【wtfPython】一組有趣的、微妙的、複雜的Python程式碼片段
原作者:董偉明 (Dongweiming)
原文連結:推薦wtfPython: 一組有趣的、微妙的、複雜的Python程式碼片段
本文有細微改動
wtfPython 1 就是「What the f**k Python?」的意思,這個專案列舉了一些程式碼片段,可能結果和你想到的是不一致的,並且作者會告訴你為什麼。本文將展示最有意義的一部分:
混合Tab和空格
def square(x):
sum_so_far = 0
for counter in range(x):
sum_so_far = sum_so_far + x
return sum_so_far
print (square(10))
結果是10??不是應該100麼?
其實這種錯誤的結果的原因,所有書籍和開發者都說過,就是不要混Tab和空格,原始碼你可以看專案中的mixed_tabs_and_spaces.py
字典鍵的隱式轉換
In [1]: some_dict = {}
...: some_dict[5.5] = "Ruby"
...: some_dict[5.0] = "JavaScript"
...: some_dict[5] = "Python"
...:
In [2]: some_dict[5.5]
Out[2]: 'Ruby'
In [3]: some_dict[5.0]
Out[3]: 'Python'
In [4]: some_dict[5]
Out[4]: 'Python'
這樣的原因是鍵被隱式的轉換了:
In [5]: hash(5) == hash(5.0)
Out[5]: True
生成器執行時間的差異
In [6]: array = [1, 8, 15]
...: g = (x for x in array if array.count(x) > 0)
...: array = [2, 8, 22]
...:
In [7]: print(list(g))
[8]
這種隱式的非預期結果在實際開發中是可能出現的,原因是in的操作是在申明時求值的,而if是在執行期求值的。
在字典迭代時修改該字典
In [8]: x = {0: None}
...:
...: for i in x:
...: del x[i]
...: x[i+1] = None
...: print(i)
...:
0
1
2
3
4
首先說的時候在迭代過程中是不能修改字典的長度的:
In [13]: for i in x:
...: del x[i]
...:
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
<ipython-input-13-a5c6e73be64f> in <module>()
----> 1 for i in x:
2 del x[i]
3
RuntimeError: dictionary changed size during iteration
但是刪掉一個新增一個是可以,運行了5次才結束是因為字典會定期重新設定以便接受更多的鍵,但是和專案中的執行8次是不一樣的
在列表迭代時刪除條目
In [14]: list_1 = [1, 2, 3, 4]
...: list_2 = [1, 2, 3, 4]
...: list_3 = [1, 2, 3, 4]
...: list_4 = [1, 2, 3, 4]
...:
...: for idx, item in enumerate(list_1):
...: del item
...:
...: for idx, item in enumerate(list_2):
...: list_2.remove(item)
...:
...: for idx, item in enumerate(list_3[:]):
...: list_3.remove(item)
...:
...: for idx, item in enumerate(list_4):
...: list_4.pop(idx)
...:
In [15]: list_1, list_2
Out[15]: ([1, 2, 3, 4], [2, 4])
In [16]: list_3, list_4
Out[16]: ([], [2, 4])
其中只有list_3是正確的行為。但是為什麼會出現[2, 4]的結果呢?第一次刪掉了index是0的1,就剩[2, 3, 4],然後移除index 1, 就是3,剩下了[2, 4],但是現在只有2個元素,迴圈就結束了
is
>>> a = 256
>>> b = 256
>>> a is b
True
>>> a = 257
>>> b = 257
>>> a is b
False
>>> a = 257; b = 257
>>> a is b
True
is 用來對比身份,而 == 用來對比值。通常is為True,==就是True,但是反之不一定:
>>> [] == []
True
>>> [] is [] # 2個列表使用了不同的記憶體位置
False
上面的例子中,-5 - 256由於太經常使用,所以設計成固定存在的物件:
>>> id(256)
10922528
>>> a = 256
>>> b = 256
>>> id(a)
10922528
>>> id(b)
10922528
>>> id(257)
140084850247312
>>> x = 257
>>> y = 257
>>> id(x)
140084850247440
>>> id(y)
140084850247344
is not … 和 is (not …)
>>> 'something' is not None
True
>>> 'something' is (not None)
False
其中(not None)優先執行,最後其實變成了 ‘something’ is True
迴圈中的函式也會輸出到相同的輸出
In [17]: funcs = []
...: results = []
...: for x in range(7):
...: def some_func():
...: return x
...: funcs.append(some_func)
...: results.append(some_func())
...:
...: funcs_results = [func() for func in funcs]
...:
In [18]: results, funcs_results
Out[18]: ([0, 1, 2, 3, 4, 5, 6], [6, 6, 6, 6, 6, 6, 6])
開發陷阱,閉包變數繫結,解決方法就是把迴圈的變數傳到some_func裡面去:
In [19]: funcs = []
...: for x in range(7):
...: def some_func(x=x):
...: return x
...: funcs.append(some_func)
...:
In [20]: [func() for func in funcs]
Out[20]: [0, 1, 2, 3, 4, 5, 6]
迴圈中的區域性變數洩露
>>> x = 1
>>> print([x for x in range(5)])
[0, 1, 2, 3, 4]
>>> print(x, ': x in global')
(4, ': x in global')
在Python 2中x的值在一個迴圈執行之後被改變了。不過在Python 3這個問題解決了
可變預設引數
In [1]: def some_func(default_arg=[]):
...: default_arg.append("some_string")
...: return default_arg
...:
In [2]: some_func()
Out[2]: ['some_string']
