Python之列表推导式最全汇总(下篇)
作者：Hann Yang
这篇文章主要介绍了Python之列表推导式最全汇总（下篇），本文章内容详细，通过案例可以更好的理解列表推导式的相关知识，本模块分为了三部分，本次为下篇，需要的朋友可以参考下
前言
网传的七天学Python的路线如下，我觉得可以在学过此表中前几天的内容后，就可以回头来学习一下
列表推导式：它综合了列表、for循环和条件语句。
第一天：基本概念(4小时) ： print,变量，输入，条件语句。
第二天：基本概念(5小时) ：列表，for循环，while循环，函数，导入模块。
第三天：简单编程问题(5小时) ：交换两个变量值，将摄氏度转换为华氏温度，求数字中各位数之和， 判断某数是否为素数， 生成随机数，删除列表中的重复项等等。
第四天：中级编程问题(6小时) ：反转-个字符串(回文检测)，计算最大公约数，合并两个有序数组，猜数字游戏，计算年龄等等。
第五天：数据结构(6小时) ：栈，队列，字典，元组，树，链表。
第六天：面向对象编程(OOP) (6小时) ：对象，类，方法和构造函数，面向对象编程之继承。
第七天：算法(6小时) ：搜索(线性和二分查找)、 排序(冒泡排序、 选择排序)、递归函数(阶乘、斐波那契数列)时间复杂度(线性、二次和常量)。
列表推导式
- list comprehension或译为列表解析式，是一种创建列表的简洁语法；
- 也可认为它是一个简版的for循环，但执行效率高于for循环。
- python 2.7+ 开始又引入了集合推导式、字典推导式，原理与列表推导式相近。
语法规范：
out_list = [out_express for out_express in input_list if out_express_condition]
其中，
- if 条件可有可无；
- for 循环可以嵌套多层，内外层循环的变量不可以同名；
- 推导式中也可以嵌套推导式，内外层推导式的变量互不影响，可以同名；
- 推导表达式out_express尽可能用内置函数，省得import或def function()。
进阶实例
乘法口诀表
>>> lst=['%sx%s=%s'%(j,i,i*j) for i in range(1,10) for j in range(1,i+1)] >>> lst ['1x1=1', '1x2=2', '2x2=4', '1x3=3', '2x3=6', '3x3=9', '1x4=4', '2x4=8', '3x4=12', '4x4=16', '1x5=5', '2x5=10', '3x5=15', '4x5=20', '5x5=25', '1x6=6', '2x6=12', '3x6=18', '4x6=24', '5x6=30', '6x6=36', '1x7=7', '2x7=14', '3x7=21', '4x7=28', '5x7=35', '6x7=42', '7x7=49', '1x8=8', '2x8=16', '3x8=24', '4x8=32', '5x8=40', '6x8=48', '7x8=56', '8x8=64', '1x9=9', '2x9=18', '3x9=27', '4x9=36', '5x9=45', '6x9=54', '7x9=63', '8x9=72', '9x9=81']
列印时，要注意它的项数通项公式是： An=n(n+1)/2+1
>>>
for i in range(9):
for j in range(i+1):
print(lst[i*(i+1)//2+j],end='\t' if i!=j else '\n')
1x1=1
1x2=2 2x2=4
1x3=3 2x3=6 3x3=9
1x4=4 2x4=8 3x4=12 4x4=16
1x5=5 2x5=10 3x5=15 4x5=20 5x5=25
1x6=6 2x6=12 3x6=18 4x6=24 5x6=30 6x6=36
1x7=7 2x7=14 3x7=21 4x7=28 5x7=35 6x7=42 7x7=49
1x8=8 2x8=16 3x8=24 4x8=32 5x8=40 6x8=48 7x8=56 8x8=64
1x9=9 2x9=18 3x9=27 4x9=36 5x9=45 6x9=54 7x9=63 8x9=72 9x9=81
>>> 或者用join()直接把 \t \n 插入列表拼接成字符串，然后输出：
>>> print('\n'.join(['\t'.join([f'{j}x{i}={i*j}' for j in range(1,i+1)]) for i in range(1,10)]))
1x1=1
1x2=2 2x2=4
1x3=3 2x3=6 3x3=9
1x4=4 2x4=8 3x4=12 4x4=16
1x5=5 2x5=10 3x5=15 4x5=20 5x5=25
1x6=6 2x6=12 3x6=18 4x6=24 5x6=30 6x6=36
1x7=7 2x7=14 3x7=21 4x7=28 5x7=35 6x7=42 7x7=49
1x8=8 2x8=16 3x8=24 4x8=32 5x8=40 6x8=48 7x8=56 8x8=64
1x9=9 2x9=18 3x9=27 4x9=36 5x9=45 6x9=54 7x9=63 8x9=72 9x9=81
>>>
>>> # f'{j}x{i}={i*j}' 等价于 '%sx%s=%s'%(j,i,i*j)求100以内的质数（或称素数）
>>> [k[0] for k in [(j,sum([j%i==0 for i in range(2,j)])) for j in range(2,100)] if k[1]==0] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>> >>> # 等价于： >>> [k[0] for k in [(j,sum([0 if j%i else 1 for i in range(2,j)])) for j in range(2,100)] if k[1]==0] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>>
上面两种方法都是累加sum布尔值bool的个数来计算的，可以用any() all()函数代替：
>>> [k[0] for k in [(j,[j%i==0 for i in range(2,j)]) for j in range(2,100)] if not any(k[1])] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>> >>> # 等价于： >>> [k[0] for k in [(j,[j%i!=0 for i in range(2,j)]) for j in range(2,100)] if all(k[1])] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>>
