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Replace string in specific index in list of lists python

How can i replace a string in list of lists in python but i want to apply the changes only to the specific index and not affecting the other index, here some example:
mylist = [["test_one", "test_two"], ["test_one", "test_two"]]
i want to change the word "test" to "my" so the result would be only affecting the second index:
mylist = [["test_one", "my_two"], ["test_one", "my_two"]]
I can figure out how to change both of list but i can't figure out what I'm supposed to do if only change one specific index.

Use indexing:
newlist = []
for l in mylist:
l[1] = l[1].replace("test", "my")
newlist.append(l)
print(newlist)
Or oneliner if you always have two elements in the sublist:
newlist = [[i, j.replace("test", "my")] for i, j in mylist]
print(newlist)
Output:
[['test_one', 'my_two'], ['test_one', 'my_two']]

There is a way to do this on one line but it is not coming to me at the moment. Here is how to do it in two lines.
for two_word_list in mylist:
two_word_list[1] = two_word_list.replace("test", "my")

Lists comparison in loop: print second tuple element if condition

I have the following list and nested list:
first_var = ["id1","id2","id3"]
second_var = [("id1","name1"),("id2","name2"),("id3","name3"),("id4","name4"),]
I want to check for each first element in 'second_var' that doesn't exist in 'first_var' and print the second element in the 'second_var'.
my code is:
for x in [x[0] for x in second_var]:
if x not in first_var:
print(...)
For now if i execute print(x) it prints:
id4
but i need it to print
name4
How can i achieve that?

The problem with your code is you are not iterating the original list. You are only iterating the first entry of each tuple within the list.
This is how you can adapt your code:
first_var = ["id1","id2","id3"]
second_var = [("id1","name1"),("id2","name2"),("id3","name3"),("id4","name4"),]
for x in second_var:
if x[0] not in first_var:
print(x[1])
The Pythonic solution is to convert this to a list comprehension:
values = set(first_var)
res = [x[1] for x in second_var if x[0] not in values]
for item in res:
print(item)
Or the functional version; not recommended, but another way of seeing the logic:
from operator import itemgetter
values = set(first_var)
res = map(itemgetter(1), filter(lambda x: x[0] not in values, second_var))

>>> [v[1] for v in second_var if v[0] not in first_var]
['name4']

You can use list comprehension feature.
ids = [tuple[1] for tuple in second_var if tuple[0] not in first_var]
print(ids)
Output
['name4']
The list comprehension statement above is equivalent to:
>>> result = []
for tuple in second_var:
if tuple[0] not in first_var:
result.append(tuple[1])
>>> result
['name4']

If you have a lot of data, you need to build a dictionary & use set & all those nice hashing/difference techniques already existing in Python instead of linear lookup in lists (O(1) vs O(n)).
first_var = ["id1","id2","id3"]
second_var = [("id1","name1"),("id2","name2"),("id3","name3"),("id4","name4"),]
second_d = dict(second_var) # create a dict directly from tuples
missing = set(second_d).difference(first_var)
for m in missing:
print(second_d[m])
this prints name4
missing is the difference between the dict keys and the list.

How to merge n lists together item by item for each list

I want to make one large list for entering into a database with values from 4 different lists. I want it to be like
[[list1[0], list2[0], list3[0], list4[0]], [list1[1], list2[1], list3[1], list4[1]], etc.....]
Another issue is that currently the data is received like this:
[ [ [list1[0], list1[1], [list1[3]]], [[list2[0]]], etc.....]
I've tried looping through each list using indexs and adding them to a new list based on those but it hasn't worked, I'm pretty sure it didn't work because some of the lists are different lengths (they're not meant to be but it's automated data so sometimes there's a mistake).
Anyone know what's the best way to go about this? Thanks.

First list can be constructed using zip function as follows (for 4 lists):
list1 = [1,2,3,4]
list2 = [5,6,7,8]
list3 = [9,10,11,12]
list4 = [13,14,15,16]
res = list(zip(list1,list2,list3,list4))
For arbitrtary number of lists stored in another list u can use *-notation to unpack outer list:
lists = [...]
res = list(zip(*lists))
To construct list of lists for zipping from you data in second issue use flatten concept to it and then zip:
def flatten(l):
res = []
for el in l:
if(isinstance(el, list)):
res += flatten(el)
else:
res.append(el)
return res
auto_data = [...]
res = list(zip(*[flatten(el) for el in auto_data]))
Some clarification at the end:
zip function construct results of the smallest length between all inputs, then you need to extend data in list comprehension in last code string to be one length to not lose some info.

