时间：2020-10-23 python教程 查看: 894
1、拆箱
>>> a, b, c = 1, 2, 3
>>> a, b, c
(1, 2, 3)
>>> a, b, c = [1, 2, 3]
>>> a, b, c
(1, 2, 3)
>>> a, b, c = (2 * i + 1 for i in range(3))
>>> a, b, c
(1, 3, 5)
>>> a, (b, c), d = [1, (2, 3), 4]
>>> a
1
>>> b
2
>>> c
3
>>> d
42、使用拆箱进行变量交换
>>> a, b = 1, 2
>>> a, b = b, a
>>> a, b
(2, 1)3、扩展的拆箱(Python 3支持)
>>> a, *b, c = [1, 2, 3, 4, 5]
>>> a
1
>>> b
[2, 3, 4]
>>> c
54、负数索引
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[-1]
10
>>> a[-3]
85、列表切片(a[start:end])
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[2:8]
[2, 3, 4, 5, 6, 7]6、负数索引的列表切片
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[-4:-2]
[7, 8]7、带步数的列表切片(a[start:end:step])
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[::2]
[0, 2, 4, 6, 8, 10]
>>> a[::3]
[0, 3, 6, 9]
>>> a[2:8:2]
[2, 4, 6]8、负数步数的列表切片
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[::-1]
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
>>> a[::-2]
[10, 8, 6, 4, 2, 0]9、列表切片赋值
>>> a = [1, 2, 3, 4, 5]
>>> a[2:3] = [0, 0]
>>> a
[1, 2, 0, 0, 4, 5]
>>> a[1:1] = [8, 9]
>>> a
[1, 8, 9, 2, 0, 0, 4, 5]
>>> a[1:-1] = []
>>> a
[1, 5]10、切片命名(slice(start, end, step))
>>> a = [0, 1, 2, 3, 4, 5]
>>> LASTTHREE = slice(-3, None)
>>> LASTTHREE
slice(-3, None, None)
>>> a[LASTTHREE]
[3, 4, 5]11、遍历列表索引和值(enumerate)
>>> a = ["Hello", "world", "!"]
>>> for i, x in enumerate(a):
... print "{}: {}".format(i, x)
...
0: Hello
1: world
2: !12、遍历字典的KEY和VALUE(dict.iteritems)
>>> m = {"a": 1, "b": 2, "c": 3, "d": 4}
>>> for k, v in m.iteritems():
... print "{}: {}".format(k, v)
...
a: 1
c: 3
b: 2
d: 4
# 注意：Python 3中要使用dict.items13、压缩 & 解压列表和可遍历对象
>>> a = [1, 2, 3]
>>> b = ["a", "b", "c"]
>>> z = zip(a, b)
>>> z
[(1, "a"), (2, "b"), (3, "c")]
>>> zip(*z)
[(1, 2, 3), ("a", "b", "c")]14、使用zip分组相邻列表项
>>> a = [1, 2, 3, 4, 5, 6]
>>> # Using iterators
>>> group_adjacent = lambda a, k: zip(*([iter(a)] * k))
>>> group_adjacent(a, 3)
[(1, 2, 3), (4, 5, 6)]
>>> group_adjacent(a, 2)
[(1, 2), (3, 4), (5, 6)]
>>> group_adjacent(a, 1)
[(1,), (2,), (3,), (4,), (5,), (6,)]
>>> # Using slices
>>> from itertools import islice
>>> group_adjacent = lambda a, k: zip(*(islice(a, i, None, k) for i in range(k)))
>>> group_adjacent(a, 3)
[(1, 2, 3), (4, 5, 6)]
>>> group_adjacent(a, 2)
[(1, 2), (3, 4), (5, 6)]
>>> group_adjacent(a, 1)
[(1,), (2,), (3,), (4,), (5,), (6,)]15、使用zip & iterators实现推拉窗(n-grams)
>>> from itertools import islice
>>> def n_grams(a, n):
... z = (islice(a, i, None) for i in range(n))
... return zip(*z)
...
