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how to remove a string element from the list if it is already there in another string element? [closed]

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I want to keep a term which is longest in list element.
Input:
List1=['this','is','to','ask','this question']
Expected output:
List1=['is','to','ask','this question']

I assume you want to remove "this" because it's a substring of "this question". You're choosing to remove "this" and not "is" because "this" is longer than "is" right?
I would first sorted this list by length of the strings, then double for loop (i and j) to see if i is in j, then remove the last occurrence, because it is the longest.
If you confirm, that my assumption was right I'll try to code it.
EDIT :
Since you confirmed my assumption, here's the code :
List1=['this','is','to','ask','this question']
def removeSubString(myList):
sortedList = sorted(myList, key=lambda x: len(x))
for i in range(len(sortedList), -1, -1):
for j in range(i + 1, len(sortedList)):
if sortedList[i] in sortedList[j]:
myList.remove(sortedList[i])
return myList
return myList
print(removeSubString(List1))

list comprehension - how to write previous number? [closed]

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How to select the previous number in list comprehension that starts from the second one?
The formula is [w(j-1)* 5 +w for t in y]?
w(j-1) is the first number in list.

Not exactly sure what you are trying to achieve. Therefore, my own interpretation of the problem. Assume you have a list of numbers:
my_list = range(1, 10)
Now we want to iterate over this list starting from the second entry and also access the previous entry:
new_list = [my_list[i - 1]*5 + my_list[i] for i, n in enumerate(my_list) if i > 0]
print(new_list)
This way you can get access to a list element and its predecessor.
Is this what you are trying to achieve?

Efficient way of parsing string [closed]

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How would you turn a string that looks like this
7.11,8,9:00,9:15,14:30,15:00
into this dictionary entry
{7.11 : [8, (9:00, 9:15), (14:30, 15:00)]}?
Suppose that the number of time pairs (such as 9:00,9:15 and 14:30,15:00 is unknown and you want to have them all as tuple pairs.

First split the string at the commas, then zip cluster starting from the 3rd element and put it into a dictionary:
s = "7.11,8,9:00,9:15,14:30,15:00"
ss = s.split(',')
d = {ss[0]: [ss[1]] + list(zip(*[iter(ss[2:])]*2))}
Output:
{'7.11': ['8', ('9:00', '9:15'), ('14:30', '15:00')]}
If you need to convert it from string to appropiate data types (you'll have to adapt it according to your needs), then after getting the ss list:
time_list = [datetime.datetime.strptime(t,'%H:%M').time() for t in ss[2:]]
d = {float(ss[0]): [int(ss[1])] + list(zip(*[iter(time_list)]*2))}

create list of increasing number of repeated characters [closed]

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I'm trying to create this kind of output in Python
["k", "kk", "kkk", "kkkk", ...]
["rep", "reprep", "repreprep", ...]
That is a list of n elements, made of the same character (or small group of characters) repeated X times, X being increased by one for each element.
I can't find a way to do this easily, without loops..
Thanks,

Here you have a generator using itertools.count, remember the property of "multiplying" strings in python by a number, where they will be replicated and concatenated nth times, where for example "a"*3 == "aaa" :
import itertools
def genSeq(item):
yield from (item*i for i in itertools.count())
Here you have a live example

repeating_value = "k" #Assign the value which do you want to be repeated
total_times=5 #how many times do you want
expected_list=[repeating_value*i for i in range(1,total_times+1)]
print(expected_list)

character = 'k'
_range = 5
output = [k*x for k in character for x in range(1, _range + 1)]
print(output)
I would multiple my character by a specified number in the range, and then I would simply iterate through the range in a list comprehension. We add 1 to the end of the range in order to get the full range.
Here is your output:
['k', 'kk', 'kkk', 'kkkk', 'kkkkk']

The following is by far the easiest which I have built upon the comment by the user3483203 which eliminates initial empty value.
var = 'rep'
list = [var * i for i in range(1,x,1)]
print(list)

Is there a way in python to compare strings for unique character [closed]

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Suppose I have a list of strings
['elvis','elo','eels','acdc']
Is there a function that accept this string and returns a unique character for each of the strings? For example, in this case, I expect to get
['e','l','s','a']
Edit: To clarify my meaning.
I want a function that will return an identifier character or string that is based on the input list members. jme answer and bonit's one are a good example.

I think I see your point. There is no built-in for this.
Correct me if I am wrong, but it seems like you want to take the first not already taken character in each string and take one only.
Here is some code to do that
def get_first_not_know(l):
chars = []
for word in l:
for letter in word:
if letter not in chars:
chars.append(letter)
break
return chars
If you don't care about order of letters you take, you can do something quicker using sets.

Assuming Benoît Latinier's interpretation of your question is right (which, it looks like it is), then there will be some cases where a unique letter can't be found, and in these cases you might throw an exception:
def unique_chars(words):
taken = set()
uniques = []
for word in words:
for char in word:
if char not in taken:
taken.add(char)
uniques.append(char)
break
else:
raise ValueError("There are no unique letters left!")
return uniques