In [3]: some_func()
Out[3]: ['some_string', 'some_string']
In [4]: some_func([])
Out[4]: ['some_string']
In [5]: some_func()
Out[5]: ['some_string', 'some_string', 'some_string']
Python是引用傳遞,上面例子的引數是一個列表,它所指向的物件可以被修改。通用的解決辦法是在函式內判斷:
def some_func(default_arg=None):
if not default_arg:
default_arg = []
default_arg.append("some_string")
return default_arg
+和+=的差別
>>> a = [1, 2, 3, 4]
>>> b = a
>>> a = a + [5, 6, 7, 8]
>>> a, b
([1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4])
>>> a = [1, 2, 3, 4]
>>> b = a
>>> a += [5, 6, 7, 8]
>>> a, b
([1, 2, 3, 4, 5, 6, 7, 8], [1, 2, 3, 4, 5, 6, 7, 8])
通常的運算過程,區別就是a = a + X 和 a += X。這是因為 a = a + X 是重新建立一個物件a,而 a += X 是在a這個list上面做extend操作
元組賦值
In [6]: another_tuple = ([1, 2], [3, 4], [5, 6])
...:
In [7]: another_tuple[2].append(1000)
In [8]: another_tuple
Out[8]: ([1, 2], [3, 4], [5, 6, 1000])
In [9]: another_tuple[2] += [99, 999]
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-9-d07c65f24a63> in <module>()
----> 1 another_tuple[2] += [99, 999]
TypeError: 'tuple' object does not support item assignment
In [10]: another_tuple
Out[10]: ([1, 2], [3, 4], [5, 6, 1000, 99, 999])
在我們的印象裡面元組是不可變的呀?簡單的說對list的賦值成功了,但是賦值失敗了,不過由於值是引用的,所以才會出現這個執行失敗實際成功的效果
使用在範圍內未定義的變數
In [11]: a = 1
...: def some_func():
...: return a
...:
...: def another_func():
...: a += 1
...: return a
...:
In [12]: some_func()
Out[12]: 1
In [13]: another_func()
---------------------------------------------------------------------------
UnboundLocalError Traceback (most recent call last)
<ipython-input-13-703bd168975f> in <module>()
----> 1 another_func()
<ipython-input-11-cff7ceae4600> in another_func()
4
5 def another_func():
----> 6 a += 1
7 return a
UnboundLocalError: local variable 'a' referenced before assignment
這是由於在another_func中的賦值操作會把a變成一個本地變數,但是在相同範圍內並沒有初始化它。如果希望它能正確執行可以加global:
In [17]: def another_func():
...: global a
...: a += 1
...: return a
...:
In [18]: another_func()
Out[18]: 2
使用finally的return
In [19]: def some_func():
...: try:
...: return 'from_try'
...: finally:
...: return 'from_finally'
...:
In [20]: some_func()
Out[20]: 'from_finally'
try…finally這種寫法裡面,finally中的return語句永遠是最後一個執行
忽略類範圍的名稱解析
In [21]: x = 5
...: class SomeClass:
...: x = 17
...: y = (x for i in range(10))
...:
In [22]: list(SomeClass.y)[0]
Out[22]: 5
In [23]: x = 5
...: class SomeClass:
...: x = 17
...: y = [x for i in range(10)]
...:
In [24]: SomeClass.y[0]
Out[24]: 5
這是由於類範圍的名稱解析被忽略了,而生成器有它自己的本地範圍,而在Python3中列表解析也有自己的範圍,所以x的值是5。不過,第二個例子在Python2中SomeClass.y[0]的值是17
列表中的布林值
In [34]: mixed_list = [False, 1.0, "some_string", 3, True, [], False]
...: integers_found_so_far = 0
...: booleans_found_so_far = 0
...:
...: for item in mixed_list:
...: if isinstance(item, int):
...: integers_found_so_far += 1
...: elif isinstance(item, bool):
...: booleans_found_so_far += 1
...:
In [35]: booleans_found_so_far
Out[35]: 0
In [36]: integers_found_so_far
Out[36]: 4
這是由於布林也是int的子類:
In [41]: isinstance(True, int)
Out[41]: True
自引用迴圈巢狀賦值
In [42]: a, b = a[b] = {}, 5
...:
In [43]: a, b
Out[43]: ({5: ({...}, 5)}, 5)
看起來有點懵吧,我們拆一下:
In [44]: a[b] = {}, 5
In [47]: a, b
Out[47]: ({5: ({}, 5)}, 5)
這樣b是5,而a[5]的值是({}, 5),所以a是{5: ({}, 5)。接著看:
In [48]: a[b] = a, b
In [49]: a
Out[49]: {5: ({...}, 5)}
這其實是一個對自己的「自引用」,看個例子:
In [50]: a = {}
In [51]: a[5] = a
In [52]: a
Out[52]: {5: {...}}
In [53]: a[5] == a
Out[53]: True
In [54]: a[5][5][5]
Out[54]: {5: {...}}
看,a[5]就是a,這可以是一個永久迴圈,Python用…來表示了