Output will not become a string from a list - python

import os
import random
file = open('getty.txt')
filetext = file.read()
def getline(words,length):
ans=[]
total=0
while (length>total) and 0 != len(words):
word=words.pop(0)
total += len(word)+1 #add 1 for the space
ans.append(word)
#now we are one word too long
if total > length:
words.insert(0,ans.pop())
return ans
def printPara(words,length):
line = []
spaces = []
while len(words) != 0:
line.append(getline(words, length))
for z in range(0,len(line)):
for i in range(0,len(line[z])):
spaces = [[1] * len(line[i]) for i in range (len(line))]
for p in range (0,len(spaces)):
spaces[p][len(spaces[p])-1] = 0
if len(words) + len(spaces) != 0:
addSpace(line,spaces,length)
printLine(line,spaces)
else:
printLine(line,spaces)
def addSpace(line,spaces,length):
totalInt = 0
for i in range (0, len(line)):
totalInt = (len(spaces[i])-2) + len(line[i])
while length < totalInt:
num = random.randint(0, len(spaces) - 2)
spaces[num] += 1
return spaces
def printLine(line, spaces):
for i in range (len(line)):
print(str(line[i]) + (' ' * len(spaces[i])))
def main():
length = 75
textparagraph = filetext.split("\n\n")
para = [0] * len(textparagraph)
for i in range (0, len(textparagraph)):
para[i] = textparagraph[i]
words = [[0] * len(textparagraph) for i in range(len(para))]
for b in range (0,len(para)):
words[b] = para[b].split()
for z in range (0, len(para)):
printPara(words[z],length)
main()
My code outputs only lists of the separate lines and will not concatenate the two lists of words and spaces. How would I get it to output correctly?
Some exampes of output.
['Four', 'score', 'and', 'seven', 'years', 'ago', 'our', 'fathers', 'brought', 'forth', 'on', 'this']
['continent,', 'a', 'new', 'nation,', 'conceived', 'in', 'Liberty,', 'and', 'dedicated', 'to', 'the']
['proposition', 'that', 'all', 'men', 'are', 'created', 'equal.']
['Now', 'we', 'are', 'engaged', 'in', 'a', 'great', 'civil', 'war,', 'testing', 'whether', 'that', 'nation,', 'or']
['any', 'nation', 'so', 'conceived', 'and', 'so', 'dedicated,', 'can', 'long', 'endure.', 'We', 'are', 'met', 'on', 'a']
Expected output "Four score and seven years ago..."


You can use
" ".join(["hello", "world"])
and you'll get
"hello world"

