The colon, :, is what tells Python you're giving it a slice and not a regular index. That's why the idiomatic way of making a shallow copy of lists in Python 2 is
To my understanding, slice means lst[start:end], and including start, excluding end. So how would I go about finding the "rest" of a list starting from an element n?
The index or slice is passed to the methods as a single argument, and the way Python does this is by converting the slice notation, (1:10:2, in this case) to a slice object: slice(1,10,2).
Whenever some other code uses slice notation on your an instance of your class Python will call your __getitem__ method with a slice object. (Older versions of Python used a __getslice__ method instead of this approach, but that has been deprecated since 2.6, and support for it is gone completely in Python 3.)
I have a Pandas Data Frame object that has 1000 rows and 10 columns. I would simply like to slice the Data Frame and take the first 10 rows. How can I do this? I've been trying to use this: >&g...
How do I split a list of arbitrary length into equal sized chunks? See also: How to iterate over a list in chunks. To chunk strings, see Split string every nth character?.
45 Python 3 only answer (that doesn't use slicing or throw away the rest of the list, but might be good enough anyway) is use unpacking generalizations to get first and last separate from the middle:
Python strings are immutable. This means that you must create at least 1 new string in order to remove the comma, as opposed to editing the string in place in a language like C.
2017 Answer - pandas 0.20: .ix is deprecated. Use .loc See the deprecation in the docs .loc uses label based indexing to select both rows and columns. The labels being the values of the index or the columns. Slicing with .loc includes the last element. Let's assume we have a DataFrame with the following columns: foo, bar, quz, ant, cat, sat, dat.
