第一章    数据结构和算法 
1.1    将序列分解为单独的变量
p = (4, 5) x, y = p print x print y data = [ 'acme', 50, 91.1, (2012, 12, 21) ] name, shares, price, date = data print name print shares print price print date name, shares, price, (year, mon, day ) = data print year p = (4, 5) #x, y, z = p 错误!!! s = 'hello!' a, b, c, d, e, f = s print a print f data = [ 'acme', 50, 91.1, (2012, 12, 21) ] _, shares, price, _ = data print shares print price #其他数据可以丢弃了
1.2 从任意长度的可迭代对象中分解元素
from audioop import avg def drop_first_last(grades): first, *middle, last = grades return avg(middle) record = ('dave', 'dave@example.com', '777-333-2323', '234-234-2345') name, email, *phone_numbers = record print name print email print phone_numbers *trailing, current = [10, 8, 7, 2, 5] print trailing #[10, 8, 7, 2, ] print current #5 records = [ ('foo', 1, 2), ('bar', 'hello'), ('foo', 5, 3) ] def do_foo(x, y): print ('foo', x, y) def do_bar(s): print ('bar', s) for tag, *args in records: if tag == 'foo': do_foo(*args) elif tag == 'bar': do_bar(*args) line = 'asdf:fedfr234://wef:678d:asdf' uname, *fields, homedir, sh = line.split(':') print uname print homedir record = ('acme', 50, 123.45, (12, 18, 2012)) name, *_, (*_, year) = record print name print year items = [1, 10, 7, 4, 5, 9] head, *tail = items print head #1 print tail #[10, 7, 4, 5, 9] def sum(items): head, *tail = items return head + sum(tail) if tail else head sum(items)
1.3 保存最后n个元素
from _collections import deque def search(lines, pattern, history=5): previous_lines = deque(maxlen = history) for line in lines: if pattern in line: yield line, previous_lines previous_lines.append(line) # example use on a file if __name__ == '__main__': with open('somefile.txt') as f: for line, prevlines in search(f, 'python', 5): for pline in prevlines: print (pline) #print (pline, end='') print (line) #print (pline, end='') print ('-'*20) q = deque(maxlen=3) q.append(1) q.append(2) q.append(3) print q q.append(4) print q q = deque() q.append(1) q.append(2) q.append(3) print q q.appendleft(4) print q q_pop = q.pop() print q_pop print q q_popleft = q.popleft() print q_popleft print q
1.4 找到最大或最小的n个元素
import heapq nums = [1,30,6,2,36,33,46,3,23,43] print (heapq.nlargest(3, nums)) print (heapq.nsmallest(3, nums)) portfolio = [ {'name':'ibm', 'shares':100, 'price':2.4}, {'name':'a', 'shares':1040, 'price':12.4}, {'name':'s', 'shares':40, 'price':23.4}, {'name':'d', 'shares':1, 'price':2.49}, {'name':'f', 'shares':9, 'price':24} ] cheap = heapq.nsmallest(3, portfolio, key=lambda s: s['price']) expensive = heapq.nlargest(3, portfolio, key=lambda s: s['price']) print cheap print expensive nums = [1,8,2,23,7,-4,18,23,42,37,2] heap = list(nums) print heap heapq.heapify(heap) print heap print heapq.heappop(heap) print heapq.heappop(heap) print heapq.heappop(heap)
1.5 实现优先级队列
import heapq class priorityqueue: def __init__(self): self._queue = [] self._index = 0 def push(self, item, priority): heapq.heappush(self._queue, (-priority, self._index, item)) self._index += 1 def pop(self): return heapq.heappop(self._queue)[-1] #example class item: def __init__(self, name): self.name = name def __repr__(self): return 'item({!r})'.format(self.name) q = priorityqueue() q.push(item('foo'), 1) q.push(item('spam'), 4) q.push(item('bar'), 5) q.push(item('grok'), 1) print q.pop() print q.pop() print q.pop() a = item('foo') b = item('bar') #a < b error a = (1, item('foo')) b = (5, item('bar')) print a < b c = (1, item('grok')) #a < c error a = (1, 0, item('foo')) b = (5, 1, item('bar')) c = (1, 2, item('grok')) print a < b print a < c
1.6 在字典中将建映射到多个值上
d = { 'a' : [1, 2, 3], 'b' : [4, 5] } e = { 'a' : {1, 2, 3}, 'b' : {4, 5} } from collections import defaultdict d = defaultdict(list) d['a'].append(1) d['a'].append(2) d['a'].append(3) print d d = defaultdict(set) d['a'].add(1) d['a'].add(2) d['a'].add(3) print d d = {} d.setdefault('a', []).append(1) d.setdefault('a', []).append(2) d.setdefault('b', []).append(3) print d d = {} for key, value in d:#pairs: if key not in d: d[key] = [] d[key].append(value) d = defaultdict(list) for key, value in d:#pairs: d[key].append(value)
