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python進階

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木子寶

def count():
fs = []
for i in range(1, 4):
def f():
return i*i
fs.append(f())
return fs

f1, f2, f3 = count()
print f1, f2, f3

往數組添加內容由f變成f()
打印由f()改成f，答案居然就正確了

1 2-9 python中閉包

2020-03-01

powerlei

其實很簡單，無論你嵌套多少層，關鍵看你的裝修函數最后被那一層調用就在那一層用functools

0 2-14 python中完善decorator

2020-02-29

RayleneWang

2-4這一節的示例中的輸出結果 4的平方寫成了10 ~

0 2-4 python中map()函數

2020-02-29

weixin_慕萊塢6456397

def count():
fs = []
a=iter([i for i in range(1,4)])
for i in range(1, 4):
def pow2():
return next(a)**2
fs.append(pow2)
return fs

f1, f2, f3 = count()
print f1(), f2(), f3()

0 2-9 python中閉包

2020-02-29

weixin_慕萊塢6456397

def count():
fs = []
for i in range(1, 4):
def f(i):
return i*i
fs.append(f(i))
return fs

f1, f2, f3 = count()
print f1(), f2(), f3()

0 2-9 python中閉包

2020-02-28

馬克約瑟

def count():
return [ (lambda j: lambda: j*j)(i) for i in range(1,4) ]

f1, f2, f3 = count()
print f1(), f2(), f3()

0 2-9 python中閉包

2020-02-24

馬克約瑟

不是的，各位！實際上是循環時前面的函數 f 被后面的最后的函數 f 給替換了，請看以下 Python 3 代碼：
def count():
fs = []
for i in range(1, 4):
def f():
return i*i
print(f())
fs.append(f)
return fs

f1, f2, f3 = count() # 依次打印1，4，9
print(f1()) # 9
print(f2()) # 9
print(f3()) # 9

0 2-9 python中閉包

2020-02-24

馬克約瑟

def calc_prod(lst):
return lambda : reduce(lambda x,y:x*y, lst)

f = calc_prod([1, 2, 3, 4])
print f()

1 2-8 python中返回函數

2020-02-24

慕絲7286668

import time
def performance(f):
def print_time(*args):
t1=time.time()
result = f(*args)
t2=time.time()
print str(t2-t1)
return f
return print_time
@performance
def factorial(n):
return reduce(lambda x,y: x*y, range(1, n+1))
print factorial(10)
報錯

0 2-12 python中編寫無參數decorator

2020-02-24