I'm trying to go through a list to find letter combinations that don't exist in English. After a fair amount of arguing, I have a word list that I can mess with. Each word is listed as 'word\n' since each word is on a line. If I wanted to find, say, the word 'winter', if in works but only if I'm looking for 'winter\n'. I can't look just for 'winter' so I can't find individual letter pairs which is the goal.
There's over a quarter million items, so I can't cycle through the list every time, it would take ages. I don't care about index, I just need a true/false of if a letter pair is anywhere in the list.
Sorry if this was a bit rambly, I hope I got my point across. Thanks!
Assuming you don't want to alter your wordlist, it sounds like you're looking for something like this:
def search(word_list, word): # word_list is your list of words, word is the word you're searching for
for w in word_list: # iterate over the list
if w.startswith(word): # check if any of them start with the word you're looking for
return True # return true if a match is found
return False # return false if no matches are found
If you instead want to find a substring anywhere in a word instead of at the beginning, replace w.startswith(word) with word in w.
There are several ways to do that but the easiest one is like this:
flag = True
STRING = 'YOUR STRING'
def check(letter):
for k in range(33 ,127):
if chr(k) == letter:
return True
return False
for i in STRING:
if not check(i):
break
flag = False
The reason for 33 and 127 in for loop is that they are the ascii code for English words and other things(such as: ?,!,*,(,), etc)
Notice: This code is just for one string!
And also you can use regex library to do that.
You can create a variable like pattern like this:
pattern = '[A-Za-z]'
this pattern is for all of the English letters.
And then:
new_string = re.sub(pattern,STRING,'')
if new_string == '':
flag = True
else:
flag = False
sub method is just like replace and you give a pattern, a string and the replace for pattern in string.
So we replace all of the English letters in a string with '' and when there is nothing left on your string it means that your string is made of English letters.
But I'm not sure about syntax for re. You have to take look at doc.
If you are looking for a fast algorithm, DO NOT USE THE FIRST WAY! BECAUSE THE ORDER OF CODE IS O(2) FOR A SINGLE STRING(NOT A LIST)
Related
I'm newbie in Python so that I have a question. I want to change letter in word if the first letter appears more than once. Moreover I want to use input to get the word from user. I'll present the problem using an example:
word = 'restart'
After changes the word should be like this:
word = 'resta$t'
I was trying couple of ideas but always I got stuck. Is there any simple sollutions for this?
Thanks in advance.
EDIT: In response to Simas Joneliunas
It's not my homework. I'm just finished reading some basic Python tutorials and I found some questions that I couldn't solve on my own. My first thought was to separate word into a single letters and then to find out the place of the letter I want to replace by "$". I have wrote that code but I couldn't came up with sollution how to get to specific place and replace it.
word = 'restart'
how_many = {}
for x in word:
how_many=+1
else:
how_many=1
for y in how_many:
if how_many[y] > 0:
print(y,how_many[y])
Using str.replace:
s = "restart"
new_s = s[0] + s[1:].replace(s[0], "$")
Output:
'resta$t'
Try:
"".join([["$" if ch in word[:i] else ch for i, ch in enumerate(word)])
enumerate iterates through the string (i.e. a list of characters) and keeps a running index of the iteration
word[:i] checks the list of chars until the current index, i.e. previously appeared characters
"$" if ch in word[:i] else ch means replace the character at existing position with $ if it appears before others keep the character
"".join() joins the list of characters into a single string.
This is where the python console is handy and lets you experiment. Since you have to keep track of number of letters, for a good visual I would list the alphabet in a list. Then in the loop remove from the list the current letter. If letter does not exist in the list replace the letter with $.
So check if it exists first thing in the loop, if it exists, remove it, if it doesn’t exist replace it from example above.
I'm trying to solve this problem were they give me a set of strings where to count how many times a certain word appears within a string like 'code' but the program also counts any variant where the 'd' changes like 'coze' but something like 'coz' doesn't count this is what I made:
def count(word):
count=0
for i in range(len(word)):
lo=word[i:i+4]
if lo=='co': # this is what gives me trouble
count+=1
return count
Test if the first two characters match co and the 4th character matches e.
def count(word):
count=0
for i in range(len(word)-3):
if word[i:i+1] == 'co' and word[i+3] == 'e'
count+=1
return count
The loop only goes up to len(word)-3 so that word[i+3] won't go out of range.
