct jump out (at line 140).