Ultimately it probably doesn't have a safe .get method because a dict is an associative collection (values are associated with names) where it is inefficient to check if a key is present (and return its value) without throwing an exception, while it is super trivial to avoid exceptions accessing list elements (as the len method is very fast). The .get method allows you to query the value associated with a name, not directly access the 37th item in the dictionary (which would be more like what you're asking of your list).

Of course, you can easily implement this yourself:

def safe_list_get (l, idx, default):
  try:
    return l[idx]
  except IndexError:
    return default

You could even monkeypatch it onto the __builtins__.list constructor in __main__, but that would be a less pervasive change since most code doesn't use it. If you just wanted to use this with lists created by your own code you could simply subclass list and add the get method.

You can use chain

from itertools import chain
l = [[1,2,3],[1,2,3]]
len(list(chain(*l))) # give you 6

the expression list(chain(*l)) give you flat list: [1, 2, 3, 1, 2, 3]

I'd start by not calling it list, since that's the name of the constructor for Python's built in list type.

But once you've renamed it to cities or something, you'd do:

print(cities[0][0], cities[1][0])
print(cities[0][1], cities[1][1])

You can access the elements in a list-of-lists by first specifying which list you're interested in and then specifying which element of that list you want. For example, 17 is element 2 in list 0, which is list1[0][2]:

>>> list1 = [[10,13,17],[3,5,1],[13,11,12]]
>>> list1[0][2]
17

So, your example would be

50 - list1[0][0] + list1[0][1] - list1[0][2]

You can iterate over your data, remove the parentheses using slicing and split the string by comma to create a list, which will be appended to your output payload:

data = ['(0.412169, mississippi)', '(0.412180, NY)']
extracted_values = []
for d in data:
    extracted_values += d[1:-1].split(",")
print(extracted_values) 
# output: ['0.412169', ' mississippi', '0.412180', ' NY']

There is a slight bit of difference in their performance as can be seen from the following:

squares1 = [x**2 for x in range(1, 11)]

3.07 µs ± 70 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

squares2 = list(x**2 for x in range(1, 11))

3.65 µs ± 35.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

this can primarily be because in case 1 you are iterating and generating values for the list at the same time.

In case 2 you are generating values while iterating and then at the end of it converting the same to a list and this is then stored as a given variable.

Lists are a mutable type - in order to create a copy (rather than just passing the same list around), you need to do so explicitly:

listoflists.append((list[:], list[0]))

However, list is already the name of a Python built-in - it'd be better not to use that name for your variable. Here's a version that doesn't use list as a variable name, and makes a copy:

listoflists = []
a_list = []
for i in range(0,10):
    a_list.append(i)
    if len(a_list)>3:
        a_list.remove(a_list[0])
        listoflists.append((list(a_list), a_list[0]))
print listoflists

Note that I demonstrated two different ways to make a copy of a list above: [:] and list().

The first, [:], is creating a slice (normally often used for getting just part of a list), which happens to contain the entire list, and thus is effectively a copy of the list.

The second, list(), is using the actual list type constructor to create a new list which has contents equal to the first list. (I didn't use it in the first example because you were overwriting that name in your code - which is a good example of why you don't want to do that!)

Ah, the incomprehensible "nested" comprehensions. Loops unroll in the same order as in the comprehension.

[leaf for branch in tree for leaf in branch]

It helps to think of it like this.

for branch in tree:
    for leaf in branch:
        yield leaf

The PEP202 asserts this syntax with "the last index varying fastest" is "the Right One", notably without an explanation of why.

As for your first question: "if item is in my_list:" is perfectly fine and should work if item equals one of the elements inside my_list. The item must exactly match an item in the list. For instance, "abc" and "ABC" do not match. Floating point values in particular may suffer from inaccuracy. For instance, 1 - 1/3 != 2/3.

As for your second question: There's actually several possible ways if "finding" things in lists.

Checking if something is inside

This is the use case you describe: Checking whether something is inside a list or not. As you know, you can use the in operator for that:

3 in [1, 2, 3] # => True

Filtering a collection

That is, finding all elements in a sequence that meet a certain condition. You can use list comprehension or generator expressions for that:

matches = [x for x in lst if fulfills_some_condition(x)]
matches = (x for x in lst if x > 6)

The latter will return a generator which you can imagine as a sort of lazy list that will only be built as soon as you iterate through it. By the way, the first one is exactly equivalent to

matches = filter(fulfills_some_condition, lst)

in Python 2. Here you can see higher-order functions at work. In Python 3, filter doesn't return a list, but a generator-like object.

Finding the first occurrence

If you only want the first thing that matches a condition (but you don't know what it is yet), it's fine to use a for loop (possibly using the else clause as well, which is not really well-known). You can also use

next(x for x in lst if ...)

which will return the first match or raise a StopIteration if none is found. Alternatively, you can use

next((x for x in lst if ...), [default value])

Finding the location of an item

For lists, there's also the index method that can sometimes be useful if you want to know where a certain element is in the list:

[1,2,3].index(2) # => 1
[1,2,3].index(4) # => ValueError

However, note that if you have duplicates, .index always returns the lowest index:......

[1,2,3,2].index(2) # => 1

If there are duplicates and you want all the indexes then you can use enumerate() instead:

[i for i,x in enumerate([1,2,3,2]) if x==2] # => [1, 3]

The len() function can be used with several different types in Python - both built-in types and library types. For example:

>>> len([1, 2, 3])
3

The problem that you're asking about is that you're inheriting from list, even though you're not trying to act like a list. All this is doing is causing confusion. In particular, you're letting the list superclass define how your objects get displayed, and since you never do anything like self.append, only self.stack.append, that means it's always going to display like an empty list.

Once you fix that, your objects will always print something like this:

<__main__.stack at 0x11d919dd8>

If you want to customize that, you need to write a __repr__ method, and decide what you want it to look like.

class stack:
    def __init__(self):
        self.stack = []
        self.top = -1
    def __repr__(self):
        return f'<stack({self.stack})>'
    def isempty(self):
        return self.stack == []
    def push(self,x):
        S.top = S.top + 1
        return self.stack.append(x)

There are additional bugs in your code—you've still got a method that mutates the global S instead of self, and you're returning the result of list.append, which always returns None, and maybe more beyond—but these two changes will together solve the specific problem you're asking about.