Related

Removing accents from keyword strings

This is a word processing code for chabot, in it it removes some articles and prepositions to make it easier for the bot to read
import json
from random import choice
class ChatterMessage:
def __init__(self, raw):
self.raw = str(raw).lower()
self.processed_str = self.reduce()
self.responses = self.get_responses()
self.data = self.process_response()
self.response = choice(self.data['response'])
def remove_unwanted_chars(self, string):
list_of_chars = ['?', ".", ",", "!", "#", "[", "]", "{", "}", "#", "$", "%", "*", "&", "(", ")", "-", "_", "+", "="]
new_str = ""
for char in string:
if char not in list_of_chars:
new_str += str(char)
return new_str
def get_responses(self, response_file="info.json"):
with open(response_file, 'r') as file:
return json.loads(file.read())
def reduce(self):
stopwords = ['de', 'a', 'o', 'que', 'e', 'é', 'do', 'da', 'em', 'um', 'para', 'com', 'não', 'uma', 'os', 'no', 'se', 'na', 'por', 'mais', 'as', 'dos', 'como', 'mas', 'ao', 'ele', 'das', 'à', 'seu', 'sua', 'ou', 'quando', 'muito', 'nos', 'já', 'eu', 'também', 'só', 'pelo', 'pela', 'até', 'isso', 'ela', 'entre', 'depois', 'sem', 'mesmo', 'aos', 'seus', 'quem', 'nas', 'me', 'esse', 'eles', 'você', 'essa', 'num', 'nem', 'suas', 'meu', 'às', 'minha', 'numa', 'pelos', 'elas', 'qual', 'nós', 'lhe', 'deles', 'essas', 'esses', 'pelas', 'este', 'dele', 'tu', 'te', 'vocês', 'vos', 'lhes', 'meus', 'minhas', 'teu', 'tua', 'teus', 'tuas', 'nosso', 'nossa', 'nossos', 'nossas', 'dela', 'delas', 'esta', 'estes', 'estas', 'aquele', 'aquela', 'aqueles', 'aquelas', 'isto', 'aquilo', 'estou', 'está', 'estamos', 'estão', 'estive', 'esteve', 'estivemos', 'estiveram', 'estava', 'estávamos', 'estavam', 'estivera', 'estivéramos', 'esteja', 'estejamos', 'estejam', 'estivesse', 'estivéssemos', 'estivessem', 'estiver', 'estivermos', 'estiverem', 'hei', 'há', 'havemos', 'hão', 'houve', 'houvemos', 'houveram', 'houvera', 'houvéramos', 'haja', 'hajamos', 'hajam', 'houvesse', 'houvéssemos', 'houvessem', 'houver', 'houvermos', 'houverem', 'houverei', 'houverá', 'houveremos', 'houverão', 'houveria', 'houveríamos', 'houveriam', 'sou', 'somos', 'são', 'era', 'éramos', 'eram', 'fui', 'foi', 'fomos', 'foram', 'fora', 'fôramos', 'seja', 'sejamos', 'sejam', 'fosse', 'fôssemos', 'fossem', 'for', 'formos', 'forem', 'serei', 'será', 'seremos', 'serão', 'seria', 'seríamos', 'seriam', 'tenho', 'tem', 'temos', 'tém', 'tinha', 'tínhamos', 'tinham', 'tive', 'teve', 'tivemos', 'tiveram', 'tivera', 'tivéramos', 'tenha', 'tenhamos', 'tenham', 'tivesse', 'tivéssemos', 'tivessem', 'tiver', 'tivermos', 'tiverem', 'terei', 'terá', 'teremos', 'terão', 'teria', 'teríamos', 'teriam']
custom_filter = []
keywords_list = []
strlist = self.raw.split(" ")
for x in strlist:
if x not in stopwords and x not in custom_filter:
keywords_list.append(self.remove_unwanted_chars(x))
return keywords_list
def process_response(self):
percentage = lambda x, y: (100 * y) / x
total = sum(len(x['keywords']) for x in self.responses)
most_acc = 0
response_data = None
acc = 0
for value in self.responses:
c = 0
for x in value['keywords']:
if str(x).lower() in self.processed_str:
c += 1
if c > most_acc:
most_acc = c
acc = percentage(total, most_acc)
print(acc)
response_data = value
if acc < 6:
return {"response": "Sorry, I do not understand. Be more clear please"}
for x in self.processed_str:
if x not in response_data['keywords']:
response_data['keywords'].append(x)
return response_data
if __name__ == '__main__':
while True:
k = input("Você: ")
res = ChatterMessage(k)
.response
print("Bot:", res)
How to remove accents from keyword strings to "make it easier" for chatbot to read? I found this explanation: How to remove string accents using Python 3? But I don't know how it would be applied to this code as the bot always stops responding

