I want to have something like this code from Python
num=3
res=str(num)
but in Maple. I couldn't find any appropriate constructors for this. Are there any?
num:=3:
convert(num,string);
"3"
sprintf("%a",num);
"3"
The best way is to use convert as already exists in #acer's answer. Just to name one more possibility here is another way.
num := 3:
res := cat( "", num );
You will get "3" for res of type string. What cat here does is concatenating 3 to the empty string "", and when there exists at least one string in the arguments of cat, the output becomes a string. You can even have something like sqrt(2) instead of 3 in num, in that case res becomes this string; "2^(1/2)". But sometimes it may give you a non-string object, for example if the number in num is of the form RootOf. See the help page to read more.
Related
I have a string say example "https://www.google.com" and a count of paging say 5
how do i iterate the URL to append p/page=incrementing numbers of paging and store the result in a variable as list?
"https://www.google.com/page=1"
"https://www.google.com/page=2"
"https://www.google.com/page=3"
"https://www.google.com/page=4"
"https://www.google.com/page=5"
So the end result will look like this having a variable query_var which will hold a list of string example below
query_var:("https://www.google.com/page=1";"https://www.google.com/page=2";"https://www.google.com/page=3";"https://www.google.com/page=4";"https://www.google.com/page=5");
count query_var \\5
You can use the join function , with the each-right adverb /:
query_var: "https://www.google.com/page=" ,/: string 1_til 6
Make it a function to support a varied count of pages:
f:{"https://www.google.com/page=" ,/: string 1_til x+1}
q)10=count f[10]
1b
Q1
You don’t even need a lambda to make this a function. It’s a sequence of unary functions, so you can compose them.
q)f: "https://www.google.com/page=",/: string ::
q)f 1+til 6
"https://www.google.com/page=1"
"https://www.google.com/page=2"
"https://www.google.com/page=3"
"https://www.google.com/page=4"
"https://www.google.com/page=5"
"https://www.google.com/page=6"
Q2
q)("https://www.google.com";;"info/history")
enlist["https://www.google.com";;"info/history"]
q)"/"sv'("https://www.google.com";;"info/history")#/: string `aapl`msft`nftx
"https://www.google.com/aapl/info/history"
"https://www.google.com/msft/info/history"
"https://www.google.com/nftx/info/history"
List notation is syntactic sugar for enlist. The list with a missing item is a projection of enlist and can be iterated.
Again, a sequence of unaries is all composable without a lambda:
q)g: "/"sv'("https://www.google.com";;"info/history")#/: string ::
q)g `aapl`msft`nftx
"https://www.google.com/aapl/info/history"
"https://www.google.com/msft/info/history"
"https://www.google.com/nftx/info/history"
If you can assume a unique character that doesn't appear elsewhere in your url (e.g. #) then ssr is a simple and easily-readable approach:
ssr["https://www.google.com/page=#";"#";]each string 1+til 5
ssr["https://www.google.com/#/info/history";"#";]each string`aapl`msft`nftx
Using q’s like function, how can we achieve the following match using a single regex string regstr?
q) ("foo7"; "foo8"; "foo9"; "foo10"; "foo11"; "foo12"; "foo13") like regstr
>>> 0111110b
That is, like regstr matches the foo-strings which end in the numbers 8,9,10,11,12.
Using regstr:"foo[8-12]" confuses the square brackets (how does it interpret this?) since 12 is not a single digit, while regstr:"foo[1[0-2]|[1-9]]" returns a type error, even without the foo-string complication.
As the other comments and answers mentioned, this can't be done using a single regex. Another alternative method is to construct the list of strings that you want to compare against:
q)str:("foo7";"foo8";"foo9";"foo10";"foo11";"foo12";"foo13")
q)match:{x in y,/:string z[0]+til 1+neg(-/)z}
q)match[str;"foo";8 12]
0111110b
If your eventual goal is to filter on the matching entries, you can replace in with inter:
q)match:{x inter y,/:string z[0]+til 1+neg(-/)z}
q)match[str;"foo";8 12]
"foo8"
"foo9"
"foo10"
"foo11"
"foo12"
A variation on Cillian’s method: test the prefix and numbers separately.
q)range:{x+til 1+y-x}.
q)s:"foo",/:string 82,range 7 13 / include "foo82" in tests
q)match:{min(x~/:;in[;string range y]')#'flip count[x]cut'z}
q)match["foo";8 12;] s
00111110b
Note how unary derived functions x~/: and in[;string range y]' are paired by #' to the split strings, then min used to AND the result:
q)flip 3 cut's
"foo" "foo" "foo" "foo" "foo" "foo" "foo" "foo"
"82" ,"7" ,"8" ,"9" "10" "11" "12" "13"
q)("foo"~/:;in[;string range 8 12]')#'flip 3 cut's
11111111b
00111110b
Compositions rock.
