I have a problem like that:
list = ['a1',['b1',2],['c1',2,3],['d1',2,3,4]]
I want to get a new list like that
new_list['a1','b1','c1','d1']
I do like this:
lst = ['a1',['b1',2],['c1',2,3],['d1',2,3,4]]
for item in lst:
print(item)
result is:
a1
['b1', 2]
['c1', 2, 3]
['d1', 2, 3, 4]
But I want the first element of each result
The best answer is like this :
my_list = list()
lst = ['a1',['b1',2],['c1',2,3],['d1',2,3,4]]
for element in lst:
if type(element)==type('string'):
my_list.append(element)
else:
my_list.append(element[0])
print(my_list)
Thank you!
Do it as below:
>>> my_list = list()
>>> lst = ['a1',['b1',2],['c1',2,3],['d1',2,3,4]]
>>> for element in lst:
if type(element)==type('string'):
my_list.append(element)
else:
my_list.append(element[0])
It will produce:
>>> my_list
['a1', 'b1', 'c1', 'd1']
>>>
As you see above, first I created a list (named my_list) and then checked each elements of your list. If the element was a string, I added it to my_list and otherwise (i.e. it is a list) I added the first element of it to my_list.
I would do
res = []
for x in the_list:
if x is Array:
res.append(x[0])
else:
res.append(x)
Related
lst =["1","2","","3","4","5","","6"]
Given the list above, how do I split the list an create another list whenever I find ""(the empty quotation marks alone)?
lst =["1","2","","3","4","5","","6"]
lst2 = []
for a in lst:
if a == "":
continue
lst2.append(int(a))
print(lst2)
Output:[1, 2, 3, 4, 5, 6]
Expected output:
lst2 = [["1","2"],["3","4","5"],["6"]]
lst =["1","2","","3","4","5","","6"]
arr = []
new_lst = []
for i in lst:
if i == "":
arr.append(new_lst)
new_lst = []
else:
new_lst.append(i)
if len(new_lst) > 0:
arr.append(new_lst)
print(arr)
I've got a mixed list, containing lists with tuples (2nd dimension), looking like this:
[[(0, 500), (755, 1800)], [2600, 2900], [4900, 9000], [(11000, 17200)]]
The list should look like this
[[0, 500], [755, 1800], [2600, 2900], [4900, 9000], [11000, 17200]]
I tried it with map and the call to the list() convert function.
#Try 1: works for just the first element
experiment = map(list,cleanedSeg[0])
#Try 2: gives error int not interabel
experiment = [map(list,elem) for elem in cleanedSeg if isinstance(elem, (tuple, list))]
#Try 3:
experiment = [list(x for x in xs) for xs in cleanedSeg]
print experiment
none of them did solve my issue
mixlist = [[(0, 500), (755, 1800)], [2600, 2900], [4900, 9000], [(11000, 17200)]]
# [[0, 500], [755, 1800], [2600, 2900], [4900, 9000], [11000, 17200]]
experiment = [list(n) if isinstance(n, tuple) else [n] for sub in mixlist for n in sub]
I tried two version of the list comprehension below. The above one and another alternative where
experiment = [list(n) if isinstance(n, tuple) else list(n) for sub in mixlist for n in sub]
This expression gives the following error:
TypeError: Argument of type 'int' is not iterable.
The difference between these two expressions is using list literal, [] and list function, ().
list_literal = [n] # Gives a list literal [n]
ls = list(n) # Iterate over n's items and produce a list from that.
For example:
>>> n = (1,2,3)
>>> list_literal = [n]
>>> list_literal
[(1, 2, 3)]
>>> n = (1,2,3)
>>> list_literal = list(n)
>>> list_literal
[1, 2, 3]
If I have an input like this (1, 2, 3, 4, 5, 6)
The output has to be ... [[1, 2], [3, 4], [5, 6]].
I know how to deal with if it's one element but not two.
x=[]
for number in numbers:
x.append([number])
I'll appreciate your any help!
