Find and arrange words with same letters from list - dictionary

I am making a word game and I am wondering how to find and arrange words in a list that have the same letters. In my game, you are presented with 5 letters, and you then have to rearrange the letters tp make a word. So the word could be "acorn", but those 5 letters could also make up "narco" or "racon" which are words in my .txt dictionary (shown on the left).
Example, third line on the right:
Any tips or help would be greatly appreciated!

This is a typical anagram problem, You can store the words as Hashmap key-ed on the sorted version of the word, so a list like
clay, lacy, note, tern, tone, nest, sent
would be stored as hashmap of
acly => {clay , lacy}
enot => {tone, note}
enst => {sent, nest}
enrt => {tern}
Easy to build this, sudo algorithm is
Hashmap<String, HashSet<String>> h;
foreach word in list
String sorted = sort(word); \\so if word is sent, sorted is enst
HashSet<String> currentSet = h.get(sorted);
currentSet.add(word);

Related

Extracting dict keys from values

I am still learning about python and I face some trouble extracting data from a dict. I need to create a loop which check each values and extract the keys. So for this code I need to find the nice students. I am stuck at line 3 #blank.
How do i go about this?
Thanks in advance
class = {"James":"naughty", "Lisa":"nice", "Bryan":"nice"}
for student in class:
if #blank:
print("Hello, "+student+" students!")
else:
print("odd")
Uses dictionary methods "keys(), values(), items()":
def get_students_by_criteria(student_class, criteria):
students = []
for candidate, value in student_class.items():
if value == criteria:
students.append(candidate)
return students
my_class = {"James":"naughty", "Lisa":"nice", "Bryan":"nice"}
print(get_students_by_criteria(my_class, "nice"))
Warning to the word "class" it is a keyword reserved for python programming oriented object

Update dictionary key inside list using map function -Python

I have a dictionary of phone numbers where number is Key and country is value. I want to update the key and add country code based on value country. I tried to use the map function for this:
print('**Exmaple: Update phone book to add Country code using map function** ')
user=[{'952-201-3787':'US'},{'952-201-5984':'US'},{'9871299':'BD'},{'01632 960513':'UK'}]
#A function that takes a dictionary as arg, not list. List is the outer part
def add_Country_Code(aDict):
for k,v in aDict.items():
if(v == 'US'):
aDict[( '1+'+k)]=aDict.pop(k)
if(v == 'UK'):
aDict[( '044+'+k)]=aDict.pop(k)
if (v == 'BD'):
aDict[('001+'+k)] =aDict.pop(k)
return aDict
new_user=list(map(add_Country_Code,user))
print(new_user)
This works partially when I run, output below :
[{'1+952-201-3787': 'US'}, {'1+1+1+952-201-5984': 'US'}, {'001+9871299': 'BD'}, {'044+01632 960513': 'UK'}]
Notice the 2nd US number has 2 additional 1s'. What is causing that?How to fix? Thanks a lot.
Issue
You are mutating a dict while iterating it. Don't do this. The Pythonic convention would be:
Make a new_dict = {}
While iterating the input a_dict, assign new items to new_dict.
Return the new_dict
IOW, create new things, rather than change old things - likely the source of your woes.
Some notes
Use lowercase with underscores when defining variable names (see PEP 8).
Lookup values rather than change the input dict, e.g. a_dict[k] vs. a_dict.pop(k)
Indent the correct number of spaces (see PEP 8)

Xquery/eXist-db - iterating over a collection +/- 3 positions from current document

