Need to extract the initial character from a Korean word in MS-Excel and MS-Access.
When I use Left("한글",1) it will return the first syllable i.e 한, what I need is the initial character i.e ㅎ .
Is there a function to do this? or at least an idiom?
If you know how to get the Unicode value from the String I'd be able to work it out from there but I'm sure I'd be reinventing the wheel. (yet again)
Disclaimer: I know little about Access or VBA, but what you're having is a generic Unicode problem, it's not specific to those tools. I retagged your question to add tags related to this issue.
Access is doing the right thing by returning 한, it is indeed the first character of that two-character string. What you want here is the canonical decomposition of this hangul in its constituent jamos, also known as Normalization Form D (NFD), for “decomposed”. The NFD form is ᄒ ᅡ ᆫ, of which the first character is what you want.
Note also that as per your example, you seem to want a function to return the equivalent hangul (ㅎ) for the jamo (ᄒ) – there really are two different code points because they represent different semantic units (a full-fledged hangul syllable, or a part of a hangul). There is no pre-defined mapping from the former to the latter, you could write a small function to that effect, as the number of jamos is limited to a few dozens (the real work is done in the first function, NFD).
Adding to Arthur's excellent answer, I want to point out that extracting jamo from hangeul syllables is very straightforward from the standard. While the solution isn't specific to Excel or Access (it's a Python module), it only involves arithmetic expressions so it should be easily translated to other languages. The formulas, as can be seen, are identical to those in page 109 of the standard. The decomposition is returned as a tuple of integers encoded strings, which can be easily verified to correspond to the Hangul Jamo Code Chart.
# -*- encoding: utf-8 -*-
SBase = 0xAC00
LBase = 0x1100
VBase = 0x1161
TBase = 0x11A7
SCount = 11172
LCount = 19
VCount = 21
TCount = 28
NCount = VCount * TCount
def decompose(syllable):
global SBase, LBase, VBase, TBase, SCount, LCount, VCount, TCount, NCount
S = ord(syllable)
SIndex = S - SBase
L = LBase + SIndex / NCount
V = VBase + (SIndex % NCount) / TCount
T = TBase + SIndex % TCount
if T == TBase:
result = (L,V)
else:
result = (L,V,T)
return tuple(map(unichr, result))
if __name__ == '__main__':
test_values = u'항가있닭넓짧'
for syllable in test_values:
print syllable, ':',
for s in decompose(syllable): print s,
print
This is the output in my console:
항 : ᄒ ᅡ ᆼ
가 : ᄀ ᅡ
있 : ᄋ ᅵ ᆻ
닭 : ᄃ ᅡ ᆰ
넓 : ᄂ ᅥ ᆲ
짧 : ᄍ ᅡ ᆲ
I think what you are looking for is a Byte Array
Dim aByte() as byte
aByte="한글"
should give you the two unicode values for each character in the string
I assume you got what you needed, but it seems rather convoluted. I don't know anything about this, but recently did some investigating of handling Unicode, and looked into all the string Byte functions, such as LeftB(), RightB(), InputB(), InStrB(), LenB(), AscB(), ChrB() and MidB(), and there's also StrConv(), which has a vbUnicode argument. These are all functions that I'd think would be used in any double-byte context, but then, I don't work in that environment so might be missing something very important.
Related
I will state the obvious that I am a beginner. I should also mention that I have been coding in Zybooks, which affects things. My textbook hasn't helped me much
I tried sub_lyric= rhyme_lyric[ : ]
Zybooks should be able to input an index number can get only that part of the sentence but my book doesnt explain how to do that. If it throws a [4:7] then it would output cow. Hopefully I have exolained everything well.
You need to set there:
sub_lyric = rhyme_lyric[start_index:end_index]
The string is as a sequence of characters and you can use string slicing to extract any sub-text from the main one. As you have observed:
sub_lyric = rhyme_lyric[:]
will copy the entire content of rhyme_lyric to sub_lyric.
To select only a portion of the text, specify the start_index (strings start with index 0) to end_index (not included).
sub_lyric = rhyme_lyric[4:7]
will extract characters in rhyme_lyric from position 4 (included) to position 7 (not included) so the result will be cow.
You can check more on string slicing here: Python 3 introduction
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).
I'm extremely new to python and I have no idea why this code gives me this output. I tried searching around for an answer but couldn't find anything because I'm not sure what to search for.
An explain-like-I'm-5 explanation would be greatly appreciated
astring = "hello world"
print(astring[3:7:2])
This gives me : "l"
Also
astring = "hello world"
print(astring[3:7:3])
gives me : "lw"
I can't wrap my head around why.
This is string slicing in python.
Slicing is similar to regular string indexing, but it can return a just a section of a string.
Using two parameters in a slice, such as [a:b] will return a string of characters, starting at index a up to, but not including, index b.
For example:
"abcdefg"[2:6] would return "cdef"
Using three parameters performs a similar function, but the slice will only return the character after a chosen gap. For example [2:6:2] will return every second character beginning at index 2, up to index 5.
ie "abcdefg"[2:6:2] will return ce, as it only counts every second character.
In your case, astring[3:7:3], the slice begins at index 3 (the second l) and moves forward the specified 3 characters (the third parameter) to w. It then stops at index 7, returning lw.
In fact when using only two parameters, the third defaults to 1, so astring[2:5] is the same as astring[2:5:1].
Python Central has some more detailed explanations of cutting and slicing strings in python.
I have a feeling you are over complicating this slightly.
Since the string astring is set statically you could more easily do the following:
# Sets the characters for the letters in the consistency of the word
letter-one = "h"
letter-two = "e"
letter-three = "l"
letter-four = "l"
letter-six = "o"
letter-7 = " "
letter-8 = "w"
letter-9 = "o"
letter-10 = "r"
letter11 = "l"
lettertwelve = "d"
# Tells the python which of the character letters that you want to have on the print screen
print(letter-three + letter-7 + letter-three)
This way its much more easily readable to human users and it should mitigate your error.
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.
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')