I have a list of strings, and I'm trying to find the position of the first letter of the substring I am searching for in the list of strings. I'm using the find() method to do this, however when I try to print the position of the first letter Python returns the correct position but then throws a -1 after it, like it couldn't find the substring, but only after it could find it. I want to know how to return the position of the first letter of he substring without returning a -1 after the correct value.
Here is my code:
mylist = ["blasdactiverehu", "sdfsfgiuyremdn"]
word = "active"
if any(word in x for x in mylist) == True:
for x in mylist:
position = x.find(word)
print(position)
The output is:
5
-1
I expected the output to just be:
5
I think it may be related to the fact the loop is searching for the substring for every string in the list and after it's found the position it still searches for more but of course returns an error as there is only one occurrence of the substring "active", however I'm not sure how to stop searching after successfully finding one substring. Any help is appreciated, thank you.
Indeed your code will not work as you want it to, since given that any of the words contain the substring, it will do the check for each and every one of them.
A good way to avoid that is using a generator. More specifically, next()
default_val = '-1'
position = next((x.find(word) for x in mylist if word in x), default_val)
print(position)
It will simply give you the position of the substring "word" for the first string "x" that will qualify for the condition if word in x, in the list 'mylist'.
By the way, no need to check for == True when using any(), it already returns True/False, so you can simply do if any(): ...
Related
i get a TypeError saying can't convert 'list' object to str implicitly
# list of words to choose from and get a random word using the random module's sample method
list_of_words = ['apple', 'banana', 'watermelon', 'kiwi', 'pineapple', 'mango']
word = random.sample(list_of_words,1)
# showcase to the player the random word with first letter missing
chosen_word = word[1:]
print('The answer so far is _' + chosen_word)
I expect the output to be for example: The answer so far is _anana
random.sample picks multiple random elements from the input sequence, according to the number given for the second argument. Because it is meant to return a variable number of elements, it returns them in a list, even if you specified that you only want one.
Therefore word is a list with one element and
word[1:]
is a sliced version of that list (the empty list), but still a list. The error message tells you, that you cannot add a list to a string by relying on an implicit conversion of the list to a string representation. What you actually want to apply [1:] to is not the list, but the element in it. So you need to select it first with word[0] (take the first element of word), then you can slice the string: word[0][1:].
If you always want only one element, you can also use random.choice instead of random.sample, which doesn't take the second argument and always returns only one randomly chosen element, not as list:
word = random.choice(list_of_words)
chosen_word = word[1:]
So, I'm doing an exercise using python, and I tried to use the terminal to do step by step to understand what's happening but I didn't.
I want to understand mainly why the conditional return just the index 0.
Looking 'casino' in [Casinoville].lower() isn't the same thing?
Exercise:
Takes a list of documents (each document is a string) and a keyword.
Returns list of the index values into the original list for all documents containing the keyword.
Exercise solution
def word_search(documents, keyword):
indices = []
for i, doc in enumerate(documents):
tokens = doc.split()
normalized = [token.rstrip('.,').lower() for token in tokens]
if keyword.lower() in normalized:
indices.append(i)
return indices
My solution
def word_search(documents, keyword):
return [i for i, word in enumerate(doc_list) if keyword.lower() in word.rstrip('.,').lower()]
Run
>>> doc_list = ["The Learn Python Challenge Casino.", "They bought a car", "Casinoville"]
Expected output
>>> word_search(doc_list, 'casino')
>>> [0]
Actual output
>>> word_search(doc_list, 'casino')
>>> [0, 2]
Let's try to understand the difference.
The "result" function can be written with list-comprehension:
def word_search(documents, keyword):
return [i for i, word in enumerate(documents)
if keyword.lower() in
[token.rstrip('.,').lower() for token in word.split()]]
The problem happens with the string : "Casinoville" at index 2.
See the output:
print([token.rstrip('.,').lower() for token in doc_list[2].split()])
# ['casinoville']
And here is the matter: you try to ckeck if a word is in the list. The answer is True only if all the string matches (this is the expected output).
However, in your solution, you only check if a word contains a substring. In this case, the condition in is on the string itself and not the list.
See it:
# On the list :
print('casino' in [token.rstrip('.,').lower() for token in doc_list[2].split()])
# False
# On the string:
print('casino' in [token.rstrip('.,').lower() for token in doc_list[2].split()][0])
# True
As result, in the first case, "Casinoville" isn't included while it is in the second one.
Hope that helps !
The question is "Returns list of the index values into the original list for all documents containing the keyword".
you need to consider word only.
In "Casinoville" case, word "casino" is not in, since this case only have word "Casinoville".
When you use the in operator, the result depends on the type of object on the right hand side. When it's a list (or most other kinds of containers), you get an exact membership test. So 'casino' in ['casino'] is True, but 'casino' in ['casinoville'] is False because the strings are not equal.
When the right hand side of is is a string though, it does something different. Rather than looking for an exact match against a single character (which is what strings contain if you think of them as sequences), it does a substring match. So 'casino' in 'casinoville' is True, as would be casino in 'montecasino' or 'casino' in 'foocasinobar' (it's not just prefixes that are checked).
