python数据结构

python 数据结构

一、列表

list.

list.extend

>>> a = [66.25, 333, 333, 1, 1234.5]
>>> print a.count(333), a.count(66.25), a.count('x')
2 1 0
>>> a.insert(2, -1)
>>> a.append(333)
>>> a
[66.25, 333, -1, 333, 1, 1234.5, 333]
>>> a.index(333)
1
>>> a.remove(333)
>>> a
[66.25, -1, 333, 1, 1234.5, 333]
>>> a.reverse()
>>> a
[333, 1234.5, 1, 333, -1, 66.25]
>>> a.sort()
>>> a
[-1, 1, 66.25, 333, 333, 1234.5]
>>> a.pop()
1234.5
>>> a
[-1, 1, 66.25, 333, 333]

二、将列表用作栈

>>> stack = [3, 4, 5]
>>> stack.append(6)
>>> stack.append(7)
>>> stack
[3, 4, 5, 6, 7]
>>> stack.pop()
7
>>> stack
[3, 4, 5, 6]
>>> stack.pop()
6
>>> stack.pop()
5
>>> stack
[3, 4]

三、将列表用作队列

>>> from collections import deque
>>> queue = deque(["Eric", "John", "Michael"])
>>> queue.append("Terry")           # 尾部追加元素 Terry
>>> queue.append("Graham")          # 尾部追加元素 Graham
>>> queue.popleft()                 # 移除并返回左侧元素
'Eric'
>>> queue.popleft()                 # 移除并返回左侧元素
'John'
>>> queue
deque(['Michael', 'Terry', 'Graham'])

四、三个常用的函数

>>> def f(x): return x % 3 == 0 or x % 5 == 0   # x除以3或5后结果为0就是真
...
>>> filter(f, range(2, 25)) # 返回[2,25）范围内能满足f为真的序列
[3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24]

>>> def cube(x): return x*x*x
...
>>> map(cube, range(1, 11))
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

>>> seq = range(8)
>>> def add(x, y): return x+y
...
>>> map(add, seq, seq)
[0, 2, 4, 6, 8, 10, 12, 14]

>>>seq=range(8)
>>>sep=range(7)
>>>def add(x,y):return x+y
...
>>>map(add,seq,sep)

Trackback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 2, in add
TypeError: unsupported operand type(s) for +: 'int' and 'NoneType'

>>> def add(x,y): return x+y
...
>>> reduce(add, range(1, 11))
55

>>> def sum(seq):
...     def add(x,y): return x+y
...     return reduce(add, seq, 0)
...
>>> sum(range(1, 11))
55
>>> sum([])
0

五、递推列表

>>> squares = []
>>> for x in range(10):
...     squares.append(x**2)
...
>>> squares
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]    # 一般方法得到的列表

>>>ListSquares = [x**2 for x in range(10)]
>>>ListSquares
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]    # 递推列表方法得到的列表

>>>NewSquares = map(lambda x: x**2, range(10))
>>>NewSquares
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]    # map 方法得到的列表

>>> [(x, y) for x in [1,2,3] for y in [3,1,4] if x != y]
[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]

>>> combs = []
>>> for x in [1,2,3]:
...     for y in [3,1,4]:
...         if x != y:
...             combs.append((x, y))
...
>>> combs
[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]

>>> vec = [-4, -2, 0, 2, 4]
>>> # 以 vec 列表元素的 2 倍构造新列表
>>> [x*2 for x in vec]
[-8, -4, 0, 4, 8]

>>> # 去除 vec 列表中的负数
>>> [x for x in vec if x >= 0]
[0, 2, 4]

>>> # vec 列表中的每个元素都执行 abs(x)函数，以求其绝对值。
>>> [abs(x) for x in vec]
[4, 2, 0, 2, 4]

>>> # freshfruit 列表中每个元素都执行 strip()方法，以去除多余空格。
>>> freshfruit = ['  banana', '  loganberry ', 'passion fruit  ']
>>> [weapon.strip() for weapon in freshfruit]
['banana', 'loganberry', 'passion fruit']

>>> # 创建形如 (number, square) 的二维元组
>>> [(x, x**2) for x in range(6)]
[(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), (5, 25)]

>>> # 元组必须用括号括起来，否则会抛出语法错误
>>> [x, x**2 for x in range(6)]
File "<stdin>", line 1, in <module>
[x, x**2 for x in range(6)]
^
SyntaxError: invalid syntax

>>> # 用两个 for 创建列表
>>> vec = [[1,2,3], [4,5,6], [7,8,9]]
>>> [num for elem in vec for num in elem]
[1, 2, 3, 4, 5, 6, 7, 8, 9]

六、嵌套的递推列表

>>> matrix = [
...     [1, 2, 3, 4],
...     [5, 6, 7, 8],
...     [9, 10, 11, 12],
... ]

>>> [[row[i] for row in matrix] for i in range(4)]
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

>>> transposed = []
>>> for i in range(4):
...     transposed.append([row[i] for row in matrix])
...
>>> transposed
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

>>> transposed = []
>>> for i in range(4):
...     transposed_row = []
...     for row in matrix:
...         transposed_row.append(row[i])
...     transposed.append(transposed_row)
...
>>> transposed
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

>>> zip(*matrix)
[(1, 5, 9), (2, 6, 10), (3, 7, 11), (4, 8, 12)]

七、del语句

>>> a = [-1, 1, 66.25, 333, 333, 1234.5]
>>> del a[0]
>>> a
[1, 66.25, 333, 333, 1234.5]
>>> del a[2:4]
>>> a
[1, 66.25, 1234.5]
>>> del a[:]
>>> a
[]

>>> a = [-1, 1, 66.25, 333, 333, 1234.5]
>>> del a
>>> a
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'a' is not defined  # 未定义名称 a

八、元组和序列

>>> t = 12345, 54321, 'hello!'
>>> t[0]
12345
>>> t
(12345, 54321, 'hello!')

