對於最近的腳本,我必須遍歷分支樹中的可能流。我建立了一個包含每個項目運行狀態的字典,以便我評估。例如:減少構建字典/字符串的嵌套循環(分支樹樣式)
for a in range(0, 2):
for b in range(0, 2):
for c in range(0, 2):
for d in range(0, 2):
...
run = [a,b,c,d ...]
不幸的是,它起初很小,但發展到十幾個州。這怎麼能減少以消除所有的嵌套循環?
如果某些州有3個或4個州而不是2個州,那麼類似的答案會有所不同嗎?
同樣,如果每個循環都來自函數列表,那麼同樣的問題會受到怎樣的影響?我懷疑它會是一樣的。例如:
def leet_lookup(char):
table = {"a": ["a","A","@"],
"b": ["b", "B", "8"],
"c": ["c", "C", "<"],
"e": ["e", "E", "3"],
"i": ["i", "I", "1"],
"l": ["l", "L", "1"],
"o": ["o", "O", "0"],
"t": ["t", "T", "7"] }
try:
result = table[char.lower()]
except KeyError:
result = [char.lower(), char.upper()]
return result
result = []
# V o l l e y b a l l = 10 chars
for c1 in leet_lookup('v'):
for c2 in leet_lookup('o'):
for c3 in leet_lookup('l'):
for c4 in leet_lookup('l'):
for c5 in leet_lookup('e'):
for c6 in leet_lookup('y'):
for c7 in leet_lookup('b'):
for c8 in leet_lookup('a'):
for c9 in leet_lookup('l'):
for c10 in leet_lookup('l'):
result.append("%s%s%s%s%s%s%s%s%s%s" % (c1, c2, c3, c4, c5, c6, c7, c8, c9, c10))
很好的解決方案壽! – soupault 2014-09-18 16:12:22
優秀的答案和建議,第二個特別令人望而生畏,但您的解決方案非常出色!謝謝 – 2014-09-18 17:53:46