使用filter()和map()函数：
[i for i in filter(lambda x:all(map(lambda p:x%p,range(2,x))), range(2,100))] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>>
条件反过来就是100以内的合数：
>>> [k[0] for k in [(j,[j%i==0 for i in range(2,j)]) for j in range(2,100)] if any(k[1])] [4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99] >>> [k[0] for k in [(j,[j%i!=0 for i in range(2,j)]) for j in range(2,100)] if not all(k[1])] [4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99] >>> [i for i in filter(lambda x:not all(map(lambda p:x%p,range(2,x))), range(2,100))] [4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99] >>>
求1000以内的质回文数（即是质数又是回文数）
>>> Pr=[str(k[0]) for k in [(j,[j%i!=0 for i in range(2,j)]) for j in range(2,1000)] if all(k[1])] >>> [int(p) for p in Pr if p[::-1] in Pr] [2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929]
求1000以内的数，满足本身和它的回文数同是质数
>>> pstr=[str(k[0]) for k in [(j,[j%i!=0 for i in range(2,j)]) for j in range(2,1000)] if all(k[1])] >>> [int(p) for p in pstr if p[::-1] in pstr] [2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 101, 107, 113, 131, 149, 151, 157, 167, 179, 181, 191, 199, 311, 313, 337, 347, 353, 359, 373, 383, 389, 701, 709, 727, 733, 739, 743, 751, 757, 761, 769, 787, 797, 907, 919, 929, 937, 941, 953, 967, 971, 983, 991] >>>
分解质因数(N<2022)
>>> P = [k[0] for k in [(j,sum([j%i==0 for i in range(2,j)])) for j in range(2,2022)] if k[1]==0]
>>>
def func(n):
res=[]
t=P[::-1]
m=t[-1]
while n>=m:
if n%m==0:
n//=m
res.append(m)
else:
t.pop()
m=t[-1]
return res
>>> i = 2021
>>> print(i,'=','x'.join([str(j) for j in func(i)]))
2021 = 43x47
>>>
>>> func(2000)
[2, 2, 2, 2, 5, 5, 5]
>>> func(3999)
[3, 31, 43]
>>> 求出字符串的所有字串（可推广到所有可切片数据类型）
>>> L='abcd' >>> [L[i:j] for i in range(len(L)) for j in range(i+1,len(L)+1)] ['a', 'ab', 'abc', 'abcd', 'b', 'bc', 'bcd', 'c', 'cd', 'd'] >>>
注：凡是可以用[i:j]来切片的“容器类”数据类型都可用此推导式。
如：
>>> L=[1,2,3,4] >>> [L[i:j] for i in range(len(L)) for j in range(i+1,len(L)+1)] [[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [2], [2, 3], [2, 3, 4], [3], [3, 4], [4]] >>>
（1）找出字符串s='aaabcddcbddba'中最长的回文字串
>>> s='aaabcddcbddba'
>>> {s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)}
{'bcddcbdd', 'ddcbddb', 'abcddcbdd', 'aabcddc', 'aabcddcbdd', 'cbddba', 'aaa',
'cddcbd', 'cbddb', 'a', 'cdd', 'bcdd', 'aabcddcbddba', 'aabcd', 'abcddcbddb',
'aaabc', 'ab', 'cbdd', 'cddcbdd', 'aaabcddcb', 'aabcddcbd', 'aaabcdd', 'ba',
'ddcbddba', 'dba', 'db', 'cddcbddba', 'cbd', 'aaabcddcbddb', 'aaabcd', 'ddb',
'dcbdd', 'abcddc', 'abcd', 'abcdd', 'bcddcb', 'aaabcddcbdd', 'abcddcbddba',
'aabc', 'bcddc', 'bdd', 'cb', 'bcddcbddba', 'c', 'dcbddb', 'ddba', 'dcbd',
'b', 'aaab', 'dd', 'd', 'ddcbd', 'bcd', 'aa', 'abcddcbd', 'bcddcbddb',
'aaabcddcbd', 'cddc', 'ddcb', 'dc', 'abc', 'bddb', 'ddc', 'bcddcbd', 'bc',
'aabcdd', 'aab', 'aaabcddcbddba', 'cddcb', 'abcddcb', 'cd', 'bddba', 'aabcddcbddb',
'bd', 'ddcbdd', 'aaabcddc', 'dcb', 'dcbddba', 'aabcddcb', 'cddcbddb'}
