Python - ビルトインオブジェクト(基本的な操作、対話型コマンドライン)

初めてのPython 第3版
Mark Lutz (著)

開発環境

• OS X Lion - Apple(OS)
• BBEdit - Bare Bones Software, Inc.(Text Editor)
• Script言語: Python

『初めてのPython 第3版』(Mark Lutz 著、夏目 大 訳、オライリー・ジャパン、2009年、ISBN978-4-87311-393-7) のII部(ビルトインオブジェクト)のまとめ演習1.(基本的な操作)を解いてみる。

その他参考書籍

• Pythonチュートリアル 第2版
• Python クックブック 第2版

1.(基本的な操作)

入出力結果(Terminal)

$ python
Python 3.3.0 (default, Sep 29 2012, 08:16:08) 
[GCC 4.2.1 Compatible Apple Clang 3.1 (tags/Apple/clang-318.0.58)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> 2 ** 16 # 
65536
>>> pow(2, 16)
65536
>>> 2 / 5 # 0.4
0.4
>>> 2/5.0 # 0.4
0.4
>>> 2 // 5 # 0 python2から3での変更点
0
>>> "spam" + "eggs" # spameggs
'spameggs'
>>> S = "ham"
>>> "eggs " + S # eggs ham
'eggs ham'
>>> S * 5 # hamhamhamhamham
'hamhamhamhamham'
>>> s[:0] # "" 空文字
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 's' is not defined
>>> S[:0] # "" 空文字
''
>>> "green %s and %s" % ("eggs", S) # green eggs and ham
'green eggs and ham'
>>> "green {0:s} and {1:s}".format("eggs", S)
'green eggs and ham'
>>> ('x', )[0] # x
'x'
>>> ('x', 'y')[1] # y
'y'
>>> L = [1,2,3] + [4,5,6]
>>> L # [1,2,3]
[1, 2, 3, 4, 5, 6]
>>> L # [1,2,3, 4, 5, 6]
[1, 2, 3, 4, 5, 6]
>>> L[:] # [1,2,3,4,5,6]
[1, 2, 3, 4, 5, 6]
>>> L[:0] # []
[]
>>> L[-2] # 5
5
>>> L[-2:] # [5,6]
[5, 6]
>>> ([1,2,3] + [4,5,6])[2:4] # [3,4]
[3, 4]
>>> [L[2], L[3]] # [3,4]
[3, 4]
>>> L.reverse(); L # [6,5,4,3,2,1]
[6, 5, 4, 3, 2, 1]
>>> L.sort(); L # [1,2,3,4,5,6]
[1, 2, 3, 4, 5, 6]
>>> L.index(4) # 5
3
>>> L.index(4) # 3
3
>>> {'a':1, 'b':2}['b'] # 2
2
>>> D = ['x':1,'y':2,'z':3}
  File "<stdin>", line 1
    D = ['x':1,'y':2,'z':3}
            ^
SyntaxError: invalid syntax
>>> D = {'x':1,'y':2,'z':3}
>>> D['w'] = 0
>>> D['x'] + D['w'] # 1
1
>>> D[(1,2,3)] = 4
>>> D.keys() # ['x', 'y', 'z', 'w', (1,2,3)]
dict_keys(['w', (1, 2, 3), 'z', 'x', 'y'])
>>> type(D.keys())
<class 'dict_keys'>
>>> D.keys() # dict_keys(['x','y','z','w',(1,2,3)]) (順番はわからない)
dict_keys(['w', (1, 2, 3), 'z', 'x', 'y'])
>>> type(D.values())
<class 'dict_values'>
>>> D.values() # dict_values([1,2,3,0,4]) (順番は分からない)
dict_values([0, 4, 3, 1, 2])
>>> D.has_key((1,2,3)) # True
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'dict' object has no attribute 'has_key'
>>> dir(D)
['__class__', '__contains__', '__delattr__', '__delitem__', '__dir__', '__doc__', '__eq__', '__format__', '__ge__', '__getattribute__', '__getitem__', '__gt__', '__hash__', '__init__', '__iter__', '__le__', '__len__', '__lt__', '__ne__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__setitem__', '__sizeof__', '__str__', '__subclasshook__', 'clear', 'copy', 'fromkeys', 'get', 'items', 'keys', 'pop', 'popitem', 'setdefault', 'update', 'values']
>>> # dict.has_keyはpython3.3では無くなったみたい
... # 代わりにinで調べてみる
... (1,2,3) in D.keys()
True
>>> [[]] # [[]]
[[]]
>>> ["",[],(),{},None]
['', [], (), {}, None]
>>> quit()
$