>>> a = [1, 2, 3, 4, 5, 6]
>>> n_grams(a, 3)
[(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6)]
>>> n_grams(a, 2)
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
>>> n_grams(a, 4)
[(1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 6)]16、使用zip反相字典对象
>>> m = {"a": 1, "b": 2, "c": 3, "d": 4}
>>> m.items()
[("a", 1), ("c", 3), ("b", 2), ("d", 4)]
>>> zip(m.values(), m.keys())
[(1, "a"), (3, "c"), (2, "b"), (4, "d")]
>>> mi = dict(zip(m.values(), m.keys()))
>>> mi
{1: "a", 2: "b", 3: "c", 4: "d"}17、合并列表
>>> a = [[1, 2], [3, 4], [5, 6]]
>>> list(itertools.chain.from_iterable(a))
[1, 2, 3, 4, 5, 6]
>>> sum(a, [])
[1, 2, 3, 4, 5, 6]
>>> [x for l in a for x in l]
[1, 2, 3, 4, 5, 6]
>>> a = [[[1, 2], [3, 4]], [[5, 6], [7, 8]]]
>>> [x for l1 in a for l2 in l1 for x in l2]
[1, 2, 3, 4, 5, 6, 7, 8]
>>> a = [1, 2, [3, 4], [[5, 6], [7, 8]]]
>>> flatten = lambda x: [y for l in x for y in flatten(l)] if type(x) is list else [x]
>>> flatten(a)
[1, 2, 3, 4, 5, 6, 7, 8]
Note: according to Python"s documentation on sum, itertools.chain.from_iterable is the preferred method for this.18、生成器
>>> g = (x ** 2 for x in xrange(10))
>>> next(g)
0
>>> next(g)
1
>>> next(g)
4
>>> next(g)
9
>>> sum(x ** 3 for x in xrange(10))
2025
>>> sum(x ** 3 for x in xrange(10) if x % 3 == 1)
40819、字典解析
>>> m = {x: x ** 2 for x in range(5)}
>>> m
{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}
>>> m = {x: "A" + str(x) for x in range(10)}
>>> m
{0: "A0", 1: "A1", 2: "A2", 3: "A3", 4: "A4", 5: "A5", 6: "A6", 7: "A7", 8: "A8", 9: "A9"}20、使用字典解析反相字典对象
>>> m = {"a": 1, "b": 2, "c": 3, "d": 4}
>>> m
{"d": 4, "a": 1, "b": 2, "c": 3}
>>> {v: k for k, v in m.items()}
{1: "a", 2: "b", 3: "c", 4: "d"}21、命名的tuples(collections.namedtuple)
>>> Point = collections.namedtuple("Point", ["x", "y"])
>>> p = Point(x=4.0, y=2.0)
>>> p
Point(x=4.0, y=2.0)
>>> p.x
4.0
>>> p.y
2.022、继承命名tuples
>>> class Point(collections.namedtuple("PointBase", ["x", "y"])):
... __slots__ = ()
... def __add__(self, other):
... return Point(x=self.x + other.x, y=self.y + other.y)
...
>>> p = Point(x=4.0, y=2.0)
>>> q = Point(x=2.0, y=3.0)
>>> p + q
Point(x=6.0, y=5.0)23、Set & Set运算
>>> A = {1, 2, 3, 3}
>>> A
set([1, 2, 3])
>>> B = {3, 4, 5, 6, 7}
>>> B
set([3, 4, 5, 6, 7])
>>> A | B
set([1, 2, 3, 4, 5, 6, 7])
>>> A & B
set([3])
>>> A - B
set([1, 2])
>>> B - A
set([4, 5, 6, 7])
>>> A ^ B
set([1, 2, 4, 5, 6, 7])
>>> (A ^ B) == ((A - B) | (B - A))
True24、Multisets运算(collections.Counter)
>>> A = collections.Counter([1, 2, 2])
>>> B = collections.Counter([2, 2, 3])
>>> A
Counter({2: 2, 1: 1})
>>> B
Counter({2: 2, 3: 1})
>>> A | B
Counter({2: 2, 1: 1, 3: 1})
>>> A & B
Counter({2: 2})
>>> A + B
Counter({2: 4, 1: 1, 3: 1})
>>> A - B
Counter({1: 1})
>>> B - A
Counter({3: 1})25、列表中出现最多的元素(collections.Counter)
>>> A = collections.Counter([1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 6, 7])
>>> A
Counter({3: 4, 1: 2, 2: 2, 4: 1, 5: 1, 6: 1, 7: 1})
>>> A.most_common(1)
[(3, 4)]
>>> A.most_common(3)
[(3, 4), (1, 2), (2, 2)]26、双向队列(collections.deque)
>>> Q = collections.deque()
>>> Q.append(1)
>>> Q.appendleft(2)
>>> Q.extend([3, 4])
>>> Q.extendleft([5, 6])
>>> Q
deque([6, 5, 2, 1, 3, 4])
>>> Q.pop()
4
>>> Q.popleft()
6
>>> Q
deque([5, 2, 1, 3])
>>> Q.rotate(3)
>>> Q
deque([2, 1, 3, 5])
>>> Q.rotate(-3)
>>> Q
deque([5, 2, 1, 3])27、限制长度的双向队列(collections.deque)
>>> last_three = collections.deque(maxlen=3)
>>> for i in xrange(10):
... last_three.append(i)
... print ", ".join(str(x) for x in last_three)
...