You could use the Python package unidecode that replaces special characters with ASCII equivalents.
from unidecode import unidecode
text = "Björn, Łukasz and Σωκράτης."
print(unidecode(text))
# ==> Bjorn, Lukasz and Sokrates.
You could apply this to both the input and keywords.
# In the function definition of reduce(), place this line of code after
# stopwords = ['de', 'a', 'o', .....])
stopwords = [unidecode(s) for s in stopwords]
# In "__main__": replace k = input("Você: ") with the following line of code.
k = unidecode(input("Você: "))
If it makes sense, you could also force the strings to be all lowercase. This will make your string comparisons even more robust.
k = unidecode(input("Você: ").lower())
Because you requested the entire code:
import json
from random import choice
from unidecode import unidecode
class ChatterMessage:
def __init__(self, raw):
self.raw = str(raw).lower()
self.processed_str = self.reduce()
self.responses = self.get_responses()
self.data = self.process_response()
self.response = choice(self.data['response'])
def remove_unwanted_chars(self, string):
list_of_chars = ['?', ".", ",", "!", "#", "[", "]", "{", "}", "#", "$", "%", "*", "&", "(", ")", "-", "_", "+", "="]
new_str = ""
for char in string:
if char not in list_of_chars:
new_str += str(char)
return new_str
def get_responses(self, response_file="info.json"):
with open(response_file, 'r') as file:
return json.loads(file.read())
def reduce(self):
stopwords = ['de', 'a', 'o', 'que', 'e', 'é', 'do', 'da', 'em', 'um', 'para', 'com', 'não', 'uma', 'os', 'no', 'se', 'na', 'por', 'mais', 'as', 'dos', 'como', 'mas', 'ao', 'ele', 'das', 'à', 'seu', 'sua', 'ou', 'quando', 'muito', 'nos', 'já', 'eu', 'também', 'só', 'pelo', 'pela', 'até', 'isso', 'ela', 'entre', 'depois', 'sem', 'mesmo', 'aos', 'seus', 'quem', 'nas', 'me', 'esse', 'eles', 'você', 'essa', 'num', 'nem', 'suas', 'meu', 'às', 'minha', 'numa', 'pelos', 'elas', 'qual', 'nós', 'lhe', 'deles', 'essas', 'esses', 'pelas', 'este', 'dele', 'tu', 'te', 'vocês', 'vos', 'lhes', 'meus', 'minhas', 'teu', 'tua', 'teus', 'tuas', 'nosso', 'nossa', 'nossos', 'nossas', 'dela', 'delas', 'esta', 'estes', 'estas', 'aquele', 'aquela', 'aqueles', 'aquelas', 'isto', 'aquilo', 'estou', 'está', 'estamos', 'estão', 'estive', 'esteve', 'estivemos', 'estiveram', 'estava', 'estávamos', 'estavam', 'estivera', 'estivéramos', 'esteja', 'estejamos', 'estejam', 'estivesse', 'estivéssemos', 'estivessem', 'estiver', 'estivermos', 'estiverem', 'hei', 'há', 'havemos', 'hão', 'houve', 'houvemos', 'houveram', 'houvera', 'houvéramos', 'haja', 'hajamos', 'hajam', 'houvesse', 'houvéssemos', 'houvessem', 'houver', 'houvermos', 'houverem', 'houverei', 'houverá', 'houveremos', 'houverão', 'houveria', 'houveríamos', 'houveriam', 'sou', 'somos', 'são', 'era', 'éramos', 'eram', 'fui', 'foi', 'fomos', 'foram', 'fora', 'fôramos', 'seja', 'sejamos', 'sejam', 'fosse', 'fôssemos', 'fossem', 'for', 'formos', 'forem', 'serei', 'será', 'seremos', 'serão', 'seria', 'seríamos', 'seriam', 'tenho', 'tem', 'temos', 'tém', 'tinha', 'tínhamos', 'tinham', 'tive', 'teve', 'tivemos', 'tiveram', 'tivera', 'tivéramos', 'tenha', 'tenhamos', 'tenham', 'tivesse', 'tivéssemos', 'tivessem', 'tiver', 'tivermos', 'tiverem', 'terei', 'terá', 'teremos', 'terão', 'teria', 'teríamos', 'teriam']
stopwords = [unidecode(s) for s in stopwords]
custom_filter = []
keywords_list = []
strlist = self.raw.split(" ")
for x in strlist:
if x not in stopwords and x not in custom_filter:
keywords_list.append(self.remove_unwanted_chars(x))
return keywords_list
def process_response(self):
percentage = lambda x, y: (100 * y) / x
total = sum(len(x['keywords']) for x in self.responses)
most_acc = 0
response_data = None
acc = 0
for value in self.responses:
c = 0
for x in value['keywords']:
if str(x).lower() in self.processed_str:
c += 1
if c > most_acc:
most_acc = c
acc = percentage(total, most_acc)
print(acc)
response_data = value
if acc < 6:
return {"response": "Sorry, I do not understand. Be more clear please"}
for x in self.processed_str:
if x not in response_data['keywords']:
response_data['keywords'].append(x)
return response_data
if __name__ == '__main__':
while True:
k = unidecode(input("Você: "))
res = ChatterMessage(k).response
print("Bot:", res)

