I have a list of strings =
['after','second','shot','take','note','of','the','temp']
I want to strip all strings after the appearance of 'note'.
It should return
['after','second','shot','take']
There are also lists which does not have the flag word 'note'.
So in case of a list of strings =
['after','second','shot','take','of','the','temp']
it should return the list as it is.
How to do that in a fast way? I have to repeat the same thing with many lists with unequal length.
tokens = [tokens[:tokens.index(v)] if v == 'note' else v for v in tokens]
There is no need of an iteration when you can slice list:
strings[:strings.index('note')+1]
where s is your input list of strings. The end slice is exclusive, hence a +1 makes sure 'note' is part.
In case of missing data ('note'):
try:
final_lst = strings[:strings.index('note')+1]
except ValueError:
final_lst = strings
if you want to make sure the flagged word is present:
if 'note' in lst:
lst = lst[:lst.index('note')+1]
Pretty much the same as #Austin's answer above.
Related
I'm writing a program to check if a given user input is a palindrome or not. if it is the program should print "Yes", if not "no". I realize that this program is entirely too complex since I actually only needed to check the whole word using the reversed() function, but I ended up making it quite complex by splitting the word into two lists and then checking the lists against each other.
Despite that, I'm not clear why the last conditional isn't returning the expected "Yes" when I pass it "racecar" as an input. When I print the lists in line 23 and 24, I get two lists that are identical, but then when I compare them in the conditional, I always get "No" meaning they are not equal to each other. can anyone explain why this is? I've tried to convert the lists to strings but no luck.
def odd_or_even(a): # function for determining if odd or even
if len(a) % 2 == 0:
return True
else:
return False
the_string = input("How about a word?\n")
x = int(len(the_string))
odd_or_even(the_string) # find out if the word has an odd or an even number of characters
if odd_or_even(the_string) == True: # if even
for i in range(x):
first_half = the_string[0:int((x/2))] #create a list with part 1
second_half = the_string[(x-(int((x/2)))):x] #create a list with part 2
else: #if odd
for i in range(x):
first_half = the_string[:(int((x-1)/2))] #create a list with part 1 without the middle index
second_half = the_string[int(int(x-1)/2)+1:] #create a list with part 2 without the middle index
print(list(reversed(second_half)))
print(list(first_half))
if first_half == reversed(second_half): ##### NOT WORKING BUT DONT KNOW WHY #####
print("Yes")
else:
print("No")
Despite your comments first_half and second_half are substrings of your input, not lists. When you print them out, you're converting them to lists, but in the comparison, you do not convert first_half or reversed(second_half). Thus you are comparing a string to an iterator (returned by reversed), which will always be false.
So a basic fix is to do the conversion for the if, just like you did when printing the lists out:
if list(first_half) == list(reversed(second_half)):
A better fix might be to compare as strings, by making one of the slices use a step of -1, so you don't need to use reversed. Try second_half = the_string[-1:x//2:-1] (or similar, you probably need to tweak either the even or odd case by one). Or you could use the "alien smiley" slice to reverse the string after you slice it out of the input: second_half = second_half[::-1].
There are a few other oddities in your code, like your for i in range(x) loop that overwrites all of its results except the last one. Just use x - 1 in the slicing code and you don't need that loop at all. You're also calling int a lot more often than you need to (if you used // instead of /, you could get rid of literally all of the int calls).
So far, I have:
my_list = ['hello', 'oi']
comparison_list = ['this hellotext', 'this oitext']
for w in my_list:
if w in comparison_list: print('yes')
However, nothing prints because no element in my_list equals any element in comparison_list.
So how do I make this check as a subset or total occurance?
Ideal output:
yes
yes
You are checking the occurrence of the complete string in the list currently. Instead you can check for the occurrence of the string inside each comparison string and make a decision. A simple approach will be to re-write the loop as below
for w in my_list:
# Check for every comparison string. any() considers atleast 1 true value
if any([True for word in comparison_list if w in word]):
print('yes')
It's because you're comparing w to the list elements. If you wanna find w in each string in your comparison_list you can use any:
my_list = ['hello', 'oi', 'abcde']
comparison_list = ['this hellotext', 'this oitext']
for w in my_list:
if any(w in s for s in comparison_list):
print('yes')
else:
print('no')
I added a string to your list and handle the 'no' case in order to get an output for each element
Output:
yes
yes
no
Edited Solution:
Apologies for older solution, I was confused somehow.
Using re module , we can use re.search to determine if the string is present in the list of items. To do this we can create an expression using str.join to concatenate all the strings using |. Once the expression is created we can iterate through the list of comparison to be done. Note | means 'Or', so any of the searched strings if present should return bool value of True. Note I am assuming there are no special characters present in the my_list.
import re
reg = '|'.join(my_list)
for item in comparison_list:
print(bool(re.search(reg, item)))
I'm trying to write a function that returns True if a string contains exactly two instance of a substring and False if it doesn't.