For your problem, you want exact matches to whole words only. The reference solution uses str.split to separate words (the with no argument it splits on any kind of whitespace). It then cleans up the words a bit (stripping off punctuation marks), then does an in match against the list of strings.
Your code never splits the strings you are passed. So when you do an in test, you're doing a substring match on the whole document, and you'll get false positives when you match part of a larger word.
I am building a puzzle word game in Python. I have the correct puzzle word, and the guessed puzzle word. I want to build a third string which shows the correct letters in the guessed puzzle in the correct puzzle word, and _ at the position of the incorrect letters.
For example, say the correct word is APPLE and the guessed word is APTLE
then i want to have a third string: AP_L_
The guessed word and correct word are guaranteed to be 3 to 5 characters long, but the guessed word is not guaranteed to be the same length as the correct word
For example, correct word is TEA and the guessed word is TEAKO, then the third string should be TEA__ because the players guessed the last two letters incorrectly.
Another example, correct word is APPLE and guessed word is POP, the third string should be:
_ _ P_ _ (without space separation)
I can successfully get the matched indexes of the correct and guessed word; however, I am having problems building the third string. I just learned that strings in Python are immutable and that i cannot assign something like str1[index] = str2[index]
I have tried many things, including using lists, but i am not getting the correct answer. The attached code is my most recent attempt, would you please help me solve this?
Thank you
find the match between puzzle_word and guess
def matcher(str_a, str_b):
#find indexes where letters overlap
matched_indexes = [i for i, (a, b) in enumerate(zip(str_a, str_b)) if a == b]
result = []
for i in str_a:
result.append('_')
for value in matched_indexes:
result[value].replace('_', str_a[value])
print(result)
matcher("apple", "allke")
the output result right now is list of five "_"
cases:
correct word is APPLE and the guessed word is APTLE third
string: AP_L_
correct word is TEA and the guessed word is TEAKO,
third string should be TEA__
correct word is APPLE and guessed
word is POP, third string should be _ _ P_ _
You can use itertools.zip_longest here to always make sure you pad out to the longest word provided and then create a new string by joining the matching characters or otherwise a _. eg:
from itertools import zip_longest
correct_and_guess = [
('APPLE', 'APTLE'),
('TEA', 'TEAKO'),
('APPLE', 'POP')
]
for correct, guess in correct_and_guess:
# If characters in same positions match - show character otherwise `_`
new_word = ''.join(c if c == g else '_' for c, g in zip_longest(correct, guess, fillvalue='_'))
print(correct, guess, new_word)
Will print the following:
APPLE APTLE AP_LE
TEA TEAKO TEA__
APPLE POP __P__
Couple of things here.
str.replace() does not replace inline; as you noted strings are immutable, so you have to assign the result of replace:
result[value] = result[value].replace('_', str_a[value])
However, there's no point doing this since you can just assign to the list element:
result[value] = str_a[value]
And finally you can assign a list of the length of str_a without the for loop, which might be more readable:
result = ['_'] * len(str_a)
In Prolog, if I am taking a list such as [hello,this,is,a,sentence] as a parameter for a predicate, how do I get a return value Y such that it will return that list as a string with spaces? e.g. [hello,this,is,a,sentence] will return hello this is a sentence.
makesentence([H|T],Y):- % some code here
I was able to run through the list recursively and have Y return the same list input with this:
makesentence([],[]). % base case returns an empty list
makesentence([X],[X]). % one list element returns that element in a list
makesentence([H|T],Y):- % a list of more than one element
makesentence(T,Result), % recursively call function on the tail
append([H],Result,Y). % append the head to the rest of the list
But when I try to make the output without a list and with spaces I faulter. I have tried this:
makesentence([],'').
makesentence([X],X).
makesentence([H|T],Y):-
makesentence(T,Result),
append(H,Result,Y).
I think it has to do with the fact the append predicate in Prolog only deals with appending a list, but I am unsure. How would I procede? Thanks in advance.
SWI-Prolog has a specialized built-in for that: atomic_list_concat/3
?- atomic_list_concat([hello,this,is,a,sentence],' ',A).
A = 'hello this is a sentence'.
figured it out with the help of Daniel. To put a list into a string with spaces use atomics_to_string/3. In my case:
makesentence([X],X).
makesentence([H|T],Y):-
makesentence(T,Result),
atomics_to_string([H,Result],' ',Y).
In the line atoms_to_string([H,Result],' ',Y)., the first parameter is the list, the second is what I want to add in between each entry, in this case a space ' ' and the third parameter is the assignment of the output, which in my case is Y. Thanks to Daniel for pointing me in the right direction.
I am new to python, and when I search for a way to get a string length without using "len()", I found this answer:
sum([1 for _ in "your string goes here"])
Can someone help me understand this line,what's the '1' doing there for example?
This is basically equivalent to this:
lst = []
for dontCareAboutTheName in "your string goes here":
lst.append(1)
print(sum(lst))
The list comprehension basically collects the number 1 for each character it finds while looping through the string. So the list will contain exactly as many elements as the length of the string. And since all those list elements are 1, when calculating the sum of all those elements, you end up with the length of the string.