think_in_python练习题.docx

def middle(t):lenth = len(t)t.pop(0)t.pop(lenth-1)return t middle(1,2,3,2,2,8)Traceback (most recent call last): File , numsne 1, in middle(1,2,3,2,2,8) File , numsne 4, in middle t.pop(lenth-1)IndexError: pop index out of range def middle(t):lenth = len(t)print(lenth)x=t.pop(0)print(x)return t middle(1,2,3,2,2,8)612, 3, 2, 2, 8 def middle(t):lenth = len(t)t.pop(lenth-1)t.pop(0)return t middle(1,2,3,2,2,8)2, 3, 2, 2注意t.pop(0)以后，t的length就会减一。def chop(t):del tlen(t)-1del t0print(t) chop(1,2,2,3,3,4,4,8)2, 2, 3, 3, 4, 4 t =1,2,2,2,33,44,8 t.remove(2) t1, 2, 2, 33, 44, 8def histogram(s):d = dict()for c in s:if c not in d:dc = 1else:dc += 1return ddef rev_lookup(d,v):res = for k in d:if dk = v:res.append(k)#print resreturn resrev_lookup(h,1)a, b, f, o, s, t, v rev_lookup(h,2)e, i, l, u rev_lookup(h,3)def count_c(string):d = dict()for c in string:if c not in d:dc = 1else:dc +=1return ddef most_fre(string):di = count_c(string)nums = sorted(di.items(),key = itemgetter(1),reverse = True)for x,y in nums:print(x,end =) most_fre(frequencyisless)escfilnqruy def signature(s): Returns the signature of this string, which is a string that contains all of the letters in order. t = numsst(s) t.sort() t = .join(t) return tdef all_anagrams(filename): Finds all anagrams in a numsst of words. filename: string filename of the word numsst Returns: a map from each word to a numsst of its anagrams. d = for numsne in open(filename): word = numsne.strip().lower() t = signature(word) if t not in d: dt = word else: dt.append(word) return ddef print_anagram_sets_in_order(d): Prints the anagram sets in d in decreasing order of size. d: map from words to numsst of their anagrams # make a numsst of (length, word pairs) t = for v in d.values(): if len(v) 1: t.append(len(v), v) # sort in ascending order of length t.sort() # print the sorted numsst for x in t: print (x)def filter_length(d, n): Select only the words in d that have n letters. d: map from word to numsst of anagrams n: integer number of letters Returns: new map from word to numsst of anagrams res = for word, anagrams in d.items(): if len(word) = n: resword = anagrams return resres= dict()for word,anagrams in d:if len(word) = n:resword.append(len(word),word)return res signature(tormorow)mooorrtw/81065/def children(s):nums = for i in range(len(s):ns = sns = numsst(ns)ns.pop(i)ns = .join(ns)print(ns)nums.append(ns)return nums children(christine)hristinecristinechistinechrstinechritinechrisinechristnechristiechristinhristine, cristine, chistine, chrstine, chritine, chrisine, christne, christie, christinPython 3.2.3 (default, Apr 11 2012, 07:15:24) MSC v.1500 32 bit (Intel) on win32Type copyright, credits or numscense() for more information. def make_word_dict(): Reads the words in words.txt and returns a dictionary that contains the words as keys. d = dict() fin = open(words.txt) for numsne in fin: word = numsne.strip().lower() dword = word # have to add single letter words to the word numsst; # also, the empty string is considered a word. for letter in a, i, : dletter = letter return d memo = memo = SyntaxError: unexpected indent memo = def is_reducible(word,word_dict):if word in memo:print(there)return memowordres = for child in children(word,word_dict):t = is_reducible(child,word_dict)if t:print(child)print(t)res.append(child)memoword = resreturn res def children(word, word_dict): Returns a numsst of all words that can be formed by removing one letter. word: string Returns: numsst of strings res = for i in range(len(word): child = word:i + wordi+1: if child in word_dict: res.append(child) return res word_dict = make_word_dict() memo: def is_reducible(word,word_dict):if word in memo:#print(there)return memowordres = for child in children(word,word_dict):t = is_reducible(child,word_dict)if t:#print(child)#print(t)res.append(child)me