Suppose, we have input strings in a form I have/had alot/none/zero money.
I would like to have a set of output strings as follows (example 1):
I have alot money
I have none money
I have zero money
I had alot money
I had none money
I had zero money
But then, real task here, is to be able to choose one or more, or none input substrings to ignore. So, the output strings would look like this:
I money
or
first example
or
I alot money
I none money
I zero money
or
I
or
money
I hope you got the point.
How can i do this in the way, friendliest to cpu cycles ?
Ok, to break the ice, this is what im Not willing to do, but considering until brighter ideas:
generate all the output strings (mentioned, example 1).
iterating through strings, i filter out ones that meet my criteria, replace unwanted substrings with "".
put the resulting string into final output array only if it is not already there.
Also, the answer to why do i care for cpu cycles, is simple : the longer this task takes, the longer it will block worker thread.
This simplest way is to find all the spaces and / character and put individual word into a two-level list, you then get a structure like this:
I
have, had
a lot, none, zero
money
.
Now you just loop thru the tree and generate a result string.
To push it to an overkill, you may also write a token parser instead of doing strchr.
How can i do this in the way, friendliest to cpu cycles ?
Why do you care?
Related
I am working with a messy manually maintained "database" that has a column containing a string with name,value pairs. I am trying to parse the entire column with regexp to pull out the values. The column is huge (>100,000 entries). As a proxy for my actual data, let's use this code:
line1={'''thing1'': ''-583'', ''thing2'': ''245'', ''thing3'': ''246'', ''morestuff'':, '''''};
line2={'''thing1'': ''617'', ''thing2'': ''239'', ''morestuff'':, '''''};
line3={'''thing1'': ''unexpected_string(with)parens5'', ''thing2'': 245, ''thing3'':''246'', ''morestuff'':, '''''};
mycell=vertcat(line1,line2,line3);
This captures the general issues encountered in the database. I want to extract what thing1, thing2, and thing3 are in each line using cellfun to output a scalar cell array. They should normally be 3 digit numbers, but sometimes they have an unexpected form. Sometimes thing3 is completely missing, without the name even showing up in the line. Sometimes there are minor formatting inconsistencies, like single quotes missing around the value, spaces missing, or dashes showing up in front of the three digit value. I have managed to handle all of these, except for the case where thing3 is completely missing.
My general approach has been to use expressions like this:
expr1='(?<=thing1''):\s?''?-?([\w\d().]*?)''?,';
expr2='(?<=thing2''):\s?''?-?([\w\d().]*?)''?,';
expr3='(?<=thing3''):\s?''?-?([\w\d().]*?)''?,';
This looks behind for thingX' and then tries to match : followed by zero or one spaces, followed by 0 or 1 single quote, followed by zero or one dash, followed by any combination of letters, numbers, parentheses, or periods (this is defined as the token), using a lazy match, until zero or one single quote is encountered, followed by a comma. I call regexp as regexp(___,'tokens','once') to return the matching token.
The problem is that when there is no match, regexp returns an empty array. This prevents me from using, say,
out=cellfun(#(x) regexp(x,expr3,'tokens','once'),mycell);
unless I call it with 'UniformOutput',false. The problem with that is twofold. First, I need to then manually find the rows where there was no match. For example, I can do this:
emptyout=cellfun(#(x) isempty(x),out);
emptyID=find(emptyout);
backfill=cell(length(emptyID),1);
[backfill{:}]=deal('Unknown');
out(emptyID)=backfill;
In this example, emptyID has a length of 1 so this code is overkill. But I believe this is the correct way to generalize for when it is longer. This code will change every empty cell array in out with the string Unknown. But this leads to the second problem. I've now got a 'messy' cell array of non-scalar values. I cannot, for example, check unique(out) as a result.
Pardon the long-windedness but I wanted to give a clear example of the problem. Now my actual question is in a few parts:
Is there a way to accomplish what I'm trying to do without using 'UniformOutput',false? For example, is there a way to have regexp pass a custom string if there is no match (e.g. pass 'Unknown' if there is no match)? I can think of one 'cheat', which would be to use the | operator in the expression, and if the first token is not matched, look for something that is ALWAYS found. I would then still need to double back through the output and change every instance of that result to 'Unknown'.