>>> # 使用字典推导式可去掉相同子串
>>>
>>> [i for i in {s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)} if i==i[::-1]]
['aaa', 'a', 'bcddcb', 'c', 'b', 'dd', 'd', 'aa', 'cddc', 'bddb']
>>>
>>> [(len(i),i) for i in {i for i in [s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)} if i==i[::-1]]]
[(3, 'aaa'), (1, 'a'), (6, 'bcddcb'), (1, 'c'), (1, 'b'), (2, 'dd'), (1, 'd'),
(2, 'aa'), (4, 'cddc'), (4, 'bddb')]
>>>
>>> max([(len(i),i) for i in [i for i in {s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)} if i==i[::-1]]])[1]
'bcddcb'
>>>（2）给定 L=[2, -3, 3, 50, 5, 0, -1]，输出其子序列中各元素合计数最大的子序列
>>> L = [2, -3, 3, 50, 5, 0, -1] >>> [L[i:j] for i in range(len(L)) for j in range(i+1,len(L)+1)] [[2], [2, -3], [2, -3, 3], [2, -3, 3, 50], [2, -3, 3, 50, 5], [2, -3, 3, 50, 5, 0], [2, -3, 3, 50, 5, 0, -1], [-3], [-3, 3], [-3, 3, 50], [-3, 3, 50, 5], [-3, 3, 50, 5, 0], [-3, 3, 50, 5, 0, -1], [3], [3, 50], [3, 50, 5], [3, 50, 5, 0], [3, 50, 5, 0, -1], [50], [50, 5], [50, 5, 0], [50, 5, 0, -1], [5], [5, 0], [5, 0, -1], [0], [0, -1], [-1]] >>> [sum(j) for j in [L[j:i] for i in range(len(L),0,-1) for j in range(len(L))]] [56, 54, 57, 54, 4, -1, -1, 57, 55, 58, 55, 5, 0, 0, 57, 55, 58, 55, 5, 0, 0, 52, 50, 53, 50, 0, 0, 0, 2, 0, 3, 0, 0, 0, 0, -1, -3, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0] >>> # 合并为一行代码： >>> max(sum(j) for j in [L[j:i] for i in range(len(L),0,-1) for j in range(len(L))]) 58 >>> >>> # 和最大的子序列为： >>> l = [L[i:j] for i in range(len(L)) for j in range(i+1,len(L)+1)] >>> m = max(sum(j) for j in [L[j:i] for i in range(len(L),0,-1) for j in range(len(L))]) >>> [i for i in l if sum(i)==m] [[3, 50, 5], [3, 50, 5, 0]] >>>
根据方程式画出字符图
略：见相关文章《探究“一行代码画爱心”的秘密，去向心爱的人表白吧》
EXCEL表格列号字串转整数
>>> ExcelCol2Int = lambda s:sum([(ord(n)-64)*26**i for i,n in enumerate(list(s)[::-1])])
>>> ExcelCol2Int('A')
1
>>> ExcelCol2Int('AA')
27
>>> ExcelCol2Int('AX')
50
>>> ExcelCol2Int('CV')
100
>>> ExcelCol2Int('AAA')
703
>>> ExcelCol2Int('XFD')
16384
>>>打印Gray格雷码序列
什么是格雷码，什么是卡诺图？不懂的问度娘吧
Gray = lambda n:[(i>>1)^i for i in range(2**n)] GrayB = lambda n:[bin((i>>1)^i)[2:].rjust(n,'0') for i in range(2**n)] ''' Gray(0) [0] Gray(1) [0, 1] Gray(2) [0, 1, 3, 2] Gray(3) [0, 1, 3, 2, 6, 7, 5, 4] Gray(4) [0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8] >>> GrayB(3) # 把输出排成矩阵，即三变量卡诺图 ['000', '001', '011', '010', '110', '111', '101', '100'] >>> GrayB(4) # 把输出排成方阵，即四变量卡诺图 ['0000', '0001', '0011', '0010', '0110', '0111', '0101', '0100', '1100', '1101', '1111', '1110', '1010', '1011', '1001', '1000'] '''
注：上面代码中的 (i>>1)^i 可以写成： i^i>>1 或 i^i//2，因为右移或整除的运算级别都比异或要高。验证代码如下：
>>> any([(i>>1)^i == i^i>>1 == i^i//2 for i in range(323)]) True >>>
另外，观察到Gray()输出的整数序列的规律，就想到用迭代法也能实现，并且只要2行代码：
# 用迭代法实现： def iGray(n): if n==0: return [0] return iGray(n-1)+[i+2**(n-1) for i in iGray(n-1)[::-1]]
高阶实例
杨辉三角形
方法一：公式递推
