你可能不知道的python
2015-09-24 16:43
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原文链接:http://www.cnblogs.com/yupeng/p/3434567.html
ints = ['a','b','c','d','e','f'] for idx, val in enumerate(ints): print idx, val
结果:
0 a
1 b
2 c
3 d
4 e
5 f
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2.python 中的switch 替代语法,使用字典
xdef test(par): print par def test1(): print 1000 def test3(par): print par def f(x): return { 'a': test, 'b': test1, }.get(x, test3) print f('a')(100) print f(4)(100)
结果:
100
100
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3.如何在非创建全局变量的时候使用全局变量,使用global关键字
globvar = 0 def test4(): global globvar globvar = 10 def test5(): print globvar test4() test5()
结果为10
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4.为什么是string.join(list)而不是list.join(string)
my_list = ["Hello", "world"]
print "-".join(my_list)
#为什么不是 my_list.join("-") 。。。。这个....
因为所有可迭代对象都可以被连接,而不只是列表,但是连接者总是字符串
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5.如何反向输出一个字符串?
'test'[::-1]
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6.如何随机生成大写字母和数字组成的字符串?
import string, random
''.join(random.choice(string.ascii_uppercase + string.digits) for x in range(N))
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7.如何将一个字符串转成字典?
使用内建模块 ast.literal_eval
import ast
print ast.literal_eval("{'muffin' : 'lolz', 'foo' : 'kitty'}")
结果为:{'muffin': 'lolz', 'foo': 'kitty'}
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8.如何填充字符串
n = '4'
print n.zfill(3)
n = 4
print '%03d' % n
print "{0:03d}".format(4) # python >= 2.6
结果:
004
004
004
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9.如何判断一个列表为空?
[code]if not a: print "List is empty" #不要用len(a)来判断
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10.列表append 和extend 的区别?
[code]>>> x = [1, 2] >>> x.append(3) >>> x [1, 2, 3] >>> x.append([4,5]) >>> x [1, 2, 3, [4, 5]] >>> >>> x = [1, 2, 3] >>> x.extend([4, 5]) >>> x [1, 2, 3, 4, 5]
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11.如何随机的从列表中获得变量?
[code]foo = ['a', 'b', 'c', 'd', 'e'] from random import choice print choice(foo)
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12.如何将列表切成长度相同的序列?
[code]def chunks(l, n): return [l[i:i+n] for i in range(0, len(l), n)]
[/code]
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13. 列表如何去重?
def f7(seq):
seen = set()
seen_add = seen.add
return [ x for x in seq if x not in seen and not seen_add(x)]
print f7([1,2,3,4,3,2,4,5,6,5,4,3])
结果:[1, 2, 3, 4, 5, 6]
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14.遍历list的时候删除某些元素?
[code]#在原有对象上进行修改 somelist[:] = [x for x in somelist if determine(x)]
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15.如何扁平一个二维数组?
[code]l = [[1,2,3],[4,5,6], [7], [8,9]] 变为[1, 2, 3, 4, 5, 6, 4, 5, 6, 7, 8, 9]
[code]1.[item for sublist in l for item in sublist]
2.import itertools
list2d = [[1,2,3],[4,5,6], [7], [8,9]]
merged = list(itertools.chain(*list2d))
print merged
# python >= 2.6
import itertools
list2d = [[1,2,3],[4,5,6], [7], [8,9]]
merged = list(itertools.chain.from_iterable(list2d))
print merged
3.print sum(list2d,[])
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16.如何获得list元素的下标?
[code]["foo","bar","baz"].index('bar') 1
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17.如何将一个字典换成object
def dict2obj(args): '把字典递归转化为类' class obj(object): def __init__(self, d): for a, b in d.items(): if isinstance(b, (list, tuple)): setattr(self, a, [obj(x) if isinstance(x, dict) else x for x in b]) else: setattr(self, a, obj(b) if isinstance(b, dict) else b) return obj(args) d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]} x = dict2obj(d) print x.a
1
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18.如何合并2字典?
[code]>>> x = {'a':1, 'b': 2} >>> y = {'b':10, 'c': 11} >>> z = x.update(y) >>> print z None >>> x {'a': 1, 'b': 10, 'c': 11}
我想要最终合并结果在z中,不是x,我要怎么做?
回答
这种情况下,可以使用
[code]z = dict(x.items() + y.items())
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19.如何映射2列表成一个字典?
两个列表
[code]keys = ('name', 'age', 'food') values = ('Monty', 42, 'spam')
如何得到
[code]dict = {'name' : 'Monty', 'age' : 42, 'food' : 'spam'}
使用zip
[code]>>> keys = ['a', 'b', 'c'] >>> values = [1, 2, 3] >>> dictionary = dict(zip(keys, values)) >>> print dictionary {'a': 1, 'b': 2, 'c': 3}
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20.排序一个列表所有dict,根据dict的值
lists = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10},{'name':'Art', 'age':40}]
from operator import itemgetter
newlist = sorted(lists, key=itemgetter('name'))
print newlist
结果为:[{'age': 40, 'name': 'Art'}, {'age': 10, 'name': 'Bart'}, {'age': 39, 'name': 'Homer'}]
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