0
0, 1
0, 1, 2
1, 2, 3
2, 3, 4
3, 4, 5
4, 5, 6
5, 6, 7
6, 7, 8
7, 8, 928、排序字典(collections.OrderedDict)
>>> m = dict((str(x), x) for x in range(10))
>>> print ", ".join(m.keys())
1, 0, 3, 2, 5, 4, 7, 6, 9, 8
>>> m = collections.OrderedDict((str(x), x) for x in range(10))
>>> print ", ".join(m.keys())
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
>>> m = collections.OrderedDict((str(x), x) for x in range(10, 0, -1))
>>> print ", ".join(m.keys())
10, 9, 8, 7, 6, 5, 4, 3, 2, 129、默认字典(collections.defaultdict)
>>> m = dict()
>>> m["a"]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
KeyError: "a"
>>>
>>> m = collections.defaultdict(int)
>>> m["a"]
0
>>> m["b"]
0
>>> m = collections.defaultdict(str)
>>> m["a"]
""
>>> m["b"] += "a"
>>> m["b"]
"a"
>>> m = collections.defaultdict(lambda: "[default value]")
>>> m["a"]
"[default value]"
>>> m["b"]
"[default value]"30、使用defaultdict代表tree
>>> import json
>>> tree = lambda: collections.defaultdict(tree)
>>> root = tree()
>>> root["menu"]["id"] = "file"
>>> root["menu"]["value"] = "File"
>>> root["menu"]["menuitems"]["new"]["value"] = "New"
>>> root["menu"]["menuitems"]["new"]["onclick"] = "new();"
>>> root["menu"]["menuitems"]["open"]["value"] = "Open"
>>> root["menu"]["menuitems"]["open"]["onclick"] = "open();"
>>> root["menu"]["menuitems"]["close"]["value"] = "Close"
>>> root["menu"]["menuitems"]["close"]["onclick"] = "close();"
>>> print json.dumps(root, sort_keys=True, indent=4, separators=(",", ": "))
{
"menu": {
"id": "file",
"menuitems": {
"close": {
"onclick": "close();",
"value": "Close"
},
"new": {
"onclick": "new();",
"value": "New"
},
"open": {
"onclick": "open();",
"value": "Open"
}
},
"value": "File"
}
}
# 查看更多：https://gist.github.com/hrldcpr/201225031、映射对象到唯一的计数数字(collections.defaultdict)
>>> import itertools, collections
>>> value_to_numeric_map = collections.defaultdict(itertools.count().next)
>>> value_to_numeric_map["a"]
0
>>> value_to_numeric_map["b"]
1
>>> value_to_numeric_map["c"]
2
>>> value_to_numeric_map["a"]
0
>>> value_to_numeric_map["b"]
132、最大 & 最小元素(heapq.nlargest and heapq.nsmallest)
>>> a = [random.randint(0, 100) for __ in xrange(100)]
>>> heapq.nsmallest(5, a)
[3, 3, 5, 6, 8]
>>> heapq.nlargest(5, a)
[100, 100, 99, 98, 98]33、笛卡尔积(itertools.product)
>>> for p in itertools.product([1, 2, 3], [4, 5]):
(1, 4)
(1, 5)
(2, 4)
(2, 5)
(3, 4)
(3, 5)
>>> for p in itertools.product([0, 1], repeat=4):
... print "".join(str(x) for x in p)
...
