Python之列表推導(dǎo)式最全匯總(下篇)
前言
網(wǎng)傳的七天學(xué)Python的路線(xiàn)如下,我覺(jué)得可以在學(xué)過(guò)此表中前幾天的內(nèi)容后,就可以回頭來(lái)學(xué)習(xí)一下
列表推導(dǎo)式:它綜合了列表、for循環(huán)和條件語(yǔ)句。
第一天:基本概念(4小時(shí)) : print,變量,輸入,條件語(yǔ)句。
第二天:基本概念(5小時(shí)) :列表,for循環(huán),while循環(huán),函數(shù),導(dǎo)入模塊。
第三天:簡(jiǎn)單編程問(wèn)題(5小時(shí)) :交換兩個(gè)變量值,將攝氏度轉(zhuǎn)換為華氏溫度,求數(shù)字中各位數(shù)之和, 判斷某數(shù)是否為素?cái)?shù), 生成隨機(jī)數(shù),刪除列表中的重復(fù)項(xiàng)等等。
第四天:中級(jí)編程問(wèn)題(6小時(shí)) :反轉(zhuǎn)-個(gè)字符串(回文檢測(cè)),計(jì)算最大公約數(shù),合并兩個(gè)有序數(shù)組,猜數(shù)字游戲,計(jì)算年齡等等。
第五天:數(shù)據(jù)結(jié)構(gòu)(6小時(shí)) :棧,隊(duì)列,字典,元組,樹(shù),鏈表。
第六天:面向?qū)ο缶幊?OOP) (6小時(shí)) :對(duì)象,類(lèi),方法和構(gòu)造函數(shù),面向?qū)ο缶幊讨^承。
第七天:算法(6小時(shí)) :搜索(線(xiàn)性和二分查找)、 排序(冒泡排序、 選擇排序)、遞歸函數(shù)(階乘、斐波那契數(shù)列)時(shí)間復(fù)雜度(線(xiàn)性、二次和常量)。
列表推導(dǎo)式
- list comprehension或譯為列表解析式,是一種創(chuàng)建列表的簡(jiǎn)潔語(yǔ)法;
- 也可認(rèn)為它是一個(gè)簡(jiǎn)版的for循環(huán),但執(zhí)行效率高于for循環(huán)。
- python 2.7+ 開(kāi)始又引入了集合推導(dǎo)式、字典推導(dǎo)式,原理與列表推導(dǎo)式相近。
語(yǔ)法規(guī)范:
out_list = [out_express for out_express in input_list if out_express_condition]
其中,
- if 條件可有可無(wú);
- for 循環(huán)可以嵌套多層,內(nèi)外層循環(huán)的變量不可以同名;
- 推導(dǎo)式中也可以嵌套推導(dǎo)式,內(nèi)外層推導(dǎo)式的變量互不影響,可以同名;
- 推導(dǎo)表達(dá)式out_express盡可能用內(nèi)置函數(shù),省得import或def function()。
進(jìn)階實(shí)例
乘法口訣表
>>> 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']
列印時(shí),要注意它的項(xiàng)數(shù)通項(xiàng)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,然后輸出:
>>> 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}' 等價(jià)于 '%sx%s=%s'%(j,i,i*j)
求100以?xún)?nèi)的質(zhì)數(shù)(或稱(chēng)素?cái)?shù))
>>> [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] >>> >>> # 等價(jià)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] >>>
上面兩種方法都是累加sum布爾值bool的個(gè)數(shù)來(lái)計(jì)算的,可以用any() all()函數(shù)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] >>> >>> # 等價(jià)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()函數(shù):
[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] >>>
條件反過(guò)來(lái)就是100以?xún)?nèi)的合數(shù):
>>> [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?xún)?nèi)的質(zhì)回文數(shù)(即是質(zhì)數(shù)又是回文數(shù))
>>> 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?xún)?nèi)的數(shù),滿(mǎn)足本身和它的回文數(shù)同是質(zhì)數(shù)
>>> 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] >>>
分解質(zhì)因數(shù)(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(可推廣到所有可切片數(shù)據(jù)類(lèi)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]來(lái)切片的“容器類(lèi)”數(shù)據(jù)類(lèi)型都可用此推導(dǎo)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'中最長(zhǎng)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'} >>> # 使用字典推導(dǎo)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)給定 L=[2, -3, 3, 50, 5, 0, -1],輸出其子序列中各元素合計(jì)數(shù)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] >>> # 合并為一行代碼: >>> max(sum(j) for j in [L[j:i] for i in range(len(L),0,-1) for j in range(len(L))]) 58 >>> >>> # 和最大的子序列為: >>> 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]] >>>
根據(jù)方程式畫(huà)出字符圖
略:見(jiàn)相關(guān)文章《探究“一行代碼畫(huà)愛(ài)心”的秘密,去向心愛(ài)的人表白吧》
EXCEL表格列號(hào)字串轉(zhuǎn)整數(shù)
>>> 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格雷碼序列
什么是格雷碼,什么是卡諾圖?不懂的問(wèn)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) # 把輸出排成矩陣,即三變量卡諾圖 ['000', '001', '011', '010', '110', '111', '101', '100'] >>> GrayB(4) # 把輸出排成方陣,即四變量卡諾圖 ['0000', '0001', '0011', '0010', '0110', '0111', '0101', '0100', '1100', '1101', '1111', '1110', '1010', '1011', '1001', '1000'] '''
