I'm stuck with something that usually is pretty easily in other programming languages.
I want to test whether a string is inside another one in R. For example I tried:
match("Diagnosi Prenatale,Esercizio Fisico", "Diagnosi Prenatale")
pmatch("Diagnosi Prenatale,Esercizio Fisico", "Diagnosi Prenatale")
grep("Diagnosi Prenatale,Esercizio Fisico", "Diagnosi Prenatale")
And none worked. To make it work I should fist split the first string with strsplit and extract the first element.
NOTE: I'd like to do this on a vector of strings to receive a yes/no vector, so in the function I wrote should go a vector not a single string. But of course if the single string doesn't work, image a full vector of them...
Any ideas?
Try grepl
grepl("Diagnosi Prenatale","Diagnosi Prenatale,Esercizio Fisico" )
[1] TRUE
You can also do this with character vectors, for example:
x <- c("Diagnosi Prenatale,Esercizio Fisico", "Diagnosi Prenatale")
grepl("Diagnosi Prenatale",x)
#[1] TRUE TRUE
Related
I'm trying to use pyparsing to build a parser that will match on all text within an arbitrarily nested set of brackets. If we consider a string like this:
"[A,[B,C],[D,E,F],G] Random Middle text [H,I,J]"
What I would like is for a parser to match in a way that it returns two matches:
[
"[A,[B,C],[D,E,F],G]",
"[H,I,J]"
]
I was able to accomplish a somewhat-working version of this using a barrage of originalTextFor mashed up with nestedExpr, but this breaks when your nesting is deeper than the number of OriginalTextFor expressions.
Is there a straightforward way to only match on the outermost expression grabbed by nestedExpr, or a way to modify its logic so that everything after the first paired match is treated as plaintext rather than being parsed?
update: One thing that seems to come close to what I want to accomplish is this modified version of the logic from nestedExpr:
def mynest(opener='{', closer='}'):
content = (empty.copy()+CharsNotIn(opener+closer+ParserElement.DEFAULT_WHITE_CHARS))
ret = Forward()
ret <<= ( Suppress(opener) + originalTextFor(ZeroOrMore( ret | content )) + Suppress(closer) )
return ret
This gets me most of the way there, although there's an extra level of list wrapping in there that I really don't need, and what I'd really like is for those brackets to be included in the string (without getting into an infinite recursion situation by not suppressing them).
parser = mynest("[","]")
result = parser.searchString("[A,[B,C],[D,E,F],G] Random Middle text [H,I,J]")
result.asList()
>>> [['A,[B,C],[D,E,F],G'], ['H,I,J']]
I know I could strip these out with a simple list comprehension, but it would be ideal if I could just eliminate that second, redundant level.
Not sure why this wouldn't work:
sample = "[A,[B,C],[D,E,F],G] Random Middle text [H,I,J]"
scanner = originalTextFor(nestedExpr('[',']'))
for match in scanner.searchString(sample):
print(match[0])
prints:
'[A,[B,C],[D,E,F],G]'
'[H,I,J]'
What is the situation where "this breaks when your nesting is deeper than the number of OriginalTextFor expressions"?
Suppose we are given two strings s1 and s2(both lowercase). We have two find the minimal lexographic string that can be formed by merging two strings.
At the beginning , it looks prettty simple as merge of the mergesort algorithm. But let us see what can go wrong.
s1: zyy
s2: zy
Now if we perform merge on these two we must decide which z to pick as they are equal, clearly if we pick z of s2 first then the string formed will be:
zyzyy
If we pick z of s1 first, the string formed will be:
zyyzy which is correct.
As we can see the merge of mergesort can lead to wrong answer.
Here's another example:
s1:zyy
s2:zyb
Now the correct answer will be zybzyy which will be got only if pick z of s2 first.
There are plenty of other cases in which the simple merge will fail. My question is Is there any standard algorithm out there used to perform merge for such output.
You could use dynamic programming. In f[x][y] store the minimal lexicographical string such that you've taken x charecters from the first string s1 and y characters from the second s2. You can calculate f in bottom-top manner using the update:
f[x][y] = min(f[x-1][y] + s1[x], f[x][y-1] + s2[y]) \\ the '+' here represents
\\ the concatenation of a
\\ string and a character
You start with f[0][0] = "" (empty string).