You could use regex for this, through the re module.
import re
string = 'this is a string containing the words code, coze, and coz'
re.findall(r'co.e', string)
['code', 'coze']
from there you could write a function such as:
def count(string, word):
return len(re.findall(word, string))
Regex is the answer to your question as mentioned above but what you need is a more refined regex pattern. since you are looking for certain word appears you need to search for boundary words. So your pattern should be sth. like this:
pattern = r'\bco.e\b'
this way your search will not match with the words like testcodetest or cozetest but only match with code coze coke but not leading or following characters
if you gonna test for multiple times, then it's better to use a compiled pattern, that way it'd be more memory efficient.
In [1]: import re
In [2]: string = 'this is a string containing the codeorg testcozetest words code, coze, and coz'
In [3]: pattern = re.compile(r'\bco.e\b')
In [4]: pattern.findall(string)
Out[4]: ['code', 'coze']
Hope that helps.
I am trying to capture the sentence after a specific word. Each sentences are different in my code and those sentence doesn't necessarily have to have this specific word to split by. If the word doesn't appear, I just need like blank string or list.
Example 1: working
my_string="Python is a amazing programming language"
print(my_string.split("amazing",1)[1])
programming language
Example 2:
my_string="Java is also a programming language."
print(my_string.split("amazing",1)[1]) # amazing word doesn't appear in the sentence.
Error: IndexError: list index out of range
Output needed :empty string or list ..etc.
I tried something like this, but it still fails.
my_string.split("amazing",1)[1] if my_string.split("amazing",1)[1] == None else my_string.split("amazing",1)[1]
When you use the .split() argument you can specify what part of the list you want to use with either integers or slices. If you want to check a specific word in your string you can do is something like this:
my_str = "Python is cool"
my_str_list = my_str.split()
if 'cool' in my_str_list:
print(my_str)`
output:
"Python is cool"
Otherwise, you can run a for loop in a list of strings to check if it finds the word in multiple strings.
You have some options here. You can split and check the result:
tmp = my_string.split("amazing", 1)
result = tmp[1] if len(tmp) > 1 else ''
Or you can check for containment up front:
result = my_string.split("amazing", 1)[1] if 'amazing' in my_string else ''
The first option is more efficient if most of the sentences have matches, the second one if most don't.
Another option similar to the first is
result = my_string.split("amazing", 1)[-1]
if result == my_string:
result = ''
In all cases, consider doing something equivalent to
result = result.lstrip()
Instead of calling index 1, call index -1. This calls the last item in the list.
my_string="Java is also a programming language."
print(my_string.split("amazing",1)[1])
returns ' programming language.'
I'm working on PythonChallenge #3. I've got a huge block of text that I have to sort through. I am trying to find a sequence in which the first and last three letters are caps, and the middle one is lowercase.
My function loops through the text. The variable block stores the seven letters that are currently being looped through. There's a variable, toPrint, which gets turned on and off based on whether the letters in block correspond to my pattern (AAAaAAA). Based on the last block printed according to my function, my loop stops early in my text. I have no idea why this is happening and if you could help me figure this out, that would be great.
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
words = []
for i in text:
toPrint = True
block = text[text.index(i):text.index(i)+7]
for b in block[:3]:
if b.isupper() == False:
toPrint = False
for b in block[3]:
if b.islower() == False:
toPrint = False
for b in block[4:]:
if b.isupper() == False:
toPrint = False
if toPrint == True and block not in words:
words.append(block)
print (block)
print (words)
With Regex:
This is a really good time to use regex, it's super fast, more clear, and doesn't require a bunch of nested if statements.
import re
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
print(re.search(r"[A-Z]{3}[a-z][A-Z]{3}", text).group(0))
Explanation of regex:
[A-Z]{3] ---> matches any 3 uppercase letters
[a-z] -------> matches a single lowercase letter
[A-Z]{3] ---> matches 3 more uppercase letters
Without Regex:
If you really don't want to use regex this is how you could do it:
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
for i, _ in enumerate(text[:-6]): #loop through index of each char (not including last 6)
sevenCharacters = text[i:i+7] #create chunk of seven characters
shouldBeCapital = sevenCharacters[0:3] + sevenCharacters[4:7] #combine all the chars that should be cap into list
if (all(char.isupper() for char in shouldBeCapital)): #make sure all those characters are indeeed capital
if(sevenCharacters[3].islower()): #make sure middle character is lowercase
print(sevenCharacters)
I think your first problem is that you are using str.index(). Like find(), the .index() method of a string returns the index of the first match that is found.
