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python 日常练习

和并两个有序列表，并且保持合并后的两个列表有序

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In [11]: l1 = [1,2,3,4]
In [12]: l2 = [2,3,7,9]
In [13]: l3 = []
In [14]: for x in l1:
    ...:     while len(l2) > 0 :
    ...:         if x > l2[0]:
    ...:             l3.append(l2.pop(0))
    ...:         else:
    ...:             l3.append(x)
    ...:             break
    ...:     if len(l2)  <= 0:
    ...:         l3.append(x)
    ...: l3.append(l2)
    ...:
    ...:
In [15]:
In [15]: l3
Out[15]: [1, 2, 2, 3, 3, 4, [7, 9]]

按单词反转字符串， 如’I love Linux’ 反转为 ‘Linux love I’

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In [30]: ' '.join(s.split()[::-1])
Out[30]: 'Linux love I'
In [31]: s = "I love Linux"
In [32]: ' '.join(s.split()[::-1])
Out[32]: 'Linux love I'
In [33]: s.split()[::-1]
Out[33]: ['Linux', 'love', 'I']

找出一个列表中只出现了一次的数字，并且保持原来的次序，例如[1,2,1,3,2,5]，结果为[3,5]

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Number（数字）
String（字符串）
List（列表）
Tuple（元组）
Set（集合）
Dictionary（字典）

有bug的代码

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# 例如，当1 出现三次时，会把1也记录下来
In [59]: lst_set
Out[59]: [1, 2, 3, 5]
 
In [60]: lst =  [1,1,2,1,3,2,5]
In [61]: ret = []
In [62]: for x in lst:
    ...:     if x in ret:
    ...:         ret.remove(x)
    ...:     else:
    ...:         ret.append(x)
    ...:
In [63]: ret
Out[63]: [1, 3, 5]

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# 去除bug后
In [51]: lst =  [1,1,2,1,3,2,5]
In [52]: ret = []
In [53]: for x in lst:
    ...:     if lst.count(x) == 1:
    ...:         ret.append(x)
    ...:
In [54]: ret
Out[54]: [3, 5]

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# 当重复的数过大的时，不便重复的元素每个都遍历，如10w个1重复
In [77]: lst =  [1,1,1,1,1,1,2,1,3,2,5]
In [78]: lst_set = list(set(lst))
In [79]: lst_set
Out[79]: [1, 2, 3, 5]
In [80]: ret = []
In [81]: ret
Out[81]: []
In [84]: for x in lst_set:
    ...:     if lst.count(x) == 1:
    ...:         ret.append(x)
    ...:
In [85]: ret
Out[85]: [3, 5]

取出一个列表中的最大值

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In [86]: lst = [1,3,5,8,10,101,301]
In [87]: max(lst)
Out[87]: 301

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In [93]: lst = [1,3,5,8,10,101,301]
In [94]: m = lst[0]
In [96]: for x in lst:
    ...:     if x > m:
    ...:         m = x
    ...:
In [97]: m
Out[97]: 301

写一个程序，把字符串转化为数字，不允许使用函数和模块

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In [138]: s = '123.456'
In [139]: integer, decimals=s.split('.')
In [140]: list(enumerate(integer))
Out[140]: [(0, '1'), (1, '2'), (2, '3')]
In [141]: integer_length = len(integer)
In [142]: result = 0
     ...: for i, x in enumerate(integer):
     ...:     if x == '0':
     ...:         result += 0 * 10 ** (integer_length -i -1)
     ...:     if x == '1':
     ...:         result += 1 * 10 ** (integer_length -i -1)
     ...:     if x == '2':
     ...:         result += 2 * 10 ** (integer_length -i -1)
     ...:     if x == '3':
     ...:         result += 3 * 10 ** (integer_length -i -1)
     ...:     if x == '4':
     ...:         result += 4 * 10 ** (integer_length -i -1)
     ...:     if x == '5':
     ...:         result += 5 * 10 ** (integer_length -i -1)
     ...:     if x == '6':
     ...:         result += 6 * 10 ** (integer_length -i -1)
     ...:     if x == '7':
     ...:         result += 7 * 10 ** (integer_length -i -1)
     ...:     if x == '8':
     ...:         result += 8 * 10 ** (integer_length -i -1)
     ...:     if x == '9':
     ...:         result += 9 * 10 ** (integer_length -i -1)
     ...:
In [143]: for i, x in enumerate(decimals):
     ...:     if x == '0':
     ...:         result += 0 * 10 ** (-1 * (i+1))
     ...:     if x == '1':
     ...:         result += 1 * 10 ** (-1 * (i+1))
     ...:     if x == '2':
     ...:         result += 2 * 10 ** (-1 * (i+1))
     ...:     if x == '3':
     ...:         result += 3 * 10 ** (-1 * (i+1))
     ...:     if x == '4':
     ...:         result += 4 * 10 ** (-1 * (i+1))
     ...:     if x == '5':
     ...:         result += 5 * 10 ** (-1 * (i+1))
     ...:     if x == '6':
     ...:         result += 6 * 10 ** (-1 * (i+1))
     ...:     if x == '7':
     ...:         result += 7 * 10 ** (-1 * (i+1))
     ...:     if x == '8':
     ...:         result += 8 * 10 ** (-1 * (i+1))
     ...:     if x == '9':
     ...:         result += 9 * 10 ** (-1 * (i+1))
     ...:
In [144]: result
Out[144]: 123.456
In [145]: type(result)
Out[145]: float