I want to search for a query (a string) in a subject (another string).
The query may appear in whole or in parts, but will not be rearranged. For instance, if the query is 'da', and the subject is 'dura', it is still a match.
I am not allowed to use string functions like strfind or find.
The constraints make this actually quite straightforward with a single loop. Imagine you have two indices initially pointing at the first character of both strings, now compare them - if they don't match, increment the subject index and try again. If they do, increment both. If you've reached the end of the query at that point, you've found it. The actual implementation should be simple enough, and I don't want to do all the work for you ;)
If this is homework, I suggest you look at the explanation which precedes the code and then try for yourself, before looking at the actual code.
The code below looks for all occurrences of chars of the query string within the subject string (variables m; and related ii, jj). It then tests all possible orders of those occurrences (variable test). An order is "acceptable" if it contains all desired chars (cond1) in increasing positions (cond2). The result (variable result) is affirmative if there is at least one acceptable order.
subject = 'this is a test string';
query = 'ten';
m = bsxfun(#eq, subject.', query);
%'// m: test if each char of query equals each char of subject
[ii jj] = find(m);
jj = jj.'; %'// ii: which char of query is found within subject...
ii = ii.'; %'// jj: ... and at which position
test = nchoosek(1:numel(jj),numel(query)).'; %'// test all possible orders
cond1 = all(jj(test) == repmat((1:numel(query)).',1,size(test,2)));
%'// cond1: for each order, are all chars of query found in subject?
cond2 = all(diff(ii(test))>0);
%// cond2: for each order, are the found chars in increasing positions?
result = any(cond1 & cond2); %// final result: 1 or 0
The code could be improved by using a better approach as regards to test, i.e. not testing all possible orders given by nchoosek.
Matlab allows you to view the source of built-in functions, so you could always try reading the code to see how the Matlab developers did it (although it will probably be very complex). (thanks Luis for the correction)
Finding a string in another string is a basic computer science problem. You can read up on it in any number of resources, such as Wikipedia.
Your requirement of non-rearranging partial matches recalls the bioinformatics problem of mapping splice variants to a genomic sequence.
You may solve your problem by using a sequence alignment algorithm such as Smith-Waterman, modified to work with all English characters and not just DNA bases.
Is this question actually from bioinformatics? If so, you should tag it as such.
Related
I want to check if string A is just a reordered version of string B. For example, "abc" = "bca" = "cab"...
There are other solutions here: https://www.geeksforgeeks.org/check-if-two-strings-are-permutation-of-each-other/
However, I was thinking a hash function would be an easy way of doing this, but the typical hash function takes order into consideration. Are there any hash functions that do not care about character order?
Are there any hash functions that do not care about character order?
I don't know of real-world hash functions that have this property, no. Because this is not a problem they are designed to solve.
However, in this specific case, you can make your own "hash" function (a very very bad one) that will indeed ignore order: just sum ASCII codes of characters. This works due to the commutative property of addition (a + b == b + a)
def isAnagram(self,a,b):
sum_a = 0
sum_b = 0
for c in a:
sum_a += ord(c)
for c in b:
sum_b += ord(c)
return sum_a == sum_b
To reiterate, this is absolutely a hack, that only happens to work because input strings are limited in content in the judge system (only have lowercase ASCII characters and do not contain spaces). It will not work (reliably) on arbitrary strings.
For a fast check you could use a kind af hash-funkction
Candidates are:
xor all characters of a String
add all characters of a String
multiply all characters of a String (be careful might lead to overflow for large Strings)
If the hash-value is equal, it could still be a collision of two not 'equal' strings. So you still need to make a dedicated compare. (e.g. sort the characters of each string before comparing them).
This is one of the practice problems from Problem solving section of Hackerrank. The problem statement says
Steve has a string of lowercase characters in range ascii[‘a’..’z’]. He wants to reduce the string to its shortest length by doing a series of operations. In each operation he selects a pair of adjacent lowercase letters that match, and he deletes them.
For example : 'aaabbccc' -> 'ac' , 'abba' -> ''
I have tried solving this using slicing of strings but this gives me timeout runtime error on larger strings. Is there anything else to be used?
My code:
s = list(input())
i=1
while i<len(s):
if s[i]==s[i-1]:
s = s[:i-1]+s[i+1:]
i = i-2
i+=1
if len(s)==0:
print("Empty String")
else:
print(''.join(s))
This gives me terminated due to timeout message.