Thus, in your example, whenever you search for 'x' you will get the index of the first 'x' found, etc. You cannot successfully work with any character that is not unique in the string, or that is not the first occurrence of a repeated character.
In order to keep the same structure (which isn't necessary- there is an answer posted using enumerate that I prefer myself) I implemented a queuing approach with your block variable. Each iteration, a character is dropped from the front of block, while the new character is appended to the end.
I also cleaned up some of your needless comparisons with False. You will find that this is not only inefficient, it is frequently wrong, because many of the "boolean" activities you perform will not be on actual boolean values. Get out of the habit of spelling out True/False. Just use if c or if not c.
Here's the result:
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
words = []
block = '.' + text[0:6]
for i in text[6:]:
block = block[1:] + i # Drop 1st char, append 'i'
toPrint = True
for b in block[:3]:
if not b.isupper():
toPrint = False
if not block[3].islower():
toPrint = False
for b in block[4:]:
if not b.isupper():
toPrint = False
if toPrint and block not in words:
words.append(block)
print (words)
If I understood your question, then according to my opinion there is no need of loop. My this simple code can find required sequence.
# Use this code
text = """kAewtloYgcFQaJNhHVGxXDiQmzjfcpYbzxlWrVcqsmUbCunkfxZWDZjUZMiGqhRRiUvGmYmvnJ"""
import re
print(re.findall("[A-Z]{3}[a-z][A-Z]{3}", text))
I have created a list of all possible outcomes for this specific wordgrid, doing diagonals,up,down and all the reverses too):
I have called this allWords, but when I try too find specific words I know are in the allWords the loop does not find the Hidden words. I know my problem but I do know how to go around it (sorry for terrible explanation hopefully an example below will show it better):
an Example follows: My wordList is the list of words that I know are hidden somewhere in the wordgrid. My allWords is a list of Rows,Columns,Diagonals from the wordgrid but
WordList = ['HAMMER','....']
allWords = ['ARBHAMMERTYU','...']
that HAMMER is in allWords but 'cloaked' by other characters after it so I am unable to show HAMMER is in the wordgrid.
length = len(allWords)
for i in range(length):
word = allWords[i]
if word in wordList:
print("I have found", word)
it does not find any word HAMMER in allWords.
Any help towards solving this problem would be great
You are not comparing each word in wordList to a word in allWords. The line if word in wordList compares the exact word.
i.e.
if word in wordList will return True only if the word Hammer is in wordList.
To match substring you need another loop:
for i in range(length):
word = allWords[i]
for w in WordList:
if w in word:
print("I have found ", word)
If I understand your problem correctly, you probably need to implement a function that checks if a token (e.g. 'HAMMER') is present in any of the entries in allWords. My best bet for solution would be to use regular expressions.
import re
def findWordInWordList(word, allWords):
pattern = re.compile(".*%s.*" % word)
for item in allWords:
match = pattern.search(item)
if match:
return match
This will return first occurence, if you want more then it's easy to collect them in a list.
You could try something like this:
for word in allWords:
if word in WordList:
print("I have found", word)
Ah, or maybe the error is that you wrote wordList and you really defined WordList. Hope this helps.
If I understand correctly, you are trying to find a match inside allWords and you want to iterate over WordList and determine if there is a substring match.
So, if that is correct, then your code is not exactly doing that. To go through your code step by step to correct what is happening:
length = len(allWords)
for i in range(length):
What you want to do above is not necessarily go over your allWords. You want to iterate over WordList and see if it is inside allWords. You are not doing that, instead you want to do this:
length = len(WordList)
for i in range(length):
With that in mind, that means now you want to reference WordList and not allWords, so you want to now change this:
word = allWords[i]
to this:
word = WordList[i]
Finally, here comes a new bit of information to determine if you in fact have a substring match in the strings you are matching. A method called "any". The "any" method works by returning True if at least one match of what you are looking for is found. It looks like this:
any(if "something" in word in word for words)
Then it will return True if it "something" is in word otherwise it will return False.
So, to put this all together, and run your code with your sample input, we get:
WordList = ['HAMMER','....']
allWords = ['ARBHAMMERTYU','...']
length = len(WordList)
for i in range(length):
word = WordList[i]
if any(word in w for w in allWords):
print("I have found", word)
Output:
I have found HAMMER