>>> def func(i): t=L=[1] while(i>1): i-=1 t=L+[t[n]+t[n+1] for n in range(len(t)-1)]+L return t >>> func(1) [1] >>> func(3) [1, 2, 1] >>> func(8) [1, 7, 21, 35, 35, 21, 7, 1] >>> >>> for i in range(1,10):print(func(i)) [1] [1, 1] [1, 2, 1] [1, 3, 3, 1] [1, 4, 6, 4, 1] [1, 5, 10, 10, 5, 1] [1, 6, 15, 20, 15, 6, 1] [1, 7, 21, 35, 35, 21, 7, 1] [1, 8, 28, 56, 70, 56, 28, 8, 1] >>>
方法二：组合公式的自定义函数
>>> def Combin(n,i):
m,t=min(i,n-i),1
for j in range(0,m):
t*=(n-j)/(m-j)
return t
>>> [Combin(8,i) for i in range(9)]
[1, 8.0, 28.0, 55.99999999999999, 70.0, 55.99999999999999, 28.0, 8.0, 1]
>>> [round(Combin(8,i)) for i in range(9)]
[1, 8, 28, 56, 70, 56, 28, 8, 1]
>>> [[round(Combin(j,i)) for i in range(j+1)] for j in range(10)]
[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56,
28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]]
>>>根据组合公式用阶乘来计算：
C(m,n)=math.factorial(n)//(math.factorial(m)*math.factorial(n-m))
递归法，虽然没有小数精度的问题，但也有递归次数不能太大即n值有限制的缺点：
>>> def Comb(n,i):
if i in [0,n]:
return 1
elif i==1:
return n
else:
return Comb(n-1,i-1)+Comb(n-1,i)
>>> [Comb(8,i) for i in range(9)]
[1, 8, 28, 56, 70, 56, 28, 8, 1]
>>> print('\n'.join(['\t'.join([str(Comb(j,i)) for i in range(j+1)]) for j in range(10)]))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
>>>方法三：使用现成的库函数
(1). itertools库combinations函数
>>> from itertools import combinations as comb
>>> [[len(list(comb(range(j),i))) for i in range(j+1)] for j in range(10)]
[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56,
28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]]
>>> print('\n'.join(['\t'.join([str(len(list(comb(range(j),i)))) for i in range(j+1)]) for j in range(10)]))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
>>>(2). scipy库comb函数
>>> from scipy.special import comb
>>> [[round(comb(j,i)) for i in range(j+1)] for j in range(10)]
[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56,
28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]]
>>> print('\n'.join(['\t'.join([str(round(comb(j,i))) for i in range(j+1)]) for j in range(10)]))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
>>>斐波那契数列
(1). 引入自定义函数或lambda表达式：
>>> f=lambda n:n<3 and 1 or f(n-1)+f(n-2) >>> [f(i) for i in range(1,30)] [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229] >>>
注：这个lambda函数用了递归法，在项数大于40后速度超慢。
另：类似【n<3 and 1 or 2】这种表达式中，逻辑与、或有这样的特性：
and 两边都对取后面一个表达式， or 两边都对取前面一个表达式。
>>> n=5 >>> 1 and n 5 >>> n and 1 1 >>> 0 and n 0 >>> n and 0 0 >>> 1 or n 1 >>> n or 1 5 >>> 0 or n 5 >>> n or 0 5 >>>
(2). 引入临时推导表达式：
>>> N=50 # 项数=50 >>> f=[1,1] >>> [f.append(f[n-2]+f[n-1]) for n in range(2,N)] [None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None] >>> f [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025] >>> ###合成一行，临时变量接收[None]*n列表### >>> f=[1,1];t=[f.append(f[n-2]+f[n-1]) for n in range(2,N)] >>> f [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025] >>>
真的一行代码：