For efficiency you can store the strings in f as references. That is, you can store in f the objects
class StringRef {
StringRef prev;
char c;
}
To extract what string you have at certain f[x][y] you just follow the references. To udapate you point back to either f[x-1][y] or f[x][y-1] depending on what your update step says.
It seems that the solution can be almost the same as you described (the "mergesort"-like approach), except that with special handling of equality. So long as the first characters of both strings are equal, you look ahead at the second character, 3rd, etc. If the end is reached for some string, consider the first character of the other string as the next character in the string for which the end is reached, etc. for the 2nd character, etc. If the ends for both strings are reached, then it doesn't matter from which string to take the first character. Note that this algorithm is O(N) because after a look-ahead on equal prefixes you know the whole look-ahead sequence (i.e. string prefix) to include, not just one first character.
EDIT: you look ahead so long as the current i-th characters from both strings are equal and alphabetically not larger than the first character in the current prefix.
I am trying to produce the following:The new values of x and y are -4 and 7, respectively, using the disp and num2str commands. I tried to do this disp('The new values of x and y are num2str(x) and num2str(y) respectively'), but it gave num2str instead of the appropriate values. What should I do?
Like Colin mentioned, one option would be converting the numbers to strings using num2str, concatenating all strings manually and feeding the final result into disp. Unfortunately, it can get very awkward and tedious, especially when you have a lot of numbers to print.
Instead, you can harness the power of sprintf, which is very similar in MATLAB to its C programming language counterpart. This produces shorter, more elegant statements, for instance:
disp(sprintf('The new values of x and y are %d and %d respectively', x, y))
You can control how variables are displayed using the format specifiers. For instance, if x is not necessarily an integer, you can use %.4f, for example, instead of %d.
EDIT: like Jonas pointed out, you can also use fprintf(...) instead of disp(sprintf(...)).
Try:
disp(['The new values of x and y are ', num2str(x), ' and ', num2str(y), ', respectively']);
You can actually omit the commas too, but IMHO they make the code more readable.
By the way, what I've done here is concatenated 5 strings together to form one string, and then fed that single string into the disp function. Notice that I essentially concatenated the string using the same syntax as you might use with numerical matrices, ie [x, y, z]. The reason I can do this is that matlab stores character strings internally AS numeric row vectors, with each character denoting an element. Thus the above operation is essentially concatenating 5 numeric row vectors horizontally!
One further point: Your code failed because matlab treated your num2str(x) as a string and not as a function. After all, you might legitimately want to print "num2str(x)", rather than evaluate this using a function call. In my code, the first, third and fifth strings are defined as strings, while the second and fourth are functions which evaluate to strings.
I want to calculate the frequency of each word in a string. For that I need to turn string into an array (matrix) of words.
For example take "Hello world, can I ask you on a date?" and turn it into
['Hello' 'world,' 'can' 'I' 'ask' 'you' 'on' 'a' 'date?']
Then I can go over each entry and count every appearance of a particular word.
Is there a way to make an array (matrix) of words in MATLAB, instead of array of just chars?
Here is a little simpler regexp:
words = regexp(s,'\w+','match');
\w here means any symbol that can appear in words (including underscore).
Notice that the last question mark will not be included. Do you need it for counting words actually?
Regular expressions
s = 'Hello world, can I ask you on a date?'
slist = regexp(s, '[^ ]*', 'match')
yield
slist =
'Hello' 'world,' 'can' 'I' 'ask' 'you' 'on' 'a' 'date?'
Another way to do it is like this:
s = cell(java.lang.String('Hello world, can I ask you on a date?').split('[^\w]+'));
I.e. by creating a Java String object and using its methods to do the work, then converting back to a cell array of strings. Not necessarily the best way to do a job this simple, but Java has a rich library of string handling methods & classes that can come in handy.
Matlab's ability to switch into Java at the drop of a hat can come in handy sometimes - for example, when parsing & writing XML.
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')