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+=1print(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 1d = [ 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)]
