• Python 在问答频道中刷题积累到的小技巧(七)


    1. max(),min()函数的key参数应用

    快速找出众数:

    1. >>> import random
    2. >>> r = [rnd(1,9) for _ in range(20)]
    3. >>> r
    4. [3, 3, 5, 9, 1, 7, 4, 1, 1, 9, 6, 8, 3, 9, 8, 8, 6, 4, 6, 6]
    5. >>> max(r, key=r.count)
    6. 6

    找最短密码(首个匹配到的):

    1. >>> from random import choices, randint as rnd
    2. >>> s = ''.join((map(chr,[*range(ord('A'),ord('Z')+1),*range(ord('a'),ord('z')+1),*range(ord('0'),ord('9')+1),ord('-')])))
    3. >>> p = [''.join(choices(s,k=rnd(3,10))) for _ in range(10)]
    4. >>> p
    5. ['fiKYKBl', 'HZjwZUGg', 'pUaKkKtrE8', '2qet35e1Q', 'is34n11F', 'Jxed96', '9K8TOE', 'sqymdfx', 'SRjuq', 'J9TBu']
    6. >>> min(p, key=lambda x:len(x))
    7. 'SRjuq'

    位数最多、最少的整数 :(首个匹配到的数,与最大值、最小值未必相等)

    1. >>> from random import choices, randint as rnd
    2. >>> n = [int(''.join(map(str,choices(range(10),k=rnd(3,10))))) for _ in range(10)]
    3. >>> n
    4. [9345447, 1030757133, 8630, 293949, 497, 1206340275, 172, 950651, 983, 138]
    5. >>> max(n, key=lambda x:len(str(x)))
    6. 1030757133
    7. >>> min(n, key=lambda x:len(str(x)))
    8. 497
    9. >>>
    10. >>> max(n)
    11. 1206340275
    12. >>> min(n)
    13. 138

     2. 列表任意指定长度的分组

    1. def splitlist(lst,*group):
    2. res,lst = [],lst[:]
    3. for i in group:
    4. t = []
    5. for _ in range(i):
    6. if lst: t.append(lst.pop(0))
    7. else: break
    8. if t: res.append(t)
    9. if lst: res.append(lst)
    10. return res
    11. a = [1,2,3,4,5,6,7,10,11,12,20,21,23]
    12. print(splitlist(a,8,4,2,1))
    13. # [[1, 2, 3, 4, 5, 6, 7, 10], [11, 12, 20, 21], [23]]
    14. print(splitlist(a,7,3,2,1))
    15. # [[1, 2, 3, 4, 5, 6, 7], [10, 11, 12], [20, 21], [23]]
    16. print(splitlist(a,7,3,2,2))
    17. # [[1, 2, 3, 4, 5, 6, 7], [10, 11, 12], [20, 21], [23]]
    18. print(splitlist(a,7,3,2))
    19. # [[1, 2, 3, 4, 5, 6, 7], [10, 11, 12], [20, 21], [23]]
    20. print(splitlist(a,7,3))
    21. # [[1, 2, 3, 4, 5, 6, 7], [10, 11, 12], [20, 21, 23]]
    22. print(splitlist(a,7,2,2,5))
    23. # [[1, 2, 3, 4, 5, 6, 7], [10, 11], [12, 20], [21, 23]]

    3. 用numpy解多元一次方程组

    1. import numpy as np
    2. A = [[1,-2,1],[0,2,-8],[-4,5,9]] # 系数矩阵
    3. B = [0,8,-9] # 常数矩阵
    4. x,y,z = np.linalg.inv(A).dot(B) # 求方程组的解
    5. print(f'x={x}, y={y}, z={z}') # 默认它有且只有一组解,不展开讨论

    4. 应用字典同时返回两数的和差积商余

    1. def exp(a,b):
    2. choice = { 1: lambda x, y: x + y,
    3. 2: lambda x, y: x - y,
    4. 3: lambda x, y: x * y,
    5. 4: lambda x, y: x // y,
    6. 5: lambda x, y: x % y }
    7. return tuple(choice[i+1](a,b) for i in range(5))
    8. print(exp(8,3)) # (11, 5, 24, 2, 2)
    9. print(exp(5,6)[2]) # 30

    5. 多种方法取出某字母开头的元素

    1. >>> s = ['foot','head','Face','Hair','nose','Mouth','finger','ear']
    2. >>> [w for w in s if w[0].lower()=='f']
    3. ['foot', 'Face', 'finger']
    4. >>> [w for w in s if w[0] in ('f','F')]
    5. ['foot', 'Face', 'finger']
    6. >>> [w for w in s if w.startswith(('F','f'))]
    7. ['foot', 'Face', 'finger']

    6. 输入一个整数,输错不会异常退出

    1. n = ''
    2. while not n.isnumeric(): n = input('输入一个非负整数:')
    3. n = int(n)
    1. n = ''
    2. while not (n.isnumeric() or len(n) and n[0]=='-' and n[1:].isnumeric()):
    3. n = input('输入一个整数:')
    4. n = int(n)

     

    【相关阅读】

    Python 在问答频道中刷题积累到的小技巧(一)https://hannyang.blog.csdn.net/article/details/124935045

    Python 在问答频道中刷题积累到的小技巧(二)https://hannyang.blog.csdn.net/article/details/125026881

    Python 在问答频道中刷题积累到的小技巧(三)
    https://hannyang.blog.csdn.net/article/details/125058178

    Python 在问答频道中刷题积累到的小技巧(四)
    https://hannyang.blog.csdn.net/article/details/125211774

    Python 在问答频道中刷题积累到的小技巧(五)https://hannyang.blog.csdn.net/article/details/125270812

    Python 在问答频道中刷题积累到的小技巧(六)
    https://hannyang.blog.csdn.net/article/details/125339202

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  • 原文地址:https://blog.csdn.net/boysoft2002/article/details/125439517