>>> N=50 # 项数=50 >>> t=[(f[n][0], f.append((f[n][1],f[n][0]+f[n][1]))) for f in ([[1,1]],) for n in range(N)] >>> t [(1, None), (1, None), (2, None), (3, None), (5, None), (8, None), (13, None), (21, None), (34, None), (55, None), (89, None), (144, None), (233, None), (377, None), (610, None), (987, None), (1597, None), (2584, None), (4181, None), (6765, None), (10946, None), (17711, None), (28657, None), (46368, None), (75025, None), (121393, None), (196418, None), (317811, None), (514229, None), (832040, None), (1346269, None), (2178309, None), (3524578, None), (5702887, None), (9227465, None), (14930352, None), (24157817, None), (39088169, None), (63245986, None), (102334155, None), (165580141, None), (267914296, None), (433494437, None), (701408733, None), (1134903170, None), (1836311903, None), (2971215073, None), (4807526976, None), (7778742049, None), (12586269025, None)] >>> [f[0] for f in t] [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025] >>> ###合并成一行### >>> [f[0] for f in [(f[n][0], f.append((f[n][1],f[n][0]+f[n][1]))) for f in ([[1,1]],) for n in range(N)]] [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025] >>>
曼德勃罗集(Mandelbrot Set)分形
>>> print('\n'.join([''.join(['*'if abs((lambda a:lambda z,c,n:a(a,z,c,n))(lambda s,z,c,n:s(s,z*z+c,c,n-1) if n else z)(0,0.02*x+0.05j*y,40))<2 else ' ' for x in range(-78,20)]) for y in range(-20,21)]))
*
**
***********
************
*********
* * ************ * *
****** * *************************** *
*************************************** *****
*******************************************
*** ******************************************** *
**************************************************
*******************************************************
* * *****************************************************
**** ******* * *******************************************************
***************** *******************************************************
*********************** *********************************************************
*********************** ********************************************************
**** ********************************************************************************
*******************************************************************************************
**** ********************************************************************************
*********************** ********************************************************
*********************** *********************************************************
***************** *******************************************************
**** ******* * *******************************************************
* * *****************************************************
*******************************************************
**************************************************
*** ******************************************** *
*******************************************
*************************************** *****
****** * *************************** *
* * ************ * *
*********
************
***********
**
*
>>>附录
到此这篇关于Python之列表推导式最全汇总(下篇)的文章就介绍到这了,关于Python之列表推导式最全汇总的所有内容也全部结束了，其他两个部分的内容(上、中篇)请搜索脚本之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持脚本之家！