So if I understand correctly, this is your input:
l = [[1.1,1.2,1.3,1.4],[2.1,2.2,2.3,2.4],[3.1,3.2,3.3,3.4],[4.1,4.2,4.3,4.4]]
and you would like to have this output
[[1.1,2.1,3.1,4.1],...]
If so, this could be done by using zip
zip(*l)

Make a for loop which only gives you the counter variable. Use that variable to index the lists. Make a temporary list , fill it up with the values from the other lists. Add that list to the final one. With this you will et the desired structure.
nestedlist = []
for counter in range(0,x):
temporarylist = []
temporarylist.append(firstlist[counter])
temporarylist.append(secondlist[counter])
temporarylist.append(thirdlist[counter])
temporarylist.append(fourthlist[counter])
nestedlist.append(temporarylist)
If all the 4 lists are the same length you can use this code to make it even nicer.
nestedlist = []
for counter in range(0,len(firstlist)): #changed line
temporarylist = []
temporarylist.append(firstlist[counter])
temporarylist.append(secondlist[counter])
temporarylist.append(thirdlist[counter])
temporarylist.append(fourthlist[counter])
nestedlist.append(temporarylist)

This comprehension should work, with a little help from zip:
mylist = [i for i in zip(list1, list2, list3, list4)]
But this assumes all the list are of the same length. If that's not the case (or you're not sure of that), you can "pad" them first, to be of same length.
def padlist(some_list, desired_length, pad_with):
while len(some_list) < desired_length:
some_list.append(pad_with)
return some_list
list_of_lists = [list1, list2, list3, list4]
maxlength = len(max(list_of_lists, key=len))
list_of_lists = [padlist(l, maxlength, 0) for l in list_of_lists]
And now do the above comprehension statement, works well in my testing of it
mylist = [i for i in zip(*list_of_lists)]

If the flatten concept doesn't work, try this out:
import numpy as np
myArray = np.array([[list1[0], list2[0], list3[0], list4[0]], [list1[1], list2[1], list3[1], list4[1]]])
np.hstack(myArray)
Also that one should work:
np.concatenate(myArray, axis=1)
Just for those who will search for the solution of this problem when lists are of the same length:
def flatten(lists):
results = []
for numbers in lists:
for output in numbers:
results.append(output)
return results
print(flatten(n))

Extract substrings from a list into a list in Python

I have a Python list like:
['user#gmail.com', 'someone#hotmail.com'...]
And I want to extract only the strings after # into another list directly, such as:
mylist = ['gmail.com', 'hotmail.com'...]
Is it possible? split() doesn't seem to be working with lists.
This is my try:
for x in range(len(mylist)):
mylist[x].split("#",1)[1]
But it didn't give me a list of the output.

You're close, try these small tweaks:
Lists are iterables, which means its easier to use for-loops than you think:
for x in mylist:
#do something
Now, the thing you want to do is 1) split x at '#' and 2) add the result to another list.
#In order to add to another list you need to make another list
newlist = []
for x in mylist:
split_results = x.split('#')
# Now you have a tuple of the results of your split
# add the second item to the new list
newlist.append(split_results[1])
Once you understand that well, you can get fancy and use list comprehension:
newlist = [x.split('#')[1] for x in mylist]

That's my solution with nested for loops:
myl = ['user#gmail.com', 'someone#hotmail.com'...]
results = []
for element in myl:
for x in element:
if x == '#':
x = element.index('#')
results.append(element[x+1:])

How can I write a list without duplicate with only for, if and boolean

My professor gave me an exercise where I write a function that returns a list without the duplicate to the old list.
This is the code but I don't know how to write the method without using .remove():
def distinct(lst):
lstnew = []
c = range(len(lst))
for i in range(len(lst)):
if i in range(len(lst)) != c:
lstnew += [i]
c += 1
return lstnew
print distinct([1,3,1,2,6])
print distinct([['a','ab','a','ab']])
I forgot to write an important thing, I must preserve order in the output list.
[UPDATE]
After I read the answer of Jai Srivastav I code this:
def distinct(lst):
lstnew = []
for element in lst:
if element not in lstnew:
lstnew = lstnew + [element]
return lstnew
And It works perfectly

def distinct(lst):
dlst = []
for val in lst:
if val not in dlst:
dlst.append(val)
return dlst

Is this considered cheating?
>>> distinct = lambda lst: list(set(lst))
>>> distinct([1,3,1,2,6])
[1, 2, 3, 6]
>>> distinct(['a','ab','a','ab'])
['a', 'ab']

If order isn't important, you can cast it to a set, then back to a list
def distinct(lst):
return list(set(lst))

If you need to eliminate duplicates AND preserve order you can do this:
def distinct(lst):
seen = set()
for item in lst:
if item not in seen:
yield item
seen.add(item)
a = [1,3,1,2,6]
print(list(distinct(a)))
[1,3,2,6]
b = ['a','ab','a','ab']
print(list(distinct(b)))
['a', 'ab']
See a demo here: https://ideone.com/a2khCg

There are Excellent Solutions That I Already Applied. But my professor said us that we don't must use the methods of the list. Has anyone else got any more thoughts?