I'm getting an error:
return' outside function
I feel I'm very close but just can't quite get it, I'd appreciate being pointed in the right direction.
Should recursion be used?
s = ("xeroxes")
exes = str(x)
count = 0
def has_two_X(s):
count = s.count(exes)
for exes in s:
count = +1
if count ==2:
return True
else:
return False
if __name__ == "__main__":
print("string has :",count.s(exes))
If the code must return True if there is two or more instances of substring, you can create a dictionary and return True if value is there in dictionary or not.
mydict = {}
for letter in s:
if letter in mydict:
return True
else:
mydict[letter] = 0
return False #Since there is no substring (two substring can be formed only if it has two same letters in the string)
To find exactly if it has two substring, I would recommend the same approach where you can maintain the dictionary and the count of substring present. Add all the substring/count to the dictionary, this would take O(n^2) to generate all substrings and approx- O(n^2) hashmap memory.
After constructing hashmap, you can iterate over the hashmap to return True if it has exactly two occurences of substring.
I am writing a script - includes(word1, word2) - that takes two strings as arguments, and finds if word1 is included in word2. Word2 is a letter jumble. It should return Boolean. Also repetition of letters are allowed, I am only checking if the letters are included in the both words in the same order.
>>>includes('queen', 'qwertyuytresdftyuiokn')
True
'queen', 'QwertyUytrEsdftyuiokN'
I tried turning each word into lists so that it is easier to work with each element. My code is this:
def includes(w1, w2):
w1 = list(w1)
w2 = list(w2)
result = False
for i in w1:
if i in w2:
result = True
else:
result = False
return result
But the problem is that I need to also check if the letters of word1 comes in the same order in word2, and my code doesn't controls that. I couldn't find a way to implement that with list. Just like I couldn't do this much with strings, so I think I need to use another data structure like dictionary but I don't know much about them.
I hope I understood what is your goal.
Python is not my thing, but I think I made it pythonic:
def is_subsequence(pattern, items_to_use):
items_to_use = (x for x in items_to_use)
return all(any(x == y for y in items_to_use) for x, _ in itertools.groupby(pattern))
https://ideone.com/Saz984
Explanation:
itertools.groupby transfers pattern in such way that constitutive duplicates are discarded
all items form form grouped pattern must fulfill conditions
any uses generator items_to_use as long as it doesn't matches current item. Note that items_to_use mus be defined outside of final expression so progress on it is kept every time next item from pattern is verified.
If you are not just checking substrings:
def include(a, b):
a = "".join(set(a)) # removes duplicates
if len(a) == 1:
if a in b:
return True
else:
return False
else:
try:
pos = b.index(a[0])
return include(a[1:], b[pos:])
except:
return False
print(include('queen', 'qwertyuytresdftyuiokn'))
#True
What I want to get accomplished is an algorithm that finds the most duplicated letter from the entire list of strings. I'm new to Python so its taken me roughly two hours to get to this stage. The problem with my current code is that it returns every duplicated letter, when I'm only looking for the most duplicated letter. Additionally, I would like to know of a faster way that doesn't use two for loops.
Code:
rock_collections = ['aasdadwadasdadawwwwwwwwww', 'wasdawdasdasdAAdad', 'WaSdaWdasSwd', 'daWdAWdawd', 'QaWAWd', 'fAWAs', 'fAWDA']
seen = []
dupes = []
for words in rock_collections:
for letter in words:
if letter not in seen:
seen.append(letter)
else:
dupes.append(letter)
print(dupes)
If you are looking for the letter which appears the greatest number of times, I would recommend the following code:
def get_popular(strings):
full = ''.join(strings)
unique = list(set(full))
return max(
list(zip(unique, map(full.count, unique))), key=lambda x: x[1]
)
rock_collections = [
'aasdadwadasdadawwwwwwwwww',
'wasdawdasdasdAAdad',
'WaSdaWdasSwd',
'daWdAWdawd',
'QaWAWd',
'fAWAs',
'fAWDA'
]
print(get_popular(rock_collections)) # ('d', 19)
Let me break down the code for you:
full contains each of the strings together with without any letters between them. set(full) produces a set, meaning that it contains every unique letter only once. list(set(full)) makes this back into a list, meaning that it retains order when you iterate over the elements in the set.
map(full.count, unique) iterates over each of the unique letters and counts how many there are of them in the string. zip(unique, ...) puts those numbers with their respective letters. key=lambda x: x[1] is a way of saying, don't take the maximum value of the tuple, instead take the maximum value of the second element of the tuple (which is the number of times the letter appears). max finds the most common letter, using the aforementioned key.