>>> # 元组也可被嵌套
... u = t, (1, 2, 3, 4, 5)
>>> u
((12345, 54321, 'hello!'), (1, 2, 3, 4, 5))

>>> # 元组元素是不可被修改的
... t[0] = 88888
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'tuple' object does not support item assignment

>>> empty = ()
>>> singleton = 'hello',    # 一个元素的元组（值后跟一逗号）
>>> len(empty)
0
>>> len(singleton)
1
>>> singleton
('hello',)

t = 12345, 54321, 'hello!'

>>> x, y, z = t # 将 元组 t 解包，各个元素值依次赋给 x,y,z

九、集合

>>> basket = ['apple', 'orange', 'apple', 'pear', 'orange', 'banana']
>>> fruit = set(basket)               # 创建一无重复元素的集合
>>> fruit
set(['orange', 'pear', 'apple', 'banana'])

>>> 'orange' in fruit                 # 关系检测，'orange'是否在fruit集合内
True
>>> 'crabgrass' in fruit
False

>>> # 挑出非重复元素分别赋给 a,b
...
>>> a = set('abracadabra')
>>> b = set('alacazam')
>>> a                                  # 集合 a
set(['a', 'r', 'b', 'c', 'd'])

>>> a - b                              # 在 a 中而不在 b 中的元素
set(['r', 'd', 'b'])

>>> a | b                              # 在 a 中或在 b 中的元素
set(['a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'])

>>> a & b                              # 在 a 中且在 b 中的元素
set(['a', 'c'])

>>> a ^ b                              # 公共元素外的元素
set(['r', 'd', 'b', 'm', 'z', 'l'])

>>> a = {x for x in 'abracadabra' if x not in 'abc'}
>>> a
set(['r', 'd'])

十、字典

>>> tel = {'jack': 4098, 'sape': 4139}
>>> tel['guido'] = 4127
>>> tel
{'sape': 4139, 'guido': 4127, 'jack': 4098}

>>> tel['jack']
4098

>>> del tel['sape']
>>> tel['irv'] = 4127
>>> tel
{'guido': 4127, 'irv': 4127, 'jack': 4098}

>>> tel.keys()
['guido', 'irv', 'jack']

>>> 'guido' in tel
True

>>> dict([('sape', 4139), ('guido', 4127), ('jack', 4098)])
{'sape': 4139, 'jack': 4098, 'guido': 4127}

>>> dict(sape=4139, guido=4127, jack=4098)
{'sape': 4139, 'jack': 4098, 'guido': 4127}

>>> {x: x**2 for x in (2, 4, 6)}
{2: 4, 4: 16, 6: 36}

十一、循环技术

>>> for i, v in enumerate(['tic', 'tac', 'toe']):
...     print i, v
...
0 tic
1 tac
2 toe

>>> questions = ['name', 'quest', 'favorite color']
>>> answers = ['lancelot', 'the holy grail', 'blue']
>>> for q, a in zip(questions, answers):
...     print 'What is your {0}?  It is {1}.'.format(q, a)
...
What is your name?  It is lancelot.
What is your quest?  It is the holy grail.
What is your favorite color?  It is blue.

>>> for i in reversed(xrange(1,10,2)):
...     print i
...
9
7
5
3
1

>>> basket = ['apple', 'orange', 'apple', 'pear', 'orange', 'banana']
>>> for f in sorted(set(basket)):
...     print f
...
apple
banana
orange
pear

>>> knights = {'gallahad': 'the pure', 'robin': 'the brave'}
>>> for k, v in knights.iteritems():
...     print k, v
...
gallahad the pure
robin the brave

>>> import math
>>> raw_data = [56.2, float('NaN'), 51.7, 55.3, 52.5, float('NaN'), 47.8]
>>> filtered_data = []
>>> for value in raw_data:
...     if not math.isnan(value):
...         filtered_data.append(value)
...
>>> filtered_data
[56.2, 51.7, 55.3, 52.5, 47.8]

十二、条件

a < b == c

and

or

not

not

or

A and not B or C

(A and (not B)) or C

and

or

A

c

B

A and B and C

C

>>> string1, string2, string3 = '', 'Trondheim', 'Hammer Dance'
>>> non_null = string1 or string2 or string3
>>> non_null
'Trondheim' # string1 和 string2 运算后得到了结果，因此不再和 string3 一起运算

十三、序列比较

(1, 2, 3)              < (1, 2, 4)
[1, 2, 3]              < [1, 2, 4]
'ABC' < 'C' < 'Pascal' < 'Python'
(1, 2, 3, 4)           < (1, 2, 4)
(1, 2)                 < (1, 2, -1)
(1, 2, 3)             == (1.0, 2.0, 3.0)
(1, 2, ('aa', 'ab'))   < (1, 2, ('abc', 'a'), 4)