Python indexing a list. 1. Assigning Indexes of Two Lists Together in Python. 3. Indexing a list with ...

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 ...

Aug 10, 2010 · In Python 3, your example range(N)[::step] produces a range object, not a list. To really see what is happening, you need to coerce the range to a list, np.array, etc. – PikalaxALT

TL; DR: use np.where as it is the fastest option. Your options are np.where, itertools.compress, and list comprehension.

Dec 2, 2013 · but it output the complete first list .i.e. [0,1] Even. print data[0][:] produces the same result. My question is specifically how to accomplish what I have mentioned. And more generally, how is python handling double/nested lists?

This is the best one I have read. numpy arrays are far more efficient than Python lists. If the list is short it's no problem making a copy of it from a Python list, if it isn't then perhaps the developer should consider storing the elements in numpy array in the first place. –

May 30, 2017 · But if you need to use list anyway, you have (using NumPy solution) create np.array from both lists, use boolean indexing and finally converting array back to list with tolist() method. To be precise, you should include those objects creation into time comparison.

numpyIndexValues -> time:1.38940598 (when converted the lists to numpy arrays) numpyIndexValues -> time:0.0193445 (using numpy array instead of python list as input, and conversion code removed) mapIndexValues -> time:0.06477512099999999 getIndexValues -> time:0.06391049500000001 multipleListItemValues -> time:0.043773591 pythonLoopOverlap ...

If you want to remove them from the list, then the Pythonic way is not to necessarily remove from the list, but write a list comprehension as if you were creating a new list, and assigning back in-place using the listvar[:] on the left-hand-side: a[:] = [x for x in a if x <= 2]

Mar 1, 2013 · I found this recipe at the python mailing list. With it you can access the elements of a container using an iterator of indexes. If you need to use container[index_1, index_2] notation, this can be adapted easily using the methods outlined by Mark's post.