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如何在列表列表中搜索特定列表，如此具体：

假设我有一个包含许多其他列表的列表 grid_list，我想通过 grid_list 搜索以找到其中一个，具体我该怎么做？这是我认为应该有效但无效的代码：

import sys #sys has been imported to take in user input
import random #random has been imported to generate random numbers to store the ants food in in the grid

#grid list is a generated list used to store grid values
grid_list = []
grid_size=0 #grid size is a variable that is assigned a value in function space and used in other functions, has default value of zero
ant_position="" #ants current position in grid used in more than one function
food_location="" #foods position in grid used in more than one function

#function for setting up a grid
def space():

print("please enter the size of the square grid you want")
#grid size is assigned by user input
global grid_size 
grid_size = int(sys.stdin.readline())
#embedded for loops to set up a list that stores grid points
for count in range(0,grid_size):
    x_number= count+1 #x number is the x number of the cell in the grid
    for count in range(0,grid_size):
        y_number= count+1 # y number is the y number of the cell in the grid
        grid_list.append([x_number,y_number,0,0,0]) #this stores a list within a list, so the first two values of the second list represent the co-ordinates of the cell the last three represent pheromone, food location, ant location (both food and ant location take value of one if either are in there)

def locations():

global food_location

food_location_x = (random.randint(1,grid_size)) #gives the x value of the cell the food is stored in
food_location_y = (random.randint(1,grid_size)) #gives the y value of the cell the food is stored in
food_location = [food_location_x,food_location_y,0,0,0] #gives the full location of the food and will use this to in other functions to reference the grid list
for count in range(0,4):
    grid_list[[grid_list.index([food_location])][count]] = [food_location_x,food_location_y,0,1,0]
print(grid_list[grid_list.index(food_location)])

此代码设置一个包含许多其他列表的列表（空间函数），然后更改列表中的一个列表（grid_list）以表示它包含食物。

然而，我得到一个值错误，而不是这个工作，说 [2,1,0,0,0] 不在列表中或代码给出的任何其他网格值（在这种情况下，2 和 1 可以用任何正整数替换）。有人可以向我解释一下如何更改它以便正确搜索吗？

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3 回答 3

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我可以看到一个错误：

grid_list.index([food_location])

您正在放入food_location列表中。But已经food_location是一个列表，所以现在您在列表中有一个列表。当你这样做时，它会告诉你在.grid_list.index(that)grid_list

老实说，看起来您正在使用一种非常时髦的方法来获取grid_list. 您应该分别使用random.choice或random.sample从您的集合中获取一个或多个随机元素。

于 2013-11-07T23:30:20.623 回答
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这段代码有两个错误。roippi 之前提到的第一个是您food_location在搜索之前放入一个列表。第二个是你做了四次同样的事情。您正在搜索[x,y,0,0,0]、替换它，然后在[x,y,0,0,0]列表中搜索相同的内容。由于您刚刚替换了该元素，因此它不再存在并且您会看到您看到的错误。

正如 roippi 所指出的，获得 4 个替换位置的更好方法是使用

random.sample(grid_list, 4)

并修改返回的四个列表中的每一个，使第四个元素为 1。random.sample 将返回与 grid_list 中相同的列表，因此修改这些列表的元素也会在 grid_list 中修改它们。

于 2013-11-07T23:36:34.397 回答
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您将二维数组表示为列表列表，并尝试在其中搜索项目。你根本不能grid_list.index()用于这样的搜索。中的项目grid_list是您网格的列，因此grid_list.index(item)将项目与整个列进行比较，因此找不到匹配项。

我建议定义一个函数来进行你想要的搜索。就像是：

def find_in_grid(grid, item_to_find):
    """search for a given item in a grid

    If found, the indexes (x, y) are returned. Otherwise, returns None.
    """
    for row in range(len(grid)):
        for column in range(len(grid[row])):
            if grid[row][column] == item_to_find:
                return (row, column)
    return None # item not found in the grid

当然，最好定义一个简单的Grid类并给它一个find()orindex()方法。

于 2013-11-07T23:41:56.780 回答