day 3 list列表生成式

Posted 2020-10-15 不要被骄傲遮蔽了双眼

1.定义一个list列表，里面元素是0-33

a = []

i = 0
while i<33:
    a.append(i)
    i+=1

print(a)

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]

 

2.range （切片）

　　1）python2 版本

　　range风险：python2 版本中，有时候一次性申请很大的内存，不会给你

####  range  返回list列表   和切片功能相同
In [1]: range(10)
Out[1]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [2]: range(10,17)
Out[2]: [10, 11, 12, 13, 14, 15, 16]

In [3]: range(10,17,2)
Out[3]: [10, 12, 14, 16]

In [5]: range(0,100000)

### 运行结果
994,
 995,
 996,
 997,
 998,
 999,
 ...]

 

In [5]: range(0,1000000000)
---------------------------------------------------------------------------
MemoryError                               Traceback (most recent call last)
<ipython-input-5-30124a0b9388> in <module>()
----> 1 range(0,1000000000)

MemoryError: 
   ##range风险：python2 版本中，有时候一次性申请很大的内存，不会给你

 

　　2）python3版本：要一个数字，给你一个，不会全部一次性给

In [1]: range(0,10)
Out[1]: range(0, 10)

In [2]: range(10)
Out[2]: range(0, 10)

In [3]: range(0,100000000)
Out[3]: range(0, 100000000)

 

3.列表生成式

　　1） a = [ i for i in range(0,18) ]

In [6]: a = [i for i in range(0,18)]

In [7]: a
Out[7]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]


In [8]: a = [22 for i in range(0,18)]   #for只负责循环的次数17次

In [9]: a
Out[9]: [22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22]

 　　　　　　

 

 　　2） a =  [ i for i in range(10) if i%2==0 ]

In [10]: a = [i for i in range(10) if i%2==0]

In [11]: a
Out[11]: [0, 2, 4, 6, 8]

 

　　3） d = [ (i,j) for i in range(3) for j in range(2)]

In [15]: d = [ i for i in range(3) for j in range(2)]

In [16]: d
Out[16]: [0, 0, 1, 1, 2, 2]


In [17]: d = [ i,j for i in range(3) for j in range(2)]
  File "<ipython-input-17-0277977bdeb0>", line 1
    d = [ i,j for i in range(3) for j in range(2)]
                ^
SyntaxError: invalid syntax



In [19]: d = [ (i,j) for i in range(3) for j in range(2)]

In [20]: d
Out[20]: [(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1)]  #坐标轴

 　　　　　　

 

　　4）e = [(i,j,k) for i in range(3) for j in range(2) for k in range(2)]

In [21]: e = [(i,j,k) for i in range(3) for j in range(2) for k in range(2)]

In [22]: e
Out[22]: 
[(0, 0, 0),
 (0, 0, 1),
 (0, 1, 0),
 (0, 1, 1),
 (1, 0, 0),
 (1, 0, 1),
 (1, 1, 0),
 (1, 1, 1),
 (2, 0, 0),
 (2, 0, 1),
 (2, 1, 0),
 (2, 1, 1)]