求简单Python代码 求简洁优美的python代码例子、片段、参考资料
\u6c42\u7b80\u5355Python\u7a0b\u5e8f\u4ee3\u7801k=0
while k>=0:
if 5**(3**k)%2 == 3:
print(k)
break
k += 1
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\u3000\u3000\u697c\u4e0b\u8ba9\u4f60\u53c2\u8003PEP 20\uff0c\u5176\u5b9e\u4e0d\u7528\u53bb\u67e5\uff0c\u6807\u51c6\u5e93\u91cc\u7684this\u6a21\u5757\u5c31\u662f\u5b83\uff08\u8bd5\u8bd5import this\uff09\uff1aThe Zen of Python\uff08Python\u4e4b\u7985\uff09\u3002\u5b83\u5c31\u662f\u4e00\u6bb5\u8bdd\uff1a
s='''The Zen of Python, by Tim PetersBeautiful is better than ugly.Explicit is better than implicit.Simple is better than complex.Complex is better than complicated.Flat is better than nested.Sparse is better than dense.Readability counts.Special cases aren't special enough to break the rules.Although practicality beats purity.Errors should never pass silently.Unless explicitly silenced.In the face of ambiguity, refuse the temptation to guess.There should be one-- and preferably only one --obvious way to do it.Although that way may not be obvious at first unless you're Dutch.Now is better than never.Although never is often better than *right* now.If the implementation is hard to explain, it's a bad idea.If the implementation is easy to explain, it may be a good idea.Namespaces are one honking great idea -- let's do more of those!'''\u3000\u3000\u8ba9\u6211\u4eec\u6765\u505a\u4e2a\u5c0f\u6e38\u620f\u5427\uff1a\u7edf\u8ba1\u4e0a\u9762\u8fd9\u6bb5\u8bdd\u7684\u5355\u8bcd\u603b\u6570\u76ee\uff0c\u4ee5\u53ca\u5404\u4e2a\u5355\u8bcd\u7684\u6570\u91cf\uff08\u4e0d\u533a\u5206\u5927\u5c0f\u5199\uff09\uff0c\u7136\u540e\u6309\u5b57\u5178\u987a\u5e8f\u8f93\u51fa\u6bcf\u4e2a\u5355\u8bcd\u51fa\u73b0\u7684\u6b21\u6570\u3002\u8981\u6c42\uff0c\u4f8b\u5982it's\u548cyou're\u7b49\u8981\u62c6\u5206\u6210it is\u548cyou are\u3002\u4f60\u4f1a\u600e\u4e48\u5199\u4ee3\u7801\u5462\uff1f\u5982\u4f55\u4fdd\u6301\u7b80\u6d01\u5462\uff1f
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import rep = re.compile("(\w+)('s|'re|n't)?")wc = {}tail_map = { "'s" : 'is', "'re" : 'are', "n't": 'not'}for m in re.finditer(p, s): word = m.group(1).lower() # Get the word in lower case wc[word] = wc.get(word, 0) + 1 # Increase word count tail = m.group(2) # Get the word tail if tail is not None: # If a word tail exists, tail = tail_map[tail] # map it to its full form wc[tail] = wc.get(tail, 0)+1 # Increase word countprint ('Total word count: %d'%sum(wc.values())) # Output the total countmax_len = max(map(len, wc.keys())) # Calculate the max length of words for pretty printingfor w in sorted(wc.keys()): # Sort the words print ('%*s => %d'%(max_len, w, wc[w])) # Output
def getsubset(myset,subtract):
if len(myset)<=1:
return []
result=[]
newsubtract=subtract.copy()
for i in subtract:
result.append(myset-{i})
newsubtract=newsubtract-{i}
result.extend(getsubset(myset-{i},newsubtract))
return result
def subset(myset):
result=[set(),myset] if myset else [myset]
result.extend(getsubset(myset,myset))
return result
result=subset({'a','b','c','d'})
toprint=[sorted(x) for x in result]
for i in sorted(toprint,key=lambda x:(len(x),x)):
print(i)
虽然这对我来说的确比较简单，可也不是几分钟就能写出来的，好歹给点分吧
而且(a,b,a,c) 不是集合，集合中元素不应有重复
一个n个元素的集合的所有子集是2的n次方个吗，如是，这个程序应该就是正确的
import itertools
#集合myset中的n个元素相互组合
def get_one_subset(myset,n):
return itertools.combinations(myset,n)
def get_all_subset(mst):
for i in range(1,len(mst)):
sub_set=set()
sub_set=sub_set|set(get_one_subset(mst,i))
yield sub_set
#-----------example---------------------------
mst={'a','b','c','d'}
for st in get_all_subset(mst):
for t in st:
print(sorted(list(t)))
利用Python制作的表白代码，文字可以随意修改
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