1、基礎。輸入下列表達式,然後試着說明每種情況所產生的結果:
2 ** 16 2 / 5, 2 / 5.0 "spam" + "eggs" S = "ham" "eggs " + S S * 5 S[:0] "green %s and %s" % ("eggs", S) 'green {0} and {1}'.format('eggs', S) ('x',)[0] ('x', 'y')[1] L = [1,2,3] + [4,5,6] L, L[:], L[:0], L[-2], L[-2:] ([1,2,3] + [4,5,6])[2:4] [L[2], L[3]] L.reverse(); L L.sort(); L L.index(4) {'a':1, 'b':2}['b'] D = {'x':1, 'y':2, 'z':3} D['w'] = 0 D['x'] + D['w'] D[(1,2,3)] = 4 list(D.keys()), list(D.values()), (1,2,3) in D [[]], ["",[],(),{},None]
2、元組賦值運算。輸入下面幾行,發生了什麼:
>>> X = 'spam' >>> Y = 'eggs' >>> X, Y = Y, X #理解:元組解包
3、字典鍵
>>> D = {} >>> D[1] = 'a' >>> D[2] = 'b' #嘗試加入新的鍵值對,如下所示 >>> D[(1, 2, 3)] = 'c' >>> D {1: 'a', 2: 'b', (1, 2, 3): 'c'}
4、思考題
L=[0,1,2,3],實驗下面一些情況:
a、輸入L[4]會發生什麼?
b、輸入L[-1000:100]又會輸出怎樣的結果?
c、如果是輸入L[3:1]呢?(提示原理:當起始值大於結束值,結束值會被重寫爲起始值)