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copied!<p>(In what follows I'm assuming you're using Python 3; in Python 2 the story would be similar but we'd have to discuss the cost of memory allocation due to <code>range</code>.)</p> <p>Your second program spends its time running <em>in Python bytecode</em>. Let's disassemble it using the <a href="http://docs.python.org/3/library/dis.html" rel="nofollow"><code>dis</code></a> module:</p> <pre><code>>>> import dis >>> dis.dis('a=[]\nfor i in range(0,60000000):\n a.append(0)') 1 0 BUILD_LIST 0 3 STORE_NAME 0 (a) 2 6 SETUP_LOOP 36 (to 45) 9 LOAD_NAME 1 (range) 12 LOAD_CONST 0 (0) 15 LOAD_CONST 1 (60000000) 18 CALL_FUNCTION 2 (2 positional, 0 keyword pair) 21 GET_ITER >> 22 FOR_ITER 19 (to 44) 25 STORE_NAME 2 (i) 3 28 LOAD_NAME 0 (a) 31 LOAD_ATTR 3 (append) 34 LOAD_CONST 0 (0) 37 CALL_FUNCTION 1 (1 positional, 0 keyword pair) 40 POP_TOP 41 JUMP_ABSOLUTE 22 >> 44 POP_BLOCK >> 45 LOAD_CONST 2 (None) 48 RETURN_VALUE </code></pre> <p>The loop runs from bytes 22 to 41, and each time round the loop, Python has to decode and execute seven byte code instructions, and call a function too (the function being <code>a.append</code>). That's 420 million byte code instructions and 60 million function calls.</p> <p>Whereas your first program spends its time running <em>in native code</em>:</p> <pre><code>>>> dis.dis('a = [0] * 60000000') 1 0 LOAD_CONST 0 (0) 3 BUILD_LIST 1 6 LOAD_CONST 1 (60000000) 9 BINARY_MULTIPLY 10 STORE_NAME 0 (a) 13 LOAD_CONST 2 (None) 16 RETURN_VALUE </code></pre> <p>You can see that not only is there no loop, there are no function calls. All the work happens "under the hood" inside the <code>BINARY_MULTIPLY</code> instruction, which dispatches to <a href="http://hg.python.org/cpython/file/ae9b74c2b357/Objects/listobject.c#l509" rel="nofollow"><code>list_multiply</code> in <code>listobject.c</code></a>, and since the list <code>[0]</code> contains just one element, the result is constructed in this tight loop at lines 529–536:</p> <pre class="lang-c prettyprint-override"><code>if (Py_SIZE(a) == 1) { elem = a->ob_item[0]; for (i = 0; i < n; i++) { items[i] = elem; Py_INCREF(elem); } return (PyObject *) np; } </code></pre>

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