[英]Python recursive call inside a loop. How does it work?
我遇到了一個內部帶有遞歸函數調用的循環,其中循環的起始范圍如下所述。 該代碼輸出以下序列,如下所示。 但是,我無法概念化為什么生成此特定序列。 有人可以對其工作發表一些見識嗎? 將該遞歸函數轉換為輸出相同序列的迭代函數是多么可行。 請幫忙。
碼:
def foo(step=0):
for i in range(step, 4):
print step
foo(step+1)
foo()
輸出:
0 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3 0 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3 0 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3 0 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3
查找類似字母的相似設計的代碼:
def find_anagrams(word, step=0):
print 'step->', step
if step == len(word):
print "".join(word)
for i in range(step, len(word)):
print step, i
word_ = list(word)
word_[step], word_[i] = word_[i], word_[step]
find_anagrams(word_, step+1)
讓我嘗試:
在您的代碼段中,在每個函數調用(即foo(step + 1))中,都會創建一個稱為激活記錄或框架的結構,以存儲有關該函數調用進度的信息。 因此,當函數的執行導致嵌套函數調用時,前一個調用的執行將被掛起,並且其激活記錄將源代碼中存儲的位置存儲在源代碼中,在返回嵌套調用時,控制流應在該位置繼續。
這是主要部分:
當step == 4時,而range(4,4)==空列表,則該時間迭代不會發生,因此它將返回None。 然后它將移至上一幀,在該處停止並開始新的迭代和遞歸函數調用,直到range(4,4)。
注意:遞歸基本情況僅在step == 4時,即時間范圍(4,4)並返回None。
其他情況都需要一個基本案例,否則它將陷入無限循環。
因此,讓我們看一下遞歸跟蹤:我添加i
來區分step
和迭代增量。
# 1 def foo(step=0):
# 2 for i in range(step, 4):
# 3 print 'i: %d, step: %d' % (i,step)
# 4 foo(step+1)
# 5 foo()
line 5
line 1 foo with step=0 which is default
line 2 range(0,4) Frame: A, 0 is over, next=1
line 3 step = 0 Output: i: 0, step: 0
line 4 calling foo(0 + 1)
line 1 foo with step=1
line 2 range(1,4) Frame: B, 1 is over, next=2
line 3 step = 1 Output: i: 1, step: 1
line 4 calling foo(1 + 1)
line 1 foo with step=2
line 2 range(2,4) Frame: C, 2 is over, next=3
line 3 step = 2 Output: i: 2, step: 2
line 4 calling foo(2 + 1)
line 1 foo with step=3
line 2 range(3,4) Frame: D, 3 is over, next=4
line 3 step = 3, Output: i: 3, step: 3
line 4 calling foo(3 + 1)
line 1 foo with step=4
line 2 range(4,4) Frame: E,
This is an empty list, so it won't come inside the loop, so return None.
Come back to previous Frame: D, i=3 was used, now increment to 4. So, again range(4,4)
line 2 range(4,4) Empty list, from Frame: D, return None
Come back to previous Frame C, now i=3, step was called with value 2
line 2 range(2,4)
line 3 step = 2 Output: i: 3, step: 2, why step == 2 because the function foo was called with step=2
line 4 calling foo(2 + 1)
line 1 foo with step=3
line 2 range(3,4)
line 3 step = 3, Output : i: 3, step: 3
line 4 calling foo(3 + 1)
line 1 foo with step=4
line 2 range(4,4) Empty list again, not going inside the list, return None
line 2 range(2,4) From Frame: B, step was == 1, because the function foo was called with step=1
line 3 step: 1 Output: i: 2, step: 1, here i ==2, because this is the second iteration of Frame B.
line 4 calling foo(1 + 1)
line 1 foo with step=2
line 2 range(2,4)
line 3 step: 2 Output: i: 2, step: 2
此后,它遵循相同的遞歸方式,直到迭代范圍被廢除,即range(4,4)
請讓我知道是否有幫助。
我認為可以通過使用stdlib來重構您的字謎代碼,從而避免遞歸循環:
from itertools import permutations
def anagrams (word):
anagrams = set ()
for p in permutations (word):
anagram = ''.join (p)
anagrams |= {anagram}
return anagrams
def isAnagram (word1, word2):
return sorted (word1) == sorted (word2)
考慮for循環。 您正在遍歷range(step, 4)
。 如果step = 0
則迭代[0,1,2,3]
;如果step = 1
則迭代[1,2,3]
,依此類推。 每次調用foo(step)
都會在該范圍內進行迭代 - 但當前調用中要迭代的范圍不會更改 。 因此,對於第一個循環,您會得到一個從0到3的迭代,第二個循環是1-3,依此類推。
為什么這樣打印? 觀察。
for i in range(0,4):
print 0
for j in range(1,4):
print 1
for k in range(2,4):
print 2
for h in range(3,4):
print 3
這將具有與遞歸函數相同的輸出
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