If I take the 'UniformOutput',false approach, how can I recover a scalar cell array at the end to easily manipulate it (e.g. pass it through unique)? I will admit I'm not 100% clear on scalar vs nonscalar cell arrays.
If there is some overall different approach that I'm not thinking of, I'm also open to it.
Tangential to the main question, I also tried using a single expression to run regexp using 3 tokens to pull out the values of thing1, thing2, and thing3 in one pass. This seems to require 'UniformOutput',false even when there are no empty results from regexp. I'm not sure how to get a scalar cell array using this approach (e.g. an Nx1 cell array where each cell is a 3x1 cell).
At the end of the day, I want to build a table using these results:
mytable=table(out1,out2,out3);
Edit: Using celldisp sheds some light on the problem:
celldisp(out)
out{1}{1} =
246
out{2} =
Unknown
out{3}{1} =
246
I assume that I need to change the structure of out so that the contents of out{1}{1} and out{3}{1} are instead just out{1} and out{3}. But I'm not sure how to accomplish this if I need 'UniformOutput',false.
Note: I've not used MATLAB and this doesn't answer the "efficient" aspect, but...
How about forcing there to always be a match?
Just thinking about you really wanting a match to skip this problem, how about an empty match?
Looking on the MATLAB help page here I can see a 'emptymatch' option, perhaps this is something to try.
E.g.
the_thing_i_want_to_find|
Match "the_thing_i_want_to_find" or an empty match, note the | character.
In capture group it might look like this:
(the_thing_i_want_to_find|)
As a workaround, I have found that using regexprep can be used to find entries where thing3 is missing. For example:
replace='$1 ''thing3'': ''Unknown'', ''morestuff''';
missingexpr='(?<=thing2'':\s?)(''?-?[\w\d().]*?''?,) ''morestuff''';
regexprep(mycell{2},missingexpr,replace)
ans =
''thing1': '617', 'thing2': '239', 'thing3': 'Unknown', 'morestuff':, '''
Applying it to the entire array:
fixedcell=cellfun(#(x) regexprep(x,missingexpr,replace),mycell);
out=cellfun(#(x) regexp(x,expr3,'tokens','once'),fixedcell,'UniformOutput',false);
This feels a little roundabout, but it works.
cellfun can be replaced with a plain old for loop. Your code will either be equally fast, or maybe even faster. cellfun is implemented with a loop anyway, there is no advantage of using it other than fewer lines of code. In your explicit loop, you can then check the output of regexp, and build your output array any way you like.
For the past few days I have taken a challenge to write an algorithm in Python 2.7, to find a passphrase given a few hints.
Specifically:
An anagram of the passphrase is: "poultry outwits ants"
The MD5 hash of the secret phrase is "4624d200580677270a54ccff86b9610e"
A Wordlist
What approaches would you use to find the password? I know that brute-forcing all the possible combinations is the sure way but also the one taking way too long to complete (never, i think it is around 20^20 possible combinations).
What I have come up with is to filter the wordlist based on the letters that exist in the anagram and the words. Meaning, if I find a word that contains a letter that is not in the passphrase, I discard it. In addition, I wanted to take into consideration the frequencies of the characters. So I removed from the wordlist any word that has characters with frequency higher than the character frequency of the passphrase. The above was performed on word-level, meaning I checked each word individual. Eventually, from 90k+ words from the wordlist I narrowed them down to 1700 unique words that can be used for the passphrase.
Question 1: Is there anything more that can be done in order to reduce the number of possibilities? Is there a more clever way to take into account the frequencies of the letters?
Next, I thought that since I narrowed it down to 1.7k words maybe I could try permutations (itertools.permutations()) of these words that could possible match the md5 hash of passphrase. Since the anagram of the passphrase contains 2 space characters, I assumed that the passphrase is also three words and not just scrambled characters (maybe I am wrong, but at least I should try it first). As a result, I tried checking permutations of three words from the filtered wordlist. As it turned out, neither that kind of approach is fast enough for my laptop to get any results. The program reaches the memory limit, and the computer freezes.