why does this it say in the console Process finished with exit code 0 instead of printing the 'sen' variable? [duplicate]

This question already has answers here:
How to check if type of a variable is string? [duplicate]
(22 answers)
Closed 2 years ago.
import random
import sys
def v1_debug(v1, subject):
if v1 != str and subject != str:
sys.exit()
else:
if subject == 'He' or 'She' or 'It':
for i in v1:
if i == [len(v1)+1]:
if i == 's' or 'z' or 'x' or 'o':
v1 = v1 + 'es'
elif i == 'y':
v1 = v1 - 'y' + 'ies'
elif v1[len(v1)] == 's' and v1[len(v1)+1] == 'h':
v1 = v1 + 'es'
elif v1[len(v1)] == 'c' and v1[len(v1)+1] == 'h':
v1 = v1 + 'es'
if subject == 'I' or 'You' or 'We' or 'They':
for i in v1:
if i == v1[len(v1)+1]:
v1 = v1 + 'ing'
return ''
def default_positive_form():
try:
sbj = ['He',
'She',
'It',
'I',
'You',
'We',
'They']
v1 = ['be',
'beat',
'become',
'begin',
'bend',
'bet',
'bid',
'bite',
'blow',
'break',
'bring',
'build',
'burn',
'buy',
'catch',
'choose',
'come',
'cost',
'cut',
'dig',
'dive',
'do',
'draw',
'dream',
'drive',
'drink',
'eat',
'fall',
'feel',
'fight',
'find',
'fly',
'forget',
'forgive',
'freeze',
'get',
'give',
'go',
'grow',
'hang',
'have',
'hear',
'hide',
'hit',
'hold',
'hurt',
'keep',
'know',
'lay',
'lead',
'leave',
'lend',
'let',
'lie',
'lose',
'make',
'mean',
'meet',
'pay',
'put',
'read',
'ride',
'ring',
'rise',
'run',
'say',
'see',
'sell',
'send',
'show',
'shut',
'sing',
'sit',
'sleep',
'speak',
'spend',
'stand',
'swim',
'take',
'teach',
'tear',
'tell',
'think',
'throw',
'understand',
'wake',
'wear',
'win',
'write']
sbj = random.choice(sbj)
v1 = random.choice(v1)
verb_debug = v1_debug(v1, sbj)
sen = ''
if sbj == 'I':
sen = sbj + 'am' + verb_debug
elif sbj == 'He' or 'She' or 'It':
sen = sbj + 'is' + verb_debug
elif sbj == 'You' or 'We' or 'They':
sen = sbj + 'are' + verb_debug
print(f'{sen}')
except NameError:
print('this is bullshit')
return
default_positive_form()
this is python 3.8

sen will only consist of an empty string if none of the conditions of your if/elif/elif blocks are met. Change the print line to
print(f"sen is: {sen}")
But that's not the real problem. obj != str does not check if obj is a string, it checks to see if the object is pointing to the type constant str (Thanks Charles Duffy for the comment). Instead, use the builtin function isinstance() like so:
if not isinstance(v1, str) and not isinstance(subject, str):
print("Variables are the wrong type!")
sys.exit()

