Iterate through every sub-list in your original list and unpack it in the print call with *: a = [[1, 3, 4], [2, 5, 7]] for s in a: print(*s) The separation is by default set to ‘ ‘ so there’s no need to explicitly provide it. This prints: 1 3 4 2 5 7 … Read more
You can also use (at least v1.9.3) of rbindlist in the data.table package: library(data.table) rbindlist(mylist, fill=TRUE) ## Hit Project Year Rating Launch ID Dept Error ## 1: True Blue 2011 4 26 Jan 2012 19 1, 2, 4 NA ## 2: False NA NA NA NA NA NA Record not found ## 3: True Green … Read more
A strange behaviour indeed, but that’s only because * operator makes shallow copies, in your case – shallow copies of [0, 0, 0] list. You can use the id() function to make sure that these internal lists are actually the same: out=[[0]*3]*3 id(out[0]) >>> 140503648365240 id(out[1]) >>> 140503648365240 id(out[2]) >>> 140503648365240 Comprehensions can be used … Read more
My best guess is that using multiplication in the form [[]] * x causes Python to store a reference to a single cell…? Yes. And you can test this yourself >>> lst = [[]] * 3 >>> print [id(x) for x in lst] [11124864, 11124864, 11124864] This shows that all three references refer to the … Read more
>>> def isplit(iterable,splitters): return [list(g) for k,g in itertools.groupby(iterable,lambda x:x in splitters) if not k] >>> isplit(L,(None,)) [[1, 4], [6, 9], [3, 9, 4]] >>> isplit(L,(None,9)) [[1, 4], [6], [3], [4]] benchmark code: import timeit kabie=(“isplit_kabie”, “”” import itertools def isplit_kabie(iterable,splitters): return [list(g) for k,g in itertools.groupby(iterable,lambda x:x in splitters) if not k] “”” ) … Read more
Python variables contain pointers, or references, to objects. All values (even integers) are objects, and assignment changes the variable to point to a different object. It does not store a new value in the variable, it changes the variable to refer or point to a different object. For this reason many people say that Python … Read more
Option 2 is correct. The nested list should be inside a <li> element of the list in which it is nested. Link to the W3C Wiki on Lists (taken from comment below): HTML Lists Wiki. Link to the HTML5 W3C ul spec: HTML5 ul. Note that a ul element may contain exactly zero or more … Read more
Using generator functions can make your example easier to read and improve performance. Python 2 Using the Iterable ABC added in 2.6: from collections import Iterable def flatten(xs): for x in xs: if isinstance(x, Iterable) and not isinstance(x, basestring): for item in flatten(x): yield item else: yield x Python 3 In Python 3, basestring is … Read more
To explain “why”: The + operation adds the array elements to the original array. The array.append operation inserts the array (or any object) into the end of the original array, which results in a reference to self in that spot (hence the infinite recursion). The difference here is that the + operation acts specific when … Read more
When you write [x]*3 you get, essentially, the list [x, x, x]. That is, a list with 3 references to the same x. When you then modify this single x it is visible via all three references to it: x = [1] * 4 l = [x] * 3 print(f”id(x): {id(x)}”) # id(x): 140560897920048 print( … Read more