Something like this would work:
out = []
lst = (1,2,3,4,5,6,7,8,9,10)
for x in range(len(lst)):
if x % 2 == 0:
out.append([lst[x], lst[x+1]])
else:
continue
To use this, just set lst equal to whatever list of numbers you want. The final product is stored in out.
There is a shorter way of doing what you want:
result = []
L = (1,2,3,4,5,6,7,8,9,10)
result = [[L[i], L[i + 1]] for i in range(0, len(L) - 1, 2)]
print(result)
You can use something like this. This solution also works for list of odd length
def func(lst):
res = []
# Go through every 2nd value | 0, 2, 4, ...
for i in range(0, len(lst), 2):
# Append a slice of the list, + 2 to include the next value
res.append(lst[i : i + 2])
return res
# Output
>>> lst = [1, 2, 3, 4, 5, 6]
>>> func(lst)
[[1, 2], [3, 4], [5, 6]]
>>> lst2 = [1, 2, 3, 4, 5, 6, 7]
>>> func(lst2)
[[1, 2], [3, 4], [5, 6], [7]]
List comprehension solution
def func(lst):
return [lst[i:i+2] for i in range(0, len(lst), 2)]
Slicing is better in this case as you don't have to account for IndexError allowing it to work for odd length as well.
If you want you can also add another parameter to let you specify the desired number of inner elements.
def func(lst, size = 2): # default of 2 it none specified
return [lst[i:i+size] for i in range(0, len(lst), size)]
There's a few hurdles in this problem. You want to iterate through the list without going past the end of the list and you need to deal with the case that list has an odd length. Here's one solution that works:
def foo(lst):
result = [[x,y] for [x,y] in zip(lst[0::2], lst[1::2])]
return result
In case this seems convoluted, let's break the code down.
Index slicing:
lst[0::2] iterates through lst by starting at the 0th element and proceeds in increments of 2. Similarly lst[1::2] iterates through starting at the 1st element (colloquially the second element) and continues in increments of 2.
Example:
>>> lst = (1,2,3,4,5,6,7)
>>> print(lst[0::2])
(1,3,5,7)
>>> print(lst[1::2])
(2,4,6)
zip: zip() takes two lists (or any iterable object for that matter) and returns a list containing tuples. Example:
>>> lst1 = (10,20,30, 40)
>>> lst2 = (15,25,35)
>>> prit(zip(lst1, lst2))
[(10,15), (20,25), (30,35)]
Notice that zip(lst1, lst2) has the nice property that if one of it's arguments is longer than the other, zip() stops zipping whenever the shortest iterable is out of items.
List comprehension: python allows iteration quite generally. Consider the statement:
>>> [[x,y] for [x,y] in zip(lst1,lst2)]
The interior bit "for [x,y] in zip(lst1,lst2)" says "iterate through all pairs of values in zip, and give their values to x and y". In the rest of the statement
"[[x,y] for [x,y] ...]", it says "for each set of values x and y takes on, make a list [x,y] to be stored in a larger list". Once this statement executes, you have a list of lists, where the interior lists are all possible pairs for zip(lst1,lst2)
Very Clear solution:
l = (1, 2, 3, 4, 5, 6)
l = iter(l)
w = []
for i in l:
sub = []
sub.append(i)
sub.append(next(l))
w.append(sub)
print w
I have a list of strings stringlist = ["elementOne" , "elementTwo" , "elementThree"] and I would like to search for elements that contain the "Two" string and delete that from the list so my list will become stringlist = ["elementOne" , "elementThree"]
I managed to print them but don't really know how to delete completely from the list using del because i don't know the index or by using stringlist.remove("elementTwo") because I don't know the exact string of the element containing "Two"
My code so far:
for x in stringlist:
if "Two" in x:
print(x)
Normally when we perform list comprehension, we build a new list and assign it the same name as the old list. Though this will get the desired result, but this will not remove the old list in place.