1.7 让字典保持有序
from collections import ordereddict d = ordereddict() d['foo'] = 1 d['bar'] = 2 d['spam'] = 3 d['grol'] = 4 for key in d: print (key, d[key]) import json json.dumps(d)
1.8 与字典有关的计算问题
price = { 'acme':23.45, 'ibm':25.45, 'fb':13.45, 'io':4.45, 'java':45.45, 'av':38.38, } min_price = min( zip( price.values(), price.keys() ) ) print min_price max_price = max( zip( price.values(), price.keys() ) ) print max_price price_sorted = sorted( zip( price.values(), price.keys() ) ) print price_sorted price_and_names = zip( price.values(), price.keys() ) print (min(price_and_names)) #print (max(price_and_names)) error zip()创建了迭代器，内容只能被消费一次 print min(price) print max(price) print min(price.values()) print max(price.values()) print min(price, key = lambda k : price[k]) print max(price, key = lambda k : price[k]) min_value = price[ min(price, key = lambda k : price[k]) ] print min_value price = { 'aaa': 23, 'zzz': 23, } print min( zip( price.values(), price.keys() ) ) print max( zip( price.values(), price.keys() ) )
1.9 在两个字典中寻找相同点
a = { 'x':1, 'y':2, 'z':3 } b = { 'x':11, 'y':2, 'w':10 } print a.keys() & b.keys() #{'x','y'} print a.keys() - b.keys() #{'z'} print a.items() & b.items() #{('y', 2)} c = {key: a[key] for key in a.keys() - {'z', 'w'} } print c #{'x':1, 'y':2}
1.10 从序列中移除重复项且保持元素间顺序不变
def dedupe(items): seen = set() for item in items: if item not in seen: yield item seen.add(item) #example a = [1,5,2,1,9,1,5,10] print list(dedupe(a)) def dedupe2(items, key = none): seen = set() for item in items: val = item if key is none else key(item) if val not in seen: yield item seen.add(val) #example a = [ {'x':1, 'y':2}, {'x':1, 'y':3}, {'x':1, 'y':2}, {'x':2, 'y':4}, ] print list( dedupe2(a, key=lambda d : (d['x'], d['y']) ) ) print list( dedupe2(a, key=lambda d : (d['x']) ) ) a = [1,5,2,1,9,1,5,10] print set(a)
1.11 对切片命名
items = [0,1,2,3,4,5,6] a = slice(2,4) print items[2:4] print items[a] items[a] = [10,11] print items print a.start print a.stop print a.step
1.12 找出序列中出现次数最多的元素
words = [ 'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes', 'the', 'look' ] from collections import counter word_counts = counter(words) top_three = word_counts.most_common(3) print top_three print word_counts['look'] print word_counts['the'] morewords = ['why', 'are', 'you', 'not', 'looking', 'in', 'my', 'eyes'] for word in morewords: word_counts[word] += 1 print word_counts['eyes'] print word_counts['why'] word_counts.update(morewords) print word_counts['eyes'] print word_counts['why'] a = counter(words) b = counter(morewords) print a print b c = a + b print c d = a - b print b
1.13 通过公共键对字典列表排序
rows = [ {'fname':'brian', 'lname':'jones', 'uid':1003}, {'fname':'david', 'lname':'beazley', 'uid':1002}, {'fname':'john', 'lname':'cleese', 'uid':1001}, {'fname':'big', 'lname':'jones', 'uid':1004} ] from operator import itemgetter rows_by_fname = sorted(rows, key=itemgetter('fname')) rows_by_uid = sorted(rows, key=itemgetter('uid')) print rows_by_fname print rows_by_uid rows_by_lfname = sorted(rows, key=itemgetter('lname', 'fname')) print rows_by_lfname rows_by_fname = sorted(rows, key=lambda r: r['fname']) rows_by_lfname = sorted(rows, key=lambda r: (r['fname'], r['lname'])) print rows_by_fname print rows_by_lfname print min(rows, key=itemgetter('uid')) print max(rows, key=itemgetter('uid'))
1.14 对不原生支持比较操作的对象排序
class user: def __init__(self, user_id): self.user_id = user_id def __repr__(self): return 'user({})'.format(self.user_id) users = [user(23), user(3), user(99)] print users print sorted(users, key = lambda u: u.user_id) from operator import attrgetter print sorted(users, key=attrgetter('user_id')) print min(users, key=attrgetter('user_id')) print max(users, key=attrgetter('user_id'))
1.15 根据字段将记录分组