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
111134、组合(itertools.combinations and itertools.combinations_with_replacement)
>>> for c in itertools.combinations([1, 2, 3, 4, 5], 3):
... print "".join(str(x) for x in c)
...
123
124
125
134
135
145
234
235
245
345
>>> for c in itertools.combinations_with_replacement([1, 2, 3], 2):
... print "".join(str(x) for x in c)
...
11
12
13
22
23
3335、排列(itertools.permutations)
>>> for p in itertools.permutations([1, 2, 3, 4]):
... print "".join(str(x) for x in p)
...
1234
1243
1324
1342
1423
1432
2134
2143
2314
2341
2413
2431
3124
3142
3214
3241
3412
3421
4123
4132
4213
4231
4312
432136、链接可遍历对象(itertools.chain)
>>> a = [1, 2, 3, 4]
>>> for p in itertools.chain(itertools.combinations(a, 2), itertools.combinations(a, 3)):
... print p
...
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
(1, 2, 3)
(1, 2, 4)
(1, 3, 4)
(2, 3, 4)
>>> for subset in itertools.chain.from_iterable(itertools.combinations(a, n) for n in range(len(a) + 1))
... print subset
...
()
(1,)
(2,)
(3,)
(4,)
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
(1, 2, 3)
(1, 2, 4)
(1, 3, 4)
(2, 3, 4)
(1, 2, 3, 4)37、根据给定的KEY分组(itertools.groupby)
>>> from operator import itemgetter
>>> import itertools
>>> with open("contactlenses.csv", "r") as infile:
... data = [line.strip().split(",") for line in infile]
...
>>> data = data[1:]
>>> def print_data(rows):
... print " ".join(" ".join("{: <16}".format(s) for s in row) for row in rows)
...
>>> print_data(data)
young myope no reduced none
young myope no normal soft
young myope yes reduced none
young myope yes normal hard
young hypermetrope no reduced none
young hypermetrope no normal soft
young hypermetrope yes reduced none
young hypermetrope yes normal hard
pre-presbyopic myope no reduced none
pre-presbyopic myope no normal soft
pre-presbyopic myope yes reduced none
pre-presbyopic myope yes normal hard
pre-presbyopic hypermetrope no reduced none
pre-presbyopic hypermetrope no normal soft
pre-presbyopic hypermetrope yes reduced none
pre-presbyopic hypermetrope yes normal none
presbyopic myope no reduced none
presbyopic myope no normal none
presbyopic myope yes reduced none
presbyopic myope yes normal hard
presbyopic hypermetrope no reduced none
presbyopic hypermetrope no normal soft
presbyopic hypermetrope yes reduced none
presbyopic hypermetrope yes normal none
>>> data.sort(key=itemgetter(-1))
>>> for value, group in itertools.groupby(data, lambda r: r[-1]):
... print "-----------"
... print "Group: " + value
... print_data(group)
...
-----------
Group: hard
young myope yes normal hard
young hypermetrope yes normal hard
pre-presbyopic myope yes normal hard
presbyopic myope yes normal hard
-----------
Group: none
young myope no reduced none
young myope yes reduced none
young hypermetrope no reduced none
young hypermetrope yes reduced none
pre-presbyopic myope no reduced none
pre-presbyopic myope yes reduced none
pre-presbyopic hypermetrope no reduced none
pre-presbyopic hypermetrope yes reduced none
pre-presbyopic hypermetrope yes normal none
presbyopic myope no reduced none
presbyopic myope no normal none
presbyopic myope yes reduced none
presbyopic hypermetrope no reduced none
presbyopic hypermetrope yes reduced none
presbyopic hypermetrope yes normal none
-----------
Group: soft
young myope no normal soft
young hypermetrope no normal soft
pre-presbyopic myope no normal soft
pre-presbyopic hypermetrope no normal soft
presbyopic hypermetrope no normal soft38、在任意目录启动HTTP服务
python -m SimpleHTTPServer 5000
Serving HTTP on 0.0.0.0 port 5000 ...
39、Python之禅
>>> import this
The Zen of Python, by Tim Peters
Beautiful 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!
40、使用C风格的大括号代替Python缩进来表示作用域
>>> from __future__ import braces
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