注:上面代碼中的 (i>>1)^i 可以寫(xiě)成: i^i>>1 或 i^i//2,因?yàn)橛乙苹蛘倪\(yùn)算級(jí)別都比異或要高。驗(yàn)證代碼如下:
>>> any([(i>>1)^i == i^i>>1 == i^i//2 for i in range(323)]) True >>>
另外,觀察到Gray()輸出的整數(shù)序列的規(guī)律,就想到用迭代法也能實(shí)現(xiàn),并且只要2行代碼:
# 用迭代法實(shí)現(xiàn): def iGray(n): if n==0: return [0] return iGray(n-1)+[i+2**(n-1) for i in iGray(n-1)[::-1]]
高階實(shí)例
楊輝三角形
方法一:公式遞推
>>> 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:組合公式的自定義函數(shù)
>>> 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]] >>>
根據(jù)組合公式用階乘來(lái)計(jì)算:
C(m,n)=math.factorial(n)//(math.factorial(m)*math.factorial(n-m))
遞歸法,雖然沒(méi)有小數(shù)精度的問(wèn)題,但也有遞歸次數(shù)不能太大即n值有限制的缺點(diǎn):
>>> 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:使用現(xiàn)成的庫(kù)函數(shù)
(1). itertools庫(kù)combinations函數(shù)
>>> 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庫(kù)comb函數(shù)
>>> 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 >>>
斐波那契數(shù)列
(1). 引入自定義函數(shù)或lambda表達(dá)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:這個(gè)lambda函數(shù)用了遞歸法,在項(xiàng)數(shù)大于40后速度超慢。
另:類(lèi)似【n<3 and 1 or 2】這種表達(dá)式中,邏輯與、或有這樣的特性:
and 兩邊都對(duì)取后面一個(gè)表達(dá)式, or 兩邊都對(duì)取前面一個(gè)表達(dá)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). 引入臨時(shí)推導(dǎo)表達(dá)式:
>>> N=50 # 項(xiàng)數(shù)=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,臨時(shí)變量接收[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] >>>
真的一行代碼:
>>> N=50 # 項(xiàng)數(shù)=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] >>>
曼德勃羅集(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)])) * ** *********** ************ ********* * * ************ * * ****** * *************************** * *************************************** ***** ******************************************* *** ******************************************** * ************************************************** ******************************************************* * * ***************************************************** **** ******* * ******************************************************* ***************** ******************************************************* *********************** ********************************************************* *********************** ******************************************************** **** ******************************************************************************** ******************************************************************************************* **** ******************************************************************************** *********************** ******************************************************** *********************** ********************************************************* ***************** ******************************************************* **** ******* * ******************************************************* * * ***************************************************** ******************************************************* ************************************************** *** ******************************************** * ******************************************* *************************************** ***** ****** * *************************** * * * ************ * * ********* ************ *********** ** * >>>
附錄
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