因此，我對(duì)python非常陌生，并且我正在嘗試制作一個(gè)基于文本的類(lèi)似rpg的東西，該事物涉及具有2個(gè)出口（左側(cè)或右側(cè)）的第一個(gè)房間，之后的每行房間都有最后一個(gè)房間之和的整數(shù)量，每個(gè)變量都是從0到3（退出量）的隨機(jī)整數(shù)， 這樣：a = [2]print(a)import randomb = []for i in range(0,2): x = random.randint(0,3) b.append(x)print(b)b_sum = sum(b)import randomc = []for i in range(0,b_sum): x = random.randint(0,3) c.append(x)print(c)c_sum = sum(c)import randomd = []for i in range(0,c_sum): x = random.randint(0,3) d.append(x)print(d)d_sum = sum(d)import randome = []for i in range(0,d_sum): x = random.randint(0,3) e.append(x)print(e)e_sum = sum(e)import randomf = []for i in range(0,e_sum): x = random.randint(0,3) f.append(x)print(f)f_sum = sum(f)import randomg = []for i in range(0,f_sum): x = random.randint(0,3) g.append(x)print(g)這工作正常，但導(dǎo)航已被證明很難。rowlist = [a,b,c,d,e,f,g,h,ii]row = (rowlist[0])room = (a[0])print(room)if room == 2: door=str(input("left or right")) if door == "left": roomsum = sum(row[row < room (+1)])我在這里試圖做的是找到當(dāng)前“房間”之前的列表中的每個(gè)整數(shù)的總和。但我不知道如何做到這一點(diǎn)！任何幫助將不勝感激，非常感謝！
1 回答
慕的地10843
TA貢獻(xiàn)1785條經(jīng)驗(yàn) 獲得超8個(gè)贊
看起來(lái)你正在尋找這樣的東西：
import random
big_list = [[2]]
for _ in range(1,5): # increase to create more "rooms"
big_list.append( [random.randint(0,3) for _ in range(sum(big_list[-1]))])
total = 0
for inner in big_list:
print(inner, "before:", total)
total += sum(inner)
生成如下列表：
[2] before: 0
[3, 1] before: 2
[2, 3, 1, 1] before: 6
[1, 1, 2, 0, 2, 2, 2] before: 13
[2, 3, 0, 2, 3, 2, 2, 1, 0, 2] before: 23
由于隨機(jī)性，您還可能得到：
[2] before: 0
[0, 1] before: 2
[0] before: 3
[] before: 3
[] before: 3
列表中的部分和可以通過(guò) sum 或列表切片和 sum 中的生成器表達(dá)式獲得：
lol = [[1,2,3], [4,5,6,7,8,9,10,11,12], [13,14,15,16,17]]
idx_in_lol = 1 # [4,5,6,7,8,9,10,11,12]
idx_in_inner = 5 # [4,5,6,7,8,***9***,10,11,12]
# generator expression and enumerate
s1 = sum( i if idx < idx_in_inner else 0 for idx,i in enumerate(lol[idx_in_lol]))
# or by slicing
s2 = sum( lol[idx_in_lol][:idx_in_inner] )
print(s1, s2)
輸出：
30 30 # 4+5+6+7+8 = 30
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