To make sure the reference remains the same, you must use this:
>>> stringlist[:] = [x for x in stringlist if "Two" not in x]
>>> stringlist
['elementOne', 'elementThree']
Advantages:
Since it is assigning to a list slice, it will replace the contents with the same Python list object, so the reference remains the same, thereby preventing some bugs if it is being referenced elsewhere.
If you do this below, you will lose the reference to the original list.
>>> stringlist = [x for x in stringlist if "Two" not in x]
>>> stringlist
['elementOne', 'elementThree']
So to preserve the reference, you build the list object and assign it the list slice.
To understand the subtle difference:
Let us take a list a1 containing some elements and assign list a2 equal to a1.
>>> a1 = [1,2,3,4]
>>> a2 = a1
Approach-1:
>>> a1 = [x for x in a1 if x<2]
>>> a1
[1]
>>> a2
[1,2,3,4]
Approach-2:
>>> a1[:] = [x for x in a1 if x<2]
>>> a1
[1]
>>> a2
[1]
Approach-2 actually replaces the contents of the original a1 list whereas Approach-1 does not.
You can use enumerate to get the index when you iterate over your list (but Note that this is not a pythonic and safe way to modify your list while iterating over it):
>>> for i,x in enumerate(stringlist):
... if "Two" in x:
... print(x)
... del stringlist[i]
...
elementTwo
>>> stringlist
['elementOne', 'elementThree']
But as a more elegant and pythonic way you can use a list comprehension to preserve the elements that doesn't contains Two :
>>> stringlist = [i for i in stringlist if not "Two" in i]
>>> stringlist
['elementOne', 'elementThree']
Doing this will help you
for i,x in enumerate(stringlist):
if "Two" in x:
del stringlist[i]
or
newList = []
for x in stringlist:
if "Two" in x:
continue
else
newList.append(x)
Using regex,
import re
txt = ["SpainTwo", "StringOne"]
for i in txt:
x = re.search(r"Two", i)
if x:
temp_list = temp_list + [x.string] if "temp_list" in locals() else [x.string]
print(temp_list)
gives
['SpainTwo']
print(list(filter(lambda x: "Two" not in x, ["elementOne" , "elementTwo" , "elementThree", "elementTwo"])))
Using lambda, if you are only looking to print.
if you want to check for multiple string and delete if detected from list of string use following method
List_of_string = [ "easyapplyone", "appliedtwotime", "approachednone", "seenthreetime", "oneseen", "twoapproached"]
q = ["one","three"]
List_of_string[:] = [x for x in List_of_string if any(xs not in x for xs in q)]
print(List_of_string)
output:[ "approachednone", "seenthreetime"]
Well this was pretty simple - sorry for all the trouble
for x in stringlist:
if "Two" in x:
stringlist.remove(x)
I have a strange reaction of python (using 2.7) here. I am trying to copy a list and append something to the copy at the same time. Here is the code:
myList = [1]
>>> newList = list(myList).append(2)
>>> newList
>>> print newList
None
>>> type(newList)
<type 'NoneType'>
Why is it that I get a NoneType object instead of my appended list-copy?
I stumbled over this when I tried to take a list1 copy it as many times as a list2 and append the elements of list2 to the ones in list1.
>>> list1 = [1,2]
>>> list2 = [3,4]
>>> list3 = [list(list1).append(i) for i in list2]
>>> list3
[None, None]
I expected:
>>> list3
[[1,2,3],[1,2,4]]
Why is it None,None?
Thanks a lot!
You can do this by adding a extra line:
myList=[1]
myList.append(2);newList=myList
You can also extend (append )list directly like:
list1 = [1,2]
list2 = [3,4]
list1.extend(list2);list3=list1
If u dont want to alter then try this:
list1 = [1,2]
list2 = [3,4]
list3=list1;list3.extend(list2)
And also:
myList=[1]
newList=myList;newList.append(2)
The append function modifies a list and returns None. Newlist was None because append() modifies the list directly, rather than returning the modified list.
This code will create the new list and add to it in one step.
myList = [1]
newList = myList + [2]