Replace multiple instances of a sub-string with items in a list

I have a string like below:
e = "how are you how do you how are they how"
My expected output is:
out = "how1 are you how2 do you how3 are they how4"
I'm trying in the following way:
def givs(y,x):
tt = []
va = [i+1 for i in list(range(y.count(x)))]
for i in va:
tt.append(x+str(i))
return tt
ls = givs(e, 'how')
ls = ['how1', 'how2', 'how3', 'how4']
fg = []
for i in e.split(' '):
fg.append(i)
fg = ['how', 'are', 'you', 'how', 'do', 'you', 'how', 'are', 'they', 'how']
For every instance of 'how' in 'fg' I want to replace with items in 'ls' and finally use join function to get the required output.
expected_output = ['how1', 'are', 'you', 'how2', 'do', 'you', 'how3', 'are',
'they', 'how4']
so that I can join the items by:
' '.join(expected_output)
to get:
out = "how1 are you how2 do you how3 are they how4"

You could use itertools.count:
from itertools import count
counter = count(1)
e = "how are you how do you how are they how"
result = ' '.join([w if w != "how" else w + str(next(counter)) for w in e.split()])
print(result)
Output
how1 are you how2 do you how3 are they how4

There is no need to make your code complex, just add a counter and add it to every "how". At the end make the new string.
e = "how are you how do you how are they how"
e_ok = ""
count = 1
for i in e.split():
if i == "how":
i = i+str(count)
count += 1
e_ok += i + " "
print(e_ok)

Finding word context with regular expressions

I have created a function to search for the contexts of a given word(w) in a text, with left and right as parameters for flexibility in the number of words to record.
import re
def get_context (text, w, left, right):
text.insert (0, "*START*")
text.append ("*END*")
all_contexts = []
for i in range(len(text)):
if re.match(w,text[i], 0):
if i < left:
context_left = text[:i]
else:
context_left = text[i-left:i]
if len(text) < (i+right):
context_right = text[i:]
else:
context_right = text[i:(i+right+1)]
context = context_left + context_right
all_contexts.append(context)
return all_contexts
So for example if a have a text in the form of a list like this:
text = ['Python', 'is', 'dynamically', 'typed', 'language', 'Python',
'functions', 'really', 'care', 'about', 'what', 'you', 'pass', 'to',
'them', 'but', 'you', 'got', 'it', 'the', 'wrong', 'way', 'if', 'you',
'want', 'to', 'pass', 'one', 'thousand', 'arguments', 'to', 'your',
'function', 'then', 'you', 'can', 'explicitly', 'define', 'every',
'parameter', 'in', 'your', 'function', 'definition', 'and', 'your',
'function', 'will', 'be', 'automagically', 'able', 'to', 'handle',
'all', 'the', 'arguments', 'you', 'pass', 'to', 'them', 'for', 'you']
The function works fine for example:
get_context(text, "function",2,2)
[['language', 'python', 'functions', 'really', 'care'], ['to', 'your', 'function', 'then', 'you'], ['in', 'your', 'function', 'definition', 'and'], ['and', 'your', 'function', 'will', 'be']]
Now I am trying to build a dictionary with the contexts of every word in the text doing the following:
d = {}
for w in set(text):
d[w] = get_context(text,w,2,2)
But I am getting this error.
Traceback (most recent call last):
File "<pyshell#32>", line 2, in <module>
d[w] = get_context(text,w,2,2)
File "<pyshell#20>", line 9, in get_context
if re.match(w,text[i], 0):
File "/usr/lib/python3.4/re.py", line 160, in match
return _compile(pattern, flags).match(string)
File "/usr/lib/python3.4/re.py", line 294, in _compile
p = sre_compile.compile(pattern, flags)
File "/usr/lib/python3.4/sre_compile.py", line 568, in compile
p = sre_parse.parse(p, flags)
File "/usr/lib/python3.4/sre_parse.py", line 760, in parse
p = _parse_sub(source, pattern, 0)
File "/usr/lib/python3.4/sre_parse.py", line 370, in _parse_sub
itemsappend(_parse(source, state))
File "/usr/lib/python3.4/sre_parse.py", line 579, in _parse
raise error("nothing to repeat")
sre_constants.error: nothing to repeat
I don't understand this error. Can anyone help me with this?