rows = [ {'address':'5412 n clark', 'data':'07/01/2012'}, {'address':'5232 n clark', 'data':'07/04/2012'}, {'address':'5542 e 58ark', 'data':'07/02/2012'}, {'address':'5152 n clark', 'data':'07/03/2012'}, {'address':'7412 n clark', 'data':'07/02/2012'}, {'address':'6789 w clark', 'data':'07/03/2012'}, {'address':'9008 n clark', 'data':'07/01/2012'}, {'address':'2227 w clark', 'data':'07/04/2012'} ] from operator import itemgetter from itertools import groupby rows.sort(key=itemgetter('data')) for data, items in groupby(rows, key=itemgetter('data')): print (data) for i in items: print (' ', i) from collections import defaultdict rows_by_date = defaultdict(list) for row in rows: rows_by_date[row['data']].append(row) for r in rows_by_date['07/04/2012']: print(r)
1.16 筛选序列中的元素
mylist = [1,4,-5,10,-7,2,3,-1] print [n for n in mylist if n > 0]#列表推导式 print [n for n in mylist if n < 0] pos = (n for n in mylist if n > 0)#生成器表达式 print pos for x in pos: print(x) values = ['1', '2', '-3', '-', '4', 'n/a', '5'] def is_int(val): try: x = int(val) return true except valueerror: return false ivals = list(filter(is_int, values)) print(ivals) mylist = [1,4,-5,10,-7,2,3,-1] import math print [math.sqrt(n) for n in mylist if n > 0] clip_neg = [n if n > 0 else 0 for n in mylist] print clip_neg clip_pos = [n if n < 0 else 0 for n in mylist] print clip_pos addresses = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'] counts = [0, 3, 10, 4, 1, 7, 6, 1] from itertools import compress more5 = [n > 5 for n in counts] print more5 print list(compress(addresses, more5))
1.17 从字典中提取子集
prices = {'acne':45.23, 'aapl':612.78, 'ibm':205.55, 'hpq':37.20, 'fb':10.75} p1 = { key:value for key, value in prices.items() if value > 200 } print p1 tech_names = {'aapl', 'ibm', 'hpq'} p2 = { key:value for key, value in prices.items() if key in tech_names } print p2 p3 = dict( (key, value) for key, value in prices.items() if value > 200 ) #慢 print p3 tech_names = {'aapl', 'ibm', 'hpq'} p4 = { key:prices[key] for key in prices.keys() if key in tech_names } #慢 print p4
1.18 将名称映射到序列的元素中
from collections import namedtuple subscriber = namedtuple('subscriber', ['addr', 'joined']) sub = subscriber('wang@qq.com', '2020-10-10') print sub print sub.joined print sub.addr print len(sub) addr, joined = sub print addr print joined def compute_cost(records): total = 0.0 for rec in records: total += rec[1]*rec[2] return total stock = namedtuple('stock', ['name', 'shares', 'price']) def compute_cost2(records): total = 0.0 for rec in records: s = stock(*rec) total += s.shares * s.price return total s = stock('acme', 100, 123.45) print s #s.shares = 75 #error s = s._replace(shares=75) print s stock = namedtuple('stock', ['name', 'shares', 'price', 'date', 'time']) stock_prototype = stock('',0, 0.0, none, none) def dict_to_stock(s): return stock_prototype._replace(**s) a = {'name':'acme', 'shares':100, 'price':123.45} print dict_to_stock(a) b = {'name':'acme', 'shares':100, 'price':123.45, 'date':'12/12/2012'} print dict_to_stock(b)
1.19 同时对数据做转换和换算
nums = [1, 2, 3, 4, 5] s = sum( x*x for x in nums ) print s import os files = os.listdir('dirname') if any(name.endswith('.py') for name in files): print('there be python!') else: print('sorry, no python!') s = ('acme', 50, 123.45) print(','.join(str(x) for x in s)) portfolio = [ {'name':'goog', 'shares':50}, {'name':'yhoo', 'shares':75}, {'name':'aol', 'shares':20}, {'name':'scox', 'shares':65} ] min_shares = min(s['shares'] for s in portfolio) print min_shares min_shares = min(portfolio, key=lambda s: s['shares']) print min_shares 1.20 将多个映射合并为单个映射 java代码 a = {'x':1, 'z':3} b = {'y':2, 'z':4} #from collections import chainmap from pip._vendor.distlib.compat import chainmap c = chainmap(a, b) print(c['x']) print(c['y']) print(c['z']) #from a 第一个映射中的值 print len(c) print list(c.values()) c['z'] = 10 c['w'] = 40 del c['x'] print a #del c['y'] #error 修改映射的操作总是会作用在列表的第一个映射结构上 values = chainmap() values['x'] = 1 values = values.new_child()#add a new map values['x'] = 2 values = values.new_child() values['x'] = 3 #print values print values['x'] values = values.parents print values['x'] values = values.parents print values['x'] a = {'x':1, 'z':3} b = {'y':2, 'z':4} merged = dict(b) merged.update(a) print merged['x'] print merged['y'] print merged['z'] a['x'] = 13 print merged['x'] #不会反应到合并后的字典中 a = {'x':1, 'z':3} b = {'y':2, 'z':4} merged = chainmap(a, b) print merged['x'] a['x'] = 42 print merged['x'] #会反应到合并后的字典中