You can use the last function to get the last element of a list. As for how to remove the first and last elements, you could use (init . tail), but I don’t know how efficient that is. I think this image from Learn You A Haskell shows the list functions fairly well:
The short form is as follows (called dict comprehension, as analogy to the list comprehension, set comprehension etc.): x = { row.SITE_NAME : row.LOOKUP_TABLE for row in cursor } so in general given some _container with some kind of elements and a function _value which for a given element returns the value that you want … Read more
You are basically asking how an async for loop works over a regular loop. That you can now use such a loop in a list comprehension doesn’t make any difference here; that’s just an optimisation that avoids repeated list.append() calls, exactly like a normal list comprehension does. An async for loop then, simply awaits each … Read more
If the iteration is being done for its side effect ( as it is in your “print” example ), then a loop is clearer. If the iteration is executed in order to build a composite value, then list comprehensions are usually more readable.
{inner_k: myfunc(inner_v)} isn’t a dictionary comprehension. It’s just a dictionary. You’re probably looking for something like this instead: data = {outer_k: {inner_k: myfunc(inner_v) for inner_k, inner_v in outer_v.items()} for outer_k, outer_v in outer_dict.items()} For the sake of readability, don’t nest dictionary comprehensions and list comprehensions too much.
Since these are all python dicts and you are calling the dict.get() method on them, you can use an empty dict to chain: [m.get(“gparents”, {}).get(“parent”, {}).get(“child”) for m in M] By leaving off the default for the last .get() you fall back to None. Now, if any of the intermediary keys is not found, the … Read more
it’s typically referred to as flattening a nested structure. >>> tupleOfTuples = ((1, 2), (3, 4), (5,)) >>> [element for tupl in tupleOfTuples for element in tupl] [1, 2, 3, 4, 5] Just to demonstrate efficiency: >>> import timeit >>> it = lambda: list(chain(*tupleOfTuples)) >>> timeit.timeit(it) 2.1475738355700913 >>> lc = lambda: [element for tupl in … Read more
Yes, the list comprehension preserves the order of the original iterable (if there is one). If the original iterable is ordered (list, tuple, file, etc.), that’s the order you’ll get in the result. If your iterable is unordered (set, dict, etc.), there are no guarantees about the order of the items.
>>> [f() for f in fl] [1, 2, 3] Absolutely 🙂
Use a nested list comprehension: result = [a for tup in y for a in tup] Example: >>> x = range(10) >>> y = [(i,j**2) for i,j in zip(x,x)] >>> [a for tup in y for a in tup] [0, 0, 1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, … Read more