I'm trying to duplicate just one element of a list.
list=['a','b','c','d','e','f']
So let's say I just want to duplicate the letter 'a'.
It would look like this:
list=['a','a','b','c','d','e','f']
Try to use insert:
list=['a','b','c','d','e','f']
list.insert(1,'a')
print(list)
1 : index where you want to insert the element
You can use list.index() to find out where the element is located and then list.insert() to insert it.
For example:
lst=['a','b','c','d','e','f']
def duplicate_element(lst, elem):
lst.insert(lst.index(elem), elem)
duplicate_element(lst, 'a')
print(lst)
Prints:
['a', 'a', 'b', 'c', 'd', 'e', 'f']
Related
I am trying to split strings inside a list but I could not find any solution on the internet. This is a sample, but it should help you guys understand my problem.
array=['a','b;','c','d)','void','plasma']
for i in array:
print(i.split())
My desired output should look like this:
output: ['a','b',';','c','d',')','void','plasma']
One approach uses re.findall on each starting list term along with a list comprehension to flatten the resulting 2D list:
inp = ['a', 'b;', 'c', 'd)', 'void', 'plasma']
output = [j for sub in [re.findall(r'\w+|\W+', x) for x in inp] for j in sub]
print(output) # ['a', 'b', ';', 'c', 'd', ')', 'void', 'plasma']
Hello everyone I have a list of lists values such as :
list_of_values=[['A','B'],['A','B','C'],['D','E'],['A','C'],['I','J','K','L','M'],['J','M']]
and I would like to keep within that list, only the lists where I have the highest amount of values.
For instance in sublist1 : ['A','B'] A and B are also present in the sublist2 ['A','B','C'], so I remove the sublist1.
The same for sublist4.
the sublist6 is also removed because J and M were present in a the longer sublist5.
at the end I should get:
list_of_no_redundant_values=[['A','B','C'],['D','E'],['I','J','K','L','M']]
other exemple =
list_of_values=[['A','B'],['A','B','C'],['B','E'],['A','C'],['I','J','K','L','M'],['J','M']]
expected output :
[['A','B','C'],['B','E'],['I','J','K','L','M']]
Does someone have an idea ?
mylist=[['A','B'],['A','C'],['A','B','C'],['D','E'],['I','J','K','L','M'],['J','M']]
def remove_subsets(lists):
outlists = lists[:]
for s1 in lists:
for s2 in lists:
if set(s1).issubset(set(s2)) and (s1 is not s2):
outlists.remove(s1)
break
return outlists
print(remove_subsets(mylist))
This should result in [['A', 'B', 'C'], ['D', 'E'], ['I', 'J', 'K', 'L', 'M']]
I have 2 lists; terms and key_terms. I need to extract the before and after elements from the terms list using the elements from the key_terms list. I have tried the below and it works but it has a bug.
terms=['b','a','f','s','w','c','g']
key_terms=['a','w','g']
context_terms=[]
for kt in key_terms:
if(kt!=0):
before=terms[(terms.index(kt))-1]
if(terms.index(kt)==len(terms)-1):
context_terms.append(before)
break
else:
after=terms[(terms.index(kt))+1]
context_terms.append(before)
context_terms.append(after)
print(context_terms)
Output: ['b', 'f', 's', 'c', 'c']
The problem with the above is that if the key_terms appear twice in the terms list, the second instance is ignored.
terms=['b','a','f','s','a','c','g']
key_terms=['a','g']
context_terms=[]
for kt in key_terms:
if(kt!=0):
before=terms[(terms.index(kt))-1]
if(terms.index(kt)==len(terms)-1):
context_terms.append(before)
break
else:
after=terms[(terms.index(kt))+1]
context_terms.append(before)
context_terms.append(after)
print(context_terms)
Output: ['b', 'f', 'c']
The correct output should be ['b', 'f', 's', 'c', 'c']
After some research i noticed that i have to use a sliding window. Can someone please help me because i can't understand how i am to apply the sliding window for my case. Thank you (P.s this is my first ever question, sorry if my issue is not clear)
Try looping over terms instead of key_terms. For every element in terms which is present in key_terms, add the element prior to and next to it.
The pseudo-code would be:
for e in terms:
if e present in key_terms:
ans.add(element_to_left_of_e)
ans.add(element_to_right_of_e)
As opposed to finding indices later, the following pseudo code might prove better to iterate over indices:
for index in range(0, length of terms):
if terms[index] present in key_terms:
ans.add(terms[index-1])
ans.add(terms[index+1])
If I get your problem correctly may be following can help:
terms=['b','a','f','s','a','c','g']
key_terms=['a','g']
context_terms=[]
for k in key_terms:
indices = [i for i, item in enumerate(terms) if item == k]
for kt in indices:
before=terms[kt - 1]
if kt == len(terms)-1:
context_terms.append(before)
break
else:
after=terms[kt + 1]
context_terms.append(before)
context_terms.append(after)
print(context_terms)
Output: ['b', 'f', 's', 'c', 'c']
Given a sorted list, I would like to retrieve the first repeated item in the list using list comprehension.
So I ran the line below:
list=['1', '2', '3', 'a', 'a', 'b', 'c']
print(k for k in list if k==k+1)
I expected the output "a". But instead I got:
<generator object <genexpr> at 0x0021AB30>
I'm pretty new at this, would someone be willing to clarify why this doesn't work?
You seem to confuse the notion of list element and index.
For example the generator expression iterating over all items of list xs equal to its predecessor would look like this:
g = (xs[k] for k in range(1, len(xs)) if xs[k] == xs[k - 1])
Since you are interested only in first such item, you could write
next(xs[k] for k in range(1, len(xs)) if xs[k] == xs[k - 1])
however you'll get an exception if there is in fact no such items.
As a general advice, prefer simple readable functions over clever long one-liners,
especially when you are new to language. Your task could be accomplished as follows:
def first_duplicate(xs):
for k in range(1, len(xs)):
if xs[k] == xs[k - 1]:
return xs[k]
chars = ['1', '2', '3', 'a', 'a', 'b', 'c']
print(first_duplicate(chars)) # 'a'
P.S. Beware using list as your variable name -- you're shadowing built-in type
If you want just the first repeated item in the list you can use the next function with a generator expression that iterates through the list zipped with itself but with an offset of 1 to compare adjacent items:
next(a for a, b in zip(lst, lst[1:]) if a == b)
so that given lst = ['1', '2', '3', 'a', 'a', 'b', 'c'], the above returns: 'a'.
i have a ranked list of elements as below :
ranked_list_1 = ['G','A','M','S','D']
i need to rerank the list as below
1) Re rank as :
re_ranked_list_1 = ['A','M','D','E','G','S']
Logic : 'G' and 'S' should always be in last 2 positions and new element 'E' should also be tagged to the list just before the last 2 position.
I think this is what you want. This will put 'G' and 'S' at the end, in the order they appear in the list.
ordered_list = list()
final_terms = ['E']
for item in ranked_list_1:
if item in ['G', 'S']:
final_terms.append(item)
else:
ordered_list.append(item)
output_list = ordered_list + final_terms
print(output_list)
>>>['A', 'M', 'D', 'E', 'G', 'S']