As the comments state, regex in kdb+ is extremely limited. If the number of trailing digits is known like in the example above then the following can be used to check multiple patterns
q)str:("foo7"; "foo8"; "foo9"; "foo10"; "foo11"; "foo12"; "foo13"; "foo3x"; "foo123")
q)any str like/:("foo[0-9]";"foo[0-9][0-9]")
111111100b
Checking for a range like 8-12 is not currently possible within kdb+ regex. One possible workaround is to write a function to implement this logic. The function range checks a list of strings start with a passed string and end with a number within the range specified.
range:{
/ checking for strings starting with string y
s:((c:count y)#'x)like y;
/ convert remainder of string to long, check if within range
d:("J"$c _'x)within z;
/ find strings satisfying both conditions
s&d
}
Example use:
q)range[str;"foo";8 12]
011111000b
q)str where range[str;"foo";8 12]
"foo8"
"foo9"
"foo10"
"foo11"
"foo12"
This could be made more efficient by checking the trailing digits only on the subset of strings starting with "foo".
For your example you can pad, fill with a char, and then simple regex works fine:
("."^5$("foo7";"foo8";"foo9";"foo10";"foo11";"foo12";"foo13")) like "foo[1|8-9][.|0-2]"
I have the following string:
{\"Id\":\"135\",\"Type\":0}
The number in the Id field will vary, but will always be an integer with no comma separator. I'm not sure how to get just that value from that string given that it's string data type and not real "XML". I was toying with the replace() function, but the special characters are making it more complex than it seems it needs to be.
is there a way to convert that to XML or something that I can reference the Id value directly?
Maybe use a regular expression, e.g.
import re
txt = "{\"Id\":\"135\",\"Type\":0}"
x = re.search('"Id":"([0-9]+)"', txt)
if x:
print(x.group(1))
gives
135
It is assumed here that the ids are numeric and consist of at least one digit.
Non-regex answer as you asked
\" is an escape sequence in python.
So if {\"Id\":\"135\",\"Type\":0} is a raw string and if you put it into a python variable like
a = '{\"Id\":\"135\",\"Type\":0}'
gives
>>> a
'{"Id":"135","Type":0}'
OR
If the above string is python string which has \" which is already escaped, then do a.replace("\\","") which will give you the string without \.
Now just load this string into a dict and access element Id like below.
import json
d = json.loads(a)
d['Id']
Output :
135
Imagine that I wanted to take the characters from a string in Scala but have the toInt conversion to behave as it would on a string instead of as on a character.
To illustrate the following code behaves like so:
"0".toInt // results in 0
"000".charAt(0).toInt // results in 48
I'd like a version of the second line that would also result in 0. I have a solution like the following:
"000".charAt(0).toString.toInt // results in 0
But I wonder if there is a more direct or better way?
You can use asDigit:
val i: Int = "000".charAt(0).asDigit
You can do:
"000".substring(0, 1).toInt
But I'm not sure it's more "direct" than "000".charAt(0).toString.toInt
In SML NJ, I want to find whether a string is substring of another string and find its index. Can any one help me with this?
The Substring.position function is the only one I can find in the basis library that seems to do string search. Unfortunately, the Substring module is kind of hard to use, so I wrote the following function to use it. Just pass two strings, and it will return an option: NONE if not found, or SOME of the index if it is found:
fun index (str, substr) = let
val (pref, suff) = Substring.position substr (Substring.full str)
val (s, i, n) = Substring.base suff
in
if i = size str then
NONE
else
SOME i
end;
Well you have all the substring functions, however if you want to also know the position of it, then the easiest is to do it yourself, with a linear scan.
Basically you want to explode both strings, and then compare the first character of the substring you want to find, with each character of the source string, incrementing a position counter each time you fail. When you find a match you move to the next char in the substring as well without moving the position counter. If the substring is "empty" (modeled when you are left with the empty list) you have matched it all and you can return the position index, however if the matching suddenly fail you have to return back to when you had the first match and skip a letter (incrementing the position counter) and start all over again.
Hope this helps you get started on doing this yourself.