Placing The: Before-Value in The Same Way as a List-Item Is Placed

How to switch position of two items in a Python list?

i = ['title', 'email', 'password2', 'password1', 'first_name', 
     'last_name', 'next', 'newsletter']
a, b = i.index('password2'), i.index('password1')
i[b], i[a] = i[a], i[b]

Loop that also accesses previous and next values

This should do the trick.

foo = somevalue
previous = next_ = None
l = len(objects)
for index, obj in enumerate(objects):
    if obj == foo:
        if index > 0:
            previous = objects[index - 1]
        if index < (l - 1):
            next_ = objects[index + 1]

Here's the docs on the enumerate function.

Insert an element at a specific index in a list and return the updated list

The shortest I got: b = a[:2] + [3] + a[2:]

>>>
>>> a = [1, 2, 4]
>>> print a
[1, 2, 4]
>>> b = a[:2] + [3] + a[2:]
>>> print a
[1, 2, 4]
>>> print b
[1, 2, 3, 4]

Append integer to beginning of list in Python

>>> x = 42
>>> xs = [1, 2, 3]
>>> xs.insert(0, x)
>>> xs
[42, 1, 2, 3]

How it works:

list.insert(index, value)

Insert an item at a given position. The first argument is the index of the element before which to insert, so xs.insert(0, x) inserts at the front of the list, and xs.insert(len(xs), x) is equivalent to xs.append(x). Negative values are treated as being relative to the end of the list.

How to align list items so it does not overlap

A simple way is to add display:flex to li:

ul {  max-width: 200px;}li {  display: flex;}

<link href="https://maxcdn.bootstrapcdn.com/font-awesome/4.7.0/css/font-awesome.min.css" rel="stylesheet" integrity="sha384-wvfXpqpZZVQGK6TAh5PVlGOfQNHSoD2xbE+QkPxCAFlNEevoEH3Sl0sibVcOQVnN" crossorigin="anonymous"><ul><li>    <span class="fa fa-check"></span> line 1 line 1 line 1 line 1 line 1 line 1 line 1  </li>
  <li>    <span class="fa fa-check"></span> line 2 line 2 line 1 line 1 line 1 line 1 line 1  </li></ul>

Converting a list to a set changes element order

1. A set is an unordered data structure, so it does not preserve the insertion order.

2. This depends on your requirements. If you have an normal list, and want to remove some set of elements while preserving the order of the list, you can do this with a list comprehension:

   >>> a = [1, 2, 20, 6, 210]
   >>> b = set([6, 20, 1])
   >>> [x for x in a if x not in b]
   [2, 210]
   

   If you need a data structure that supports both fast membership tests and preservation of insertion order, you can use the keys of a Python dictionary, which starting from Python 3.7 is guaranteed to preserve the insertion order:

   >>> a = dict.fromkeys([1, 2, 20, 6, 210])
   >>> b = dict.fromkeys([6, 20, 1])
   >>> dict.fromkeys(x for x in a if x not in b)
   {2: None, 210: None}
   

   b doesn't really need to be ordered here – you could use a set as well. Note that a.keys() - b.keys() returns the set difference as a set, so it won't preserve the insertion order.

   In older versions of Python, you can use collections.OrderedDict instead:

   >>> a = collections.OrderedDict.fromkeys([1, 2, 20, 6, 210])
   >>> b = collections.OrderedDict.fromkeys([6, 20, 1])
   >>> collections.OrderedDict.fromkeys(x for x in a if x not in b)
   OrderedDict([(2, None), (210, None)])
   

Move an item inside a list?

Use the insert method of a list:

l = list(...)
l.insert(index, item)

Alternatively, you can use a slice notation:

l[index:index] = [item]

If you want to move an item that's already in the list to the specified position, you would have to delete it and insert it at the new position:

l.insert(newindex, l.pop(oldindex))

Extract elements of list at odd positions

Solution

Yes, you can:

l = L[1::2]

And this is all. The result will contain the elements placed on the following positions (0-based, so first element is at position 0, second at 1 etc.):

1, 3, 5

so the result (actual numbers) will be:

2, 4, 6

Explanation

The [1::2] at the end is just a notation for list slicing. Usually it is in the following form:

some_list[start:stop:step]

If we omitted start, the default (0) would be used. So the first element (at position 0, because the indexes are 0-based) would be selected. In this case the second element will be selected.

Because the second element is omitted, the default is being used (the end of the list). So the list is being iterated from the second element to the end.

We also provided third argument (step) which is 2. Which means that one element will be selected, the next will be skipped, and so on...

So, to sum up, in this case [1::2] means:

1. take the second element (which, by the way, is an odd element, if you judge from the index),
2. skip one element (because we have step=2, so we are skipping one, as a contrary to step=1 which is default),
3. take the next element,
4. Repeat steps 2.-3. until the end of the list is reached,

EDIT: @PreetKukreti gave a link for another explanation on Python's list slicing notation. See here: Explain Python's slice notation

Extras - replacing counter with enumerate()

In your code, you explicitly create and increase the counter. In Python this is not necessary, as you can enumerate through some iterable using enumerate():

for count, i in enumerate(L):
    if count % 2 == 1:
        l.append(i)

The above serves exactly the same purpose as the code you were using:

count = 0
for i in L:
    if count % 2 == 1:
        l.append(i)
    count += 1

More on emulating for loops with counter in Python: Accessing the index in Python 'for' loops

How to insert an item into an array at a specific index (JavaScript)

You want the splice function on the native array object.

arr.splice(index, 0, item); will insert item into arr at the specified index (deleting 0 items first, that is, it's just an insert).

In this example we will create an array and add an element to it into index 2:

var arr = [];
arr[0] = "Jani";
arr[1] = "Hege";
arr[2] = "Stale";
arr[3] = "Kai Jim";
arr[4] = "Borge";

console.log(arr.join()); // Jani,Hege,Stale,Kai Jim,Borge
arr.splice(2, 0, "Lene");
console.log(arr.join()); // Jani,Hege,Lene,Stale,Kai Jim,Borge

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