The problem is that "*START*" and "*END*" are being interpreted as regex. Also, note that inserting "*START*" and "*END*" in text in the begging of the function will cause problem. You should do it just once.
Here is a complete version of the working code:
import re
def get_context(text, w, left, right):
all_contexts = []
for i in range(len(text)):
if re.match(w,text[i], 0):
if i < left:
context_left = text[:i]
else:
context_left = text[i-left:i]
if len(text) < (i+right):
context_right = text[i:]
else:
context_right = text[i:(i+right+1)]
context = context_left + context_right
all_contexts.append(context)
return all_contexts
text = ['Python', 'is', 'dynamically', 'typed', 'language',
'Python', 'functions', 'really', 'care', 'about', 'what',
'you', 'pass', 'to', 'them', 'but', 'you', 'got', 'it', 'the',
'wrong', 'way', 'if', 'you', 'want', 'to', 'pass', 'one',
'thousand', 'arguments', 'to', 'your', 'function', 'then',
'you', 'can', 'explicitly', 'define', 'every', 'parameter',
'in', 'your', 'function', 'definition', 'and', 'your',
'function', 'will', 'be', 'automagically', 'able', 'to', 'handle',
'all', 'the', 'arguments', 'you', 'pass', 'to', 'them', 'for', 'you']
text.insert(0, "START")
text.append("END")
d = {}
for w in set(text):
d[w] = get_context(text,w,2,2)
Maybe you can replace re.match(w,text[i], 0) with w == text[i].

The whole thing can be re-written very succinctly follows,
text = 'Python is dynamically typed language Python functions really care about what you pass to them but you got it the wrong way if you want to pass one thousand arguments to your function then you can explicitly define every parameter in your function definition and your function will be automagically able to handle all the arguments you pass to them for you'
Keeping it a str, assuming context = 'function',
pat = re.compile(r'(\w+\s\w+\s)functions?(?=(\s\w+\s\w+))')
pat.findall(text)
[('language Python ', ' really care'),
('to your ', ' then you'),
('in your ', ' definition and'),
('and your ', ' will be')]
Now, minor customization will be needed in the regex to allow for, words like say, functional or functioning not only function or functions. But the important idea is to do away with indexing and go more functional.
Please comment if this doesn't work out for you, when you apply it in bulk.

At least one of the elements in text contains characters that are special in a regular expression. If you're just trying to find whether the word is in the string, just use str.startswith, i.e.
if text[i].startswith(w): # instead of re.match(w,text[i], 0):
But I don't understand why you are checking for that anyways, and not for equality.

Python find words within a user input word game

I'm trying to write a program that allows the user to input a word then find all words of length 4 or greater hidden within that word that are in a word text file. So far my code can detect the words in the user inputted word that aren't jumbled. For example if I type in houses, the output will show house, houses, ho, us, use, uses. It should also recognize hose, hoses, shoe, shoes, hues, etc.
I know itertools is the simplest solution but I want to use a different method using only loops, dictionaries, and lists.
Here is my code so far:
def main():
filename = open('dictionary.txt').readlines()
word_list = []
for line in filename:
word_list.append(line.strip())
print 'Lets Play Words within a Word!\n'
word = raw_input('Enter a word: ')
words_left = 0
for words in word_list:
letters = list(words)
if words in word:
print words
words_left += 1
else:
False
The output format I'm trying to create should look like so:
Lets play Words within a Word!
Enter a word: exams #user inputted word
exams --- 6 words are remaining
> same #user types in guess
Found! # prints 'Found!' if above word is found in the dictionary.txt file
exams --- 5 words are remaining
> exam
Found!
exams --- 4 words are remaining
> mesa
Found!
exams --- 3 words are remaining
> quit() #if they type this command in the game will end
So my question is, after entering the base word (in the ex it's EXAMS), how do I determine the total number of words within that word and if the user inputted word guesses are in the text file? Also print if the word was found.

