使用隨機數在Python中創建二維列表。

Question

import random 

rows = 3 
cols = 3 
virus = [0,0,0,0,0,1,2] 
virus_level = random.choice(virus) 
first_gen = [] 
row =[] 
for rownum in range(rows): 
    for colnum in range(cols): 
     virus_level = random.choice(virus) 
     row.append(virus_level) 
    first_gen.append(row) 

我需要使用列表中的隨機數創建一個二維列表病毒。這是我迄今爲止，數字不是隨機的順序，列表打印出9乘3，而不是3乘3。一些指向正確方向的指針會很好。

Answer 1

你需要在每個循環復位row。

將您的代碼片段保持原樣，如果您只需將row =[]一行放入外部循環中，那應該會給您正確的結果。

import random 

rows = 3 
cols = 3 
virus = [0,0,0,0,0,1,2] 
virus_level = random.choice(virus) 
first_gen = [] 
for rownum in range(rows): 
    row = []  
    for colnum in range(cols): 
     virus_level = random.choice(virus) 
     row.append(virus_level) 
    first_gen.append(row) 

print first_gen # [[2, 0, 0], [0, 0, 0], [1, 0, 0]] 

Answer 2

你分配row[]應該是最外層的循環中，像這樣：

import random 

rows = 3 
cols = 3 
virus = [0,0,0,0,0,1,2] 
virus_level = random.choice(virus) 
first_gen = [] 
for rownum in range(rows): 
    row = [] 
    for colnum in range(cols): 
     virus_level = random.choice(virus) 
     row.append(virus_level) 
    first_gen.append(row) 

不過，我更喜歡使用列表理解這些各種各樣的事情。這裏是一個另類：

from pprint import pprint 
import random 

rows = 3 
cols = 3 
virus = [0,0,0,0,0,1,2] 

first_gen = [[random.choice(virus) for _ in range(rows)] for _ in range(cols)] 
pprint (first_gen) 

Answer 3

該做的伎倆：

>>> import random 
>>> rows=3 
>>> cols=3 
>>> virus=[0,0,0,0,0,1,2] 
>>> [[random.choice(virus) for r in range(rows)] for c in range(cols)] 
[[0, 2, 2], [0, 0, 0], [0, 0, 0]]