I also thought about taking into consideration pairs of words instead of triplets and somehow match the letter frequencies in order to filter out some possibilities but I did not come up with a way to do it yet.
Question 2: Is there a way that I can get any more information about the passphrase without checking all the permutations, since this task is prohibitive for my laptop (at least for 1.7k words and above).
I tried using hashcat but I found it too complicated. I ran a couple mask attacks on the md5 hash but with no success. I tried brute-forcing it too, since it can use the GPU but it was still impossible. The main reason was that I could not understand the kind of arguments I needed to give. I know there is an extensive wiki, but as someone with close to no background about hash cracking it was not really helpful. In addition, I would prefer if there was a way to do it on my own and use other programs as less as possible.
If you have any suggestions about solving this please let me know. I am doing this for education purposes, so any input on this will be greatly appreciated.
Thanks
I want a dictionary of English words available, to pick random english words. I have a dictionary text file that I downloaded form the internet which has almost 1 million words, what's the best way to go about using this list in Clojure, given that most of the time I'll only need 1 randomly selected word?
Edit:
To answer the comments, this is for some tests which I may turn into load tests which is why I want a decent number of random words and I guess access speed is the most important thing. I do not want to use a database for this. I originally thought of a dictionary just because that's the first thing that popped into my mind but I think a random sequence of letters and numbers would be good enough, perhaps I will just use a UUID as a string.
Read all the words into a Vector and then call rand-nth , e.g.
(rand-nth all-words)
rand-nth uses the nth function on the underlying data structure and Clojure Vectors have log32N performance for index based retrieval.
Edit: This is assuming that it is for a test environment as you described in your question. A more memory efficient method would be to use RandomAccessFile and seek to a random location in the file of words, read until you find the first word delimiter (e.g. comma, EOL) and then read the following bytes until the next delimiter which will give you a random word.
As of right now, I decided to take a dictionary and iterate through the entire thing. Every time I see a newline, I make a string containing from that newline to the next newline, then I do string.find() to see if that English word is somewhere in there. This takes a VERY long time, each word taking about 1/2-1/4 a second to verify.
It is working perfectly, but I need to check thousands of words a second. I can run several windows, which doesn't affect the speed (Multithreading), but it still only checks like 10 a second. (I need thousands)
I'm currently writing code to pre-compile a large array containing every word in the English language, which should speed it up a lot, but still not get the speed I want. There has to be a better way to do this.
The strings I'm checking will look like this:
"hithisisastringthatmustbechecked"
but most of them contained complete garbage, just random letters.
I can't check for impossible compinations of letters, because that string would be thrown out because of the 'tm', in between 'thatmust'.
You can speed up the search by employing the Knuth–Morris–Pratt (KMP) algorithm.
Go through every dictionary word, and build a search table for it. You need to do it only once. Now your search for individual words will proceed at faster pace, because the "false starts" will be eliminated.
There are a lot of strategies for doing this quickly.
Idea 1
Take the string you are searching and make a copy of each possible substring beginning at some column and continuing through the whole string. Then store each one in an array indexed by the letter it begins with. (If a letter is used twice store the longer substring.
So the array looks like this:
a - substr[0] = "astringthatmustbechecked"
b - substr[1] = "bechecked"
c - substr[2] = "checked"
d - substr[3] = "d"
e - substr[4] = "echecked"
f - substr[5] = null // since there is no 'f' in it
... and so forth
Then, for each word in the dictionary, search in the array element indicated by its first letter. This limits the amount of stuff that has to be searched. Plus you can't ever find a word beginning with, say 'r', anywhere before the first 'r' in the string. And some words won't even do a search if the letter isn't in there at all.
Idea 2
Expand upon that idea by noting the longest word in the dictionary and get rid of letters from those strings in the arrays that are longer than that distance away.
So you have this in the array:
a - substr[0] = "astringthatmustbechecked"
But if the longest word in the list is 5 letters, there is no need to keep any more than:
a - substr[0] = "astri"
If the letter is present several times you have to keep more letters. So this one has to keep the whole string because the "e" keeps showing up less than 5 letters apart.
e - substr[4] = "echecked"
You can expand upon this by using the longest words starting with any particular letter when condensing the strings.
