The best source of information is the official Python tutorial on list comprehensions. List comprehensions are nearly the same as for loops (certainly any list comprehension can be written as a for-loop) but they are often faster than using a for loop. Look at this longer list comprehension from the tutorial (the if part filters … Read more
list2 = [x for ind, x in enumerate(list1) if 4 > ind > 0]
Try this: [(i, j) for i, j in enumerate(mylist)] You need to put i,j inside a tuple for the list comprehension to work. Alternatively, given that enumerate() already returns a tuple, you can return it directly without unpacking it first: [pair for pair in enumerate(mylist)] Either way, the result that gets returned is as expected: … Read more
Assuming you are using Python 3.x: print(*myList, sep=’\n’) You can get the same behavior on Python 2.x using from __future__ import print_function, as noted by mgilson in comments. With the print statement on Python 2.x you will need iteration of some kind, regarding your question about print(p) for p in myList not working, you can … Read more
Try using a list comprehension: >>> mylis = [‘this is test’, ‘another test’] >>> [item.upper() for item in mylis] [‘THIS IS TEST’, ‘ANOTHER TEST’]
TLDR; No, for loops are not blanket “bad”, at least, not always. It is probably more accurate to say that some vectorized operations are slower than iterating, versus saying that iteration is faster than some vectorized operations. Knowing when and why is key to getting the most performance out of your code. In a nutshell, … Read more
The best way to remember this is that the order of for loop inside the list comprehension is based on the order in which they appear in traditional loop approach. Outer most loop comes first, and then the inner loops subsequently. So, the equivalent list comprehension would be: [entry for tag in tags for entry … Read more
In Python, creating a new object e.g. with a list comprehension is often better than modifying an existing one: item_list = [‘item’, 5, ‘foo’, 3.14, True] item_list = [e for e in item_list if e not in (‘item’, 5)] … which is equivalent to: item_list = [‘item’, 5, ‘foo’, 3.14, True] new_list = [] for … Read more
I realize this question is quite old, but you can also create a general function to make this kind of thing easier: def catch(func, handle=lambda e : e, *args, **kwargs): try: return func(*args, **kwargs) except Exception as e: return handle(e) Then, in your comprehension: eggs = (1,3,0,3,2) [catch(lambda : 1/egg) for egg in eggs] [1, … Read more
Python’s conditional expressions were designed exactly for this sort of use-case: >>> l = [1, 2, 3, 4, 5] >>> [‘yes’ if v == 1 else ‘no’ if v == 2 else ‘idle’ for v in l] [‘yes’, ‘no’, ‘idle’, ‘idle’, ‘idle’]