something like this should work...
wordlist=[list of words]
solutionlist=[]
userword=[userword[i] for i in range(len(userword))]
for word in wordlist:
inword=True
letters=[word[j] for j in range(len(word))]
for letter in set(letters):
if letters.count(letter)>userword.count(letter):
inword=False
break
if inword:
solutionlist.append(word)
for line in solutionlist:
print line

This works:
# read from file in actual implementation
all_words = [
"foo", "bar", "baz", "hose", "hoses", "shoe", "shoes", "hues", "house",
"houses", "ho", "us", "use", "uses", "shoe", "same", "exam", "mesa", "mass"]
RETAIN_ORDERING = False
def matches(inp, word):
if inp[0] == word[0]:
return (
True if len(word) == 1 else
False if len(inp) == 1 else
matches(inp[1:], word[1:]))
else:
return matches(inp[1:], word) if len(inp) >= 2 else False
# with sorting enabled, "houses" will also match "shoe"; otherwise not
def maybe_sort(x):
return x if RETAIN_ORDERING else ''.join(sorted(x))
inp = raw_input("enter a word: ")
results = [word for word in all_words if matches(maybe_sort(inp), maybe_sort(word))]
print results
Output:
$ python matches.py
enter a word: houses
['hose', 'hoses', 'shoe', 'shoes', 'hues', 'house', 'houses', 'ho', 'us', 'use', 'uses', 'shoe']
$ python matches.py
enter a word: exams
['same', 'exam', 'mesa']
If you want to avoid matches like shoe where the ordering of letters is not the same as in the input, just set RETAIN_ORDERING = True.

A naive implementation (using collections.Counter):
>>> all_words = ['foo', 'bar', 'baz', 'hose', 'hoses', 'shoe', 'shoes', 'hues', 'house', 'houses', 'ho', 'us', 'use', 'uses', 'shoe', 'same', 'exam', 'mesa', 'mass']
>>> def find_hidden(user_input):
from collections import Counter
user_word_counts = Counter(user_input)
for word in all_words:
isvalid = True
for letter, count in Counter(word).iteritems():
if user_word_counts[letter] == 0 or user_word_counts[letter] < count:
isvalid = False
break
if isvalid: yield word
>>> list(find_hidden("houses"))
['hose', 'hoses', 'shoe', 'shoes', 'hues', 'house', 'houses', 'ho', 'us', 'use', 'uses', 'shoe']
>>> list(find_hidden("exams"))
['same', 'exam', 'mesa']
Or,
Using permutations:
>>> all_words = ['foo', 'bar', 'baz', 'hose', 'hoses', 'shoe', 'shoes', 'hues', 'house', 'houses', 'ho', 'us', 'use', 'uses', 'shoe', 'same', 'exam', 'mesa', 'mass']
>>> def permutations(s, n): # could use itertools.permutations
if n == 1:
for c in s:
yield c
for i in range(len(s)):
for p in permutation(s[:i]+s[i+1:], n-1):
yield s[i] + p
>>> def find_hidden(input_str):
for word_len in range(2, len(input_str)+1):
for word in permutations(input_str, word_len):
if word in all_words:
yield word
>>> set(find_hidden("houses"))
set(['use', 'hose', 'shoes', 'houses', 'house', 'us', 'hues', 'hoses', 'uses', 'ho', 'shoe'])
>>> set(find_hidden("exams"))
set(['mesa', 'exam', 'same'])

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