Thanks for your time :)
Interning each new immutable string can be expensive,
as it has O(N) linear cost with the length of the string.
Consider processing "aa" * int(1e6).
You will write on the order of 1e12 characters to memory
by the time you're finished.
Take a moment (well, take linear time) to
copy each character over to a mutable list element:
[c for c in giant_string]
Then you can perform dup processing by writing a tombstone
of "" to each character you wish to delete,
using just constant time.
Finally, in linear time you can scan through the survivors using "".join( ... )
One other possible solution is to use regex. The pattern ([a-z])\1 matches a duplicate lowercase letter. The implementation would involve something like this:
import re
pattern = re.compile(r'([a-z])\1')
while pattern.search(s): # While match is found
s = pattern.sub('', s) # Remove all matches from "s"
I'm not an expert at efficiency, but this seems to write fewer strings to memory than your solution. For the case of "aa" * int(1e6) that J_H mentioned, it will only write one, thanks to pattern.sub replacing all occurances at once.
Today I've finally decided to make an account, in hope for some aid in an issue I've spent the last few hours hunting. (I've spent the past couple hours hunting down a response, from Google to here to Unity Answers. Here's everything that I've found so far, which doesn't work.)
What I'm looking for, is to change a string of purely words/letters into an integer. Therefore "Hello World", would be translated into a string of numbers accordingly. This may be surprising, but this is a lot harder than it sounds. I've found a way to do essentially everything but, thus far.
Presumably the best way would be to get the ASCII value of each letter in the string, and put them all together into a single integer. (No sequences or need to separate them, but one single number.) I have no idea where to get started or how to do that, however. Really anything that you think would work, preferably as short-hand and un-bothersome as possible.
To be as clear as possible, I need to take the letter-only variable "example" and transmorph it to be a integer/only a sequence of numbers.
If you're just trying to convert an arbitrary string into a random seed, then why not try randomSeed.GetHashCode()? That will return an int value suitable for setting the seed, which would produce the same number each time the same string is entered.
You can iterate over all characters, get their charCode and chain them together. The first method splits the string into single chars and uses Array.reduce:
var str = 'qwertzuiop';
var num = parseInt(str.split('').reduce(function(a, b) {return a + b.charCodeAt(0);}, '');
The second calls Array.forEach on the string, because it has numerical indices and a length property.
var num = ''; [].forEach.call(str, function(c) {num += c.charCodeAt(0);});
num = parseInt(num);
In stoneaged browsers you have to use for-loops instead.
Say I have a string
local a = "Hello universe"
I find the substring "universe" by
a:find("universe")
Now, suppose the string is
local a = "un#verse"
The string to be searched is universe; but the substring differs by a single character.
So obviously Lua ignores it.
How do I make the function find the string even if there is a discrepancy by a single character?
If you know where the character would be, use . instead of that character: a:find("un.verse")
However, it looks like you're looking for a fuzzy string search. It is out of a scope for a Lua string library. You may want to start with this article: http://ntz-develop.blogspot.com/2011/03/fuzzy-string-search.html
As for Lua fuzzy search implementations — I haven't used any, but googing "lua fuzzy search" gives a few results. Some are based on this paper: http://web.archive.org/web/20070518080535/http://www.heise.de/ct/english/97/04/386/
Try https://github.com/ajsher/luafuzzy.
It sounds like you want something along the lines of TRE:
TRE is a lightweight, robust, and efficient POSIX compliant regexp matching library with some exciting features such as approximate (fuzzy) matching.
Approximate pattern matching allows matches to be approximate, that is, allows the matches to be close to the searched pattern under some measure of closeness. TRE uses the edit-distance measure (also known as the Levenshtein distance) where characters can be inserted, deleted, or substituted in the searched text in order to get an exact match. Each insertion, deletion, or substitution adds the distance, or cost, of the match. TRE can report the matches which have a cost lower than some given threshold value. TRE can also be used to search for matches with the lowest cost.
A Lua binding for it is available as part of lrexlib.