In eXist-db I have hundreds of documents in /db/apps/foo/resources/documents like so:
...
BNF9992-J305-1.xml
BNF9992-J305-5.xml
BNF9992-J308-9.xml
BNF9992-J310-8.xml
BNF9992-J311-1.xml
BNF9992-J312-6.xml
BNF9992-J312-7.xml
BNF9992-J315-9.xml
BNF9992-J316-2.xml
BNF9992-J317-2.xml
BNF9992-J319-3.xml
...
Imagine I want to present to the user a list of 3 documents appearing before and after a specific document (based on alpha-numeric sort). So, my 'current document' is BNF9992-J312-7.xml, and I want to show the user something like:
BNF9992-J310-8.xml
BNF9992-J311-1.xml
BNF9992-J312-6.xml
BNF9992-J312-7.xml (current document)
BNF9992-J315-9.xml
BNF9992-J316-2.xml
BNF9992-J317-2.xml
Is there a function/method in Xquery 3.1 for iterating up/down a list of documents once they've been retrieved. The most I've been able to do is a simple retrieval of document names from a collection:
for $resource in collection("/db/apps/foo/resources/documents")
let $uri := base-uri($resource)
return util:unescape-uri(replace($uri, ".+/(.+)$","$1"), "UTF-8")
But I don't know how to iterate up and down the list from a given document.
Perhaps writing the list into nodes and applying a formula to node ordinals?
Many thanks.
If this were a list of strings $list, and the "current string" is $s, then I would do
let $i := index-of($list, $s)
return subsequence($list, $i - 3, 7)
I'm not sure whether the fact that you have a list of documents (rather than strings) changes this.

In Biztalk mapper how to use split array concept

Required suggestion on below part.please any one give solution.
We have mapping from 850 to FlatFile
X12/PO1Loop1/PO1/PO109 and I need to map to field VALUE which is under record Option which is unbounded.
Split PO109 into substrings delimited by '.', foreach subsring after the first, create new Option with value=substring
So in input sample we have value like 147895632qwerqtyuui.789456123321456987
Similarly the field repeats under POLoop1.
So I need to split value based on (.) then pass a value to value field under option record(unbounded).
I tried using below code snippet
public string SplitValues(string strValue)
{
string[] arrValue = strValue.Split(".".ToCharArray());
foreach (string strDisplay in arrValue)
{
return strDisplay;
}
}
But it doesn't works, and I am not really familiar with the String methods and I am not sure if there's an easy way to do this. I have a String which contains couple of values delimited with "." .
So I need to separate values based on delimiter(.) and pass value to field.
How can I do this
As I mentioned, not too clear what is your objective, but I think you want to split a node that has some kind of delimiter into multiple nodes... if so, Try this: https://seroter.wordpress.com/2008/10/07/splitting-delimited-values-in-biztalk-maps/
He is doing exactly that. Given a node with a|b|c|d as value, output multiple nodes, each containing the value after splitted by |, so node1 = a, node2 = b, node3 = c, node4 = d.

How to extract results from asp.net regex.match?

Coming from perl, I'm I bit confused by the asp.net regex classes.
I have a simple pattern I'm trying to match: "number text number"
My code looks like:
Match results = Regex.Match(mystring, #"(\d+)\s+(Highway|Hwy|Route|Rte)\s+(\d+)",RegexOptions.IgnoreCase);
foreach (Group g in results.Groups)
{
string token = g.Value;
}
The problem is that the groups seems to contain 4 results, not the 3 I would expect - the first is the entire string that gets matched, while the next 3 are what I would expect.
Is there a simple way to directly access my 3 results?
You could use Matches:
// Define a test string.
string text = "The the quick brown fox fox jumped over the lazy dog dog.";
// Find matches.
MatchCollection matches = rx.Matches(text);
// Report the number of matches found.
Console.WriteLine("{0} matches found in:\n {1}",
matches.Count,
text);
// Report on each match.
foreach (Match match in matches)
{
...
}
var results = Regex.Match("55 Hwy 66", #"(\d+)\s+(Highway|Hwy|Route|Rte)\s+(\d+)", RegexOptions.IgnoreCase).Groups.OfType<Group>().Select((name, index) => new {name, index}).Where(x => x.index > 0).Select(x => x.name).ToList();
This is just a case of how it is designed to work, and it is just a case of ignoring the first match. I do agree that it is a strange implementation and not how I would have expected it to work.
If the regular expression engine can find a match, the first element of the GroupCollection object returned by the Groups property contains a string that matches the entire regular expression pattern.
Taken from here
I know this is an old question, but I ended up here through a search confirming my own thoughts and there was no definitive answer.

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