Idea 3
This has nothing to do with 1 and 2. Its an idea that you could use instead.
You can turn the dictionary into a sort of regular expression stored in a linked data structure. It is possible to write the regular expression too and then apply it.
Assume these are the words in the dictionary:
arun
bob
bill
billy
body
jose
Build this sort of linked structure. (Its a binary tree, really, represented in such a way that I can explain how to use it.)
a -> r -> u -> n -> *
|
b -> i -> l -> l -> *
| | |
| o -> b -> * y -> *
| |
| d -> y -> *
|
j -> o -> s -> e -> *
The arrows denote a letter that has to follow another letter. So "r" has to be after an "a" or it can't match.
The lines going down denote an option. You have the "a or b or j" possible letters and then the "i or o" possible letters after the "b".
The regular expression looks sort of like: /(arun)|(b(ill(y+))|(o(b|dy)))|(jose)/ (though I might have slipped a paren). This gives the gist of creating it as a regex.
Once you build this structure, you apply it to your string starting at the first column. Try to run the match by checking for the alternatives and if one matches, more forward tentatively and try the letter after the arrow and its alternatives. If you reach the star/asterisk, it matches. If you run out of alternatives, including backtracking, you move to the next column.
This is a lot of work but can, sometimes, be handy.
Side note I built one of these some time back by writing a program that wrote the code that ran the algorithm directly instead of having code looking at the binary tree data structure.
Think of each set of vertical bar options being a switch statement against a particular character column and each arrow turning into a nesting. If there is only one option, you don't need a full switch statement, just an if.
That was some fast character matching and really handy for some reason that eludes me today.
How about a Bloom Filter?
A Bloom filter, conceived by Burton Howard Bloom in 1970 is a
space-efficient probabilistic data structure that is used to test
whether an element is a member of a set. False positive matches are
possible, but false negatives are not; i.e. a query returns either
"inside set (may be wrong)" or "definitely not in set". Elements can
be added to the set, but not removed (though this can be addressed
with a "counting" filter). The more elements that are added to the
set, the larger the probability of false positives.
The approach could work as follows: you create the set of words that you want to check against (this is done only once), and then you can quickly run the "in/not-in" check for every sub-string. If the outcome is "not-in", you are safe to continue (Bloom filters do not give false negatives). If the outcome is "in", you then run your more sophisticated check to confirm (Bloom filters can give false positives).
It is my understanding that some spell-checkers rely on bloom filters to quickly test whether your latest word belongs to the dictionary of known words.
This code was modified from How to split text without spaces into list of words?:
from math import log
words = open("english125k.txt").read().split()
wordcost = dict((k, log((i+1)*log(len(words)))) for i,k in enumerate(words))
maxword = max(len(x) for x in words)
def infer_spaces(s):
"""Uses dynamic programming to infer the location of spaces in a string
without spaces."""
# Find the best match for the i first characters, assuming cost has
# been built for the i-1 first characters.
# Returns a pair (match_cost, match_length).
def best_match(i):
candidates = enumerate(reversed(cost[max(0, i-maxword):i]))
return min((c + wordcost.get(s[i-k-1:i], 9e999), k+1) for k,c in candidates)
# Build the cost array.
cost = [0]
for i in range(1,len(s)+1):
c,k = best_match(i)
cost.append(c)
# Backtrack to recover the minimal-cost string.
costsum = 0
i = len(s)
while i>0:
c,k = best_match(i)
assert c == cost[i]
costsum += c
i -= k
return costsum
Using the same dictionary of that answer and testing your string outputs
>>> infer_spaces("hithisisastringthatmustbechecked")
294.99768817854056
The trick here is finding out what threshold you can use, keeping in mind that using smaller words makes the cost higher (if the algorithm can't find any usable word, it returns inf, since it would split everything to single-letter words).
In theory, I think you should be able to train a Markov model and use that to decide if a string is probably a sentence or probably garbage. There's another question about doing this to recognize words, not sentences: How do I determine if a random string sounds like English?
The only difference for training on sentences is that your probability tables will be a bit larger. In my experience, though, a modern desktop computer has more than enough RAM to handle Markov matrices unless you are training on the entire Library of Congress (which is unnecessary- even 5 or so books by different authors should be enough for very accurate classification).