If you are really looking for a single character difference and do not care about performance, here is a simple approach that should work:
local a = "Hello un#verse"
local myfind = function(s,p)
local withdot = function(n)
return p:sub(1,n-1) .. '.' .. p:sub(n+1)
end
local a,b
for i=1,#s do
a,b = s:find(withdot(i))
if a then return a,b end
end
end
print(myfind(a,"universe"))
A simple roll your own approach (based on the assumption that the pattern keeps the same length):
function hammingdistance(a,b)
local ta={a:byte(1,-1)}
local tb={b:byte(1,-1)}
local res = 0
for k=1,#a do
if ta[k]~=tb[k] then
res=res+1
end
end
print(a,b,res) -- debugging/demonstration print
return res
end
function fuz(s,pat)
local best_match=10000
local best_location
for k=1,#s-#pat+1 do
local cur_diff=hammingdistance(s:sub(k,k+#pat-1),pat)
if cur_diff < best_match then
best_location = k
best_match = cur_diff
end
end
local start,ending = math.max(1,best_location),math.min(best_location+#pat-1,#s)
return start,ending,s:sub(start,ending)
end
s=[[Hello, Universe! UnIvErSe]]
print(fuz(s,'universe'))
Disclaimer: not recommended, just for fun:
If you want a better syntax (and you don't mind messing with standard type's metatables) you could use this:
getmetatable('').__sub=hammingdistance
a='Hello'
b='hello'
print(a-b)
But note that a-b does not equal b-a this way.
Is there any way to replace a character at position N in a string in Lua.
This is what I've come up with so far:
function replace_char(pos, str, r)
return str:sub(pos, pos - 1) .. r .. str:sub(pos + 1, str:len())
end
str = replace_char(2, "aaaaaa", "X")
print(str)
I can't use gsub either as that would replace every capture, not just the capture at position N.
Strings in Lua are immutable. That means, that any solution that replaces text in a string must end up constructing a new string with the desired content. For the specific case of replacing a single character with some other content, you will need to split the original string into a prefix part and a postfix part, and concatenate them back together around the new content.
This variation on your code:
function replace_char(pos, str, r)
return str:sub(1, pos-1) .. r .. str:sub(pos+1)
end
is the most direct translation to straightforward Lua. It is probably fast enough for most purposes. I've fixed the bug that the prefix should be the first pos-1 chars, and taken advantage of the fact that if the last argument to string.sub is missing it is assumed to be -1 which is equivalent to the end of the string.
But do note that it creates a number of temporary strings that will hang around in the string store until garbage collection eats them. The temporaries for the prefix and postfix can't be avoided in any solution. But this also has to create a temporary for the first .. operator to be consumed by the second.
It is possible that one of two alternate approaches could be faster. The first is the solution offered by Paŭlo Ebermann, but with one small tweak:
function replace_char2(pos, str, r)
return ("%s%s%s"):format(str:sub(1,pos-1), r, str:sub(pos+1))
end
This uses string.format to do the assembly of the result in the hopes that it can guess the final buffer size without needing extra temporary objects.
But do beware that string.format is likely to have issues with any \0 characters in any string that it passes through its %s format. Specifically, since it is implemented in terms of standard C's sprintf() function, it would be reasonable to expect it to terminate the substituted string at the first occurrence of \0. (Noted by user Delusional Logic in a comment.)
A third alternative that comes to mind is this:
function replace_char3(pos, str, r)
return table.concat{str:sub(1,pos-1), r, str:sub(pos+1)}
end
table.concat efficiently concatenates a list of strings into a final result. It has an optional second argument which is text to insert between the strings, which defaults to "" which suits our purpose here.
My guess is that unless your strings are huge and you do this substitution frequently, you won't see any practical performance differences between these methods. However, I've been surprised before, so profile your application to verify there is a bottleneck, and benchmark potential solutions carefully.
You should use pos inside your function instead of literal 1 and 3, but apart from this it looks good. Since Lua strings are immutable you can't really do much better than this.
Maybe
"%s%s%s":format(str:sub(1,pos-1), r, str:sub(pos+1, str:len())
is more efficient than the .. operator, but I doubt it - if it turns out to be a bottleneck, measure it (and then decide to implement this replacement function in C).
With luajit, you can use the FFI library to cast the string to a list of unsigned charts:
local ffi = require 'ffi'
txt = 'test'
ptr = ffi.cast('uint8_t*', txt)
ptr[1] = string.byte('o')