Since your sentences are mashed together without clear word boundaries, it's a bit tricky, but the good news is that the Markov model doesn't care about words, just about what follows what. So, you can make it ignore spaces, by first stripping all spaces from your training data. If you were going to use Alice in Wonderland as your training text, the first paragraph would, perhaps, look like so:
alicewasbeginningtogetverytiredofsittingbyhersisteronthebankandofhavingnothingtodoonceortwiceshehadpeepedintothebookhersisterwasreadingbutithadnopicturesorconversationsinitandwhatistheuseofabookthoughtalicewithoutpicturesorconversation
It looks weird, but as far as a Markov model is concerned, it's a trivial difference from the classical implementation.
I see that you are concerned about time: Training may take a few minutes (assuming you have already compiled gold standard "sentences" and "random scrambled strings" texts). You only need to train once, you can easily save the "trained" model to disk and reuse it for subsequent runs by loading from disk, which may take a few seconds. Making a call on a string would take a trivially small number of floating point multiplications to get a probability, so after you finish training it, it should be very fast.
Intro
I work in a facility where we have microscopes. These guys can be asked to generate 4D movies of a sample: they take e.g. 10 pictures at different Z position, then wait a certain amount of time (next timepoint) and take 10 slices again.
They can be asked to save a file for each slice, and they use an explicit naming pattern, something like 2009-11-03-experiment1-Z07-T42.tif. The file names are numbered to reflect the Z position and the time point
Question
Once you have all these file names, you can use a regex pattern to extract the Z and T value, if you know the backbone pattern of the file name. This I know how to do.
The question I have is: do you know a way to automatically generate regex pattern from the file name list? For instance, there is an awesome tool on the net that does similar thing: txt2re.
What algorithm would you use to parse all the file name list and generate a most likely regex pattern?
There is a Perl module called String::Diff which has the ability to generate a regular expression for two different strings. The example it gives is
my $diff = String::Diff::diff_regexp('this is Perl', 'this is Ruby');
print "$diff\n";
outputs:
this\ is\ (?:Perl|Ruby)
Maybe you could feed pairs of filenames into this kind of thing to get an initial regex. However, this wouldn't give you capturing of numbers etc. so it wouldn't be completely automatic. After getting the diff you would have to hand-edit or do some kind of substitution to get a working final regex.
First of all, you are trying to do this the hard way. I suspect that this may not be impossible but you would have to apply some artificial intelligence techniques and it would be far more complicated than it is worth. Either neural networks or a genetic algorithm system could be trained to recognize the Z numbers and T numbers, assuming that the format of Z[0-9]+ and T[0-9]+ is always used somewhere in the regex.
What I would do with this problem is to write a Python script to process all of the filenames. In this script, I would match twice against the filename, one time looking for Z[0-9]+ and one time looking for T[0-9]+. Each time I would count the matches for Z-numbers and T-numbers.
I would keep four other counters with running totals, two for Z-numbers and two for T-numbers. Each pair would represent the count of filenames with 1 match, and the ones with multiple matches. And I would count the total number of filenames processed.
At the end, I would report as follows:
nnnnnnnnnn filenames processed
Z-numbers matched only once in nnnnnnnnnn filenames.
Z-numbers matched multiple times in nnnnnn filenames.
T-numbers matched only once in nnnnnnnnnn filenames.
T-numbers matched multiple times in nnnnnn filenames.
If you are lucky, there will be no multiple matches at all, and you could use the regexes above to extract your numbers. However, if there are any significant number of multiple matches, you can run the script again with some print statements to show you example filenames that provoke a multiple match. This would tell you whether or not a simple adjustment to the regex might work.
For instance, if you have 23,768 multiple matches on T-numbers, then make the script print every 500th filename with multiple matches, which would give you 47 samples to examine.
Probably something like [ -/.=]T[0-9]+[ -/.=] would be enough to get the multiple matches down to zero, while also giving a one-time match for every filename. Or at worst, [0-9][ -/.=]T[0-9]+[ -/.=]
For Python, see this question about TemplateMaker.