比较python中的多维数组

我有一个多维数组，叫做

old_arr

，它是这样的

[[8,8,8,0,0,0,0,6,6,5,5]，…]

然后我有一个更新的多维数组

new_arr

像这样的

[[9,9,6,7,3,6,5,0,6,3,4]，…]

我想做的是更新

new\u arr

，这样如果其中的值对应于

old\u arr

中的

0

，那么该值应该是

0

，否则保留新值。因此，在上述示例中，

new_arr

看起来像这样的

[[9,9,6,7,0,0,0,0,6,4,3,4]，…]

，其中3,6,5替换为0。有什么建议吗

我还想知道，如果周围8个相邻单元中有4个单元的值也为0，是否可以将单元更新为0？与new_arr和old_arr一样，new_arr和old_arr都是多维数组（列表），表示行和列，因此它们就像下图所示的一个大表，如果old_arr中的对应单元格为0，并且其相邻单元格中的4个为0（照片中的白色单元格），则new_arr中的蓝色单元格将仅更新为零

因此，我需要检查所有相邻的8个单元（有时是6个或7个），取决于它们在中间（8）或边缘（7）或角（6）中的单元位置，如果它们为零或不计数，则如果计数为4或更多，则将单元格值设置为0。 所以如果老阿瑞是

[[8,8,8,8,0,0,0,0,6,6,5,5],
 [8,8,8,8,0,x,0,0,6,6,5,5],
 [8,8,8,8,0,0,0,0,6,6,5,5],
 [8,8,8,8,0,0,0,0,6,6,5,5],....]

Where x is a zero

而新奥尔是

[[9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],....]

[[9,9,6,7,3,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],....]

对于高亮显示的单元，新的_arr中的对应单元将为零，因为旧的_arr中高亮显示的单元为0，并且其4个以上的相邻单元也为零

更新后的新arr为

[[9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],....]

[[9,9,6,7,3,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],....]

---

假设

old_arr

和

new_arr

的长度相同，您可以执行以下操作：

old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5], [8,8,7,0,0,0,0,0,6,6,5,5]]
new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4], [9,0,6,7,4,6,5,0,6,4,3,4]]

new_arr = [[x if old[i] else 0 for i, x in enumerate(new)] for old, new in zip(old_arr, new_arr)]

print(new_arr)

first=[[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5]]
second = [[9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4]]

for first_1,second_1 in zip(first,second):
    for index,value in enumerate(first_1):
        if value==0:
            try:
                if first_1[index-1]==0 and first_1[index+1]==0:
                    second_1[index]=0
                    second_1[index-1]=0
                    second_1[index+1]=0
            except IndexError:
                pass
print(second)

哪些产出：

[[9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], [9, 0, 6, 0, 0, 0, 0, 0, 6, 4, 3, 4]]

更新：

以下是一个暴力解决方案，用于处理相邻单元：

old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5]]

new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4]]

def first_last(row, next_row, old, new):
    for i in range(len(new[row])):
        count = 0
        if old[row][i] == 0:
            if old[row][i-1] == 0:
                count += 1
            if old[row][i+1] == 0:
                count += 1
            if old[next_row][i] == 0:
                count += 1
            if old[next_row][i-1] == 0:
                count += 1
            if old[next_row][i+1] == 0:
                count += 1  

        if count > 4:
            new[row][i] = 0

def middle(old, new):
    for i, l in enumerate(new[1:-1]):
        for j in range(len(l)):
            count = 0
            if old[i][j] == 0:
                if old[i][j-1] == 0:
                    count += 1
                if old[i][j+1] == 0:
                    count += 1
                if old[i-1][j] == 0:
                    count += 1
                if old[i-1][j-1] == 0:
                    count += 1
                if old[i-1][j+1] == 0:
                    count += 1
                if old[i+1][j] == 0:
                    count += 1
                if old[i+1][j-1] == 0:
                    count += 1
                if old[i+1][j+1] == 0:
                    count += 1

            if count > 4:
                l[j] = 0

# first row
first_last(0, 1, old_arr, new_arr)

# middle rows
middle(old_arr, new_arr)

# last row
first_last(-1, -2, old_arr, new_arr)

print(new_arr)

哪些产出：

[[9, 9, 6, 7, 3, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 3, 0, 0, 0, 6, 4, 3, 4]]

---

注意：这可以做得更好，但您可以根据自己的喜好进行优化

假设

old_arr

和

new_arr

的长度相同，您可以执行以下操作：

old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5], [8,8,7,0,0,0,0,0,6,6,5,5]]
new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4], [9,0,6,7,4,6,5,0,6,4,3,4]]

new_arr = [[x if old[i] else 0 for i, x in enumerate(new)] for old, new in zip(old_arr, new_arr)]

print(new_arr)

first=[[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5]]
second = [[9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4]]

for first_1,second_1 in zip(first,second):
    for index,value in enumerate(first_1):
        if value==0:
            try:
                if first_1[index-1]==0 and first_1[index+1]==0:
                    second_1[index]=0
                    second_1[index-1]=0
                    second_1[index+1]=0
            except IndexError:
                pass
print(second)

哪些产出：

[[9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], [9, 0, 6, 0, 0, 0, 0, 0, 6, 4, 3, 4]]

更新：

以下是一个暴力解决方案，用于处理相邻单元：

old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5]]

new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4]]

def first_last(row, next_row, old, new):
    for i in range(len(new[row])):
        count = 0
        if old[row][i] == 0:
            if old[row][i-1] == 0:
                count += 1
            if old[row][i+1] == 0:
                count += 1
            if old[next_row][i] == 0:
                count += 1
            if old[next_row][i-1] == 0:
                count += 1
            if old[next_row][i+1] == 0:
                count += 1  

        if count > 4:
            new[row][i] = 0

def middle(old, new):
    for i, l in enumerate(new[1:-1]):
        for j in range(len(l)):
            count = 0
            if old[i][j] == 0:
                if old[i][j-1] == 0:
                    count += 1
                if old[i][j+1] == 0:
                    count += 1
                if old[i-1][j] == 0:
                    count += 1
                if old[i-1][j-1] == 0:
                    count += 1
                if old[i-1][j+1] == 0:
                    count += 1
                if old[i+1][j] == 0:
                    count += 1
                if old[i+1][j-1] == 0:
                    count += 1
                if old[i+1][j+1] == 0:
                    count += 1

            if count > 4:
                l[j] = 0

# first row
first_last(0, 1, old_arr, new_arr)

# middle rows
middle(old_arr, new_arr)

# last row
first_last(-1, -2, old_arr, new_arr)

print(new_arr)

哪些产出：

[[9, 9, 6, 7, 3, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 3, 0, 0, 0, 6, 4, 3, 4]]

---

注意：这可以做得更好，但您可以根据自己的喜好进行优化

具有列表理解功能的简单解决方案：

>>> old_arr=[9,9,6,7,3,6,5,0,6,4,3,4]
>>> new_arr=[8,8,8,8,0,0,0,0,6,6,5,5]
>>> [new_arr[i] if old_arr[i] else 0 for i in range(len(new_arr))]
[9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4]

---

具有列表理解功能的简单解决方案：

>>> old_arr=[9,9,6,7,3,6,5,0,6,4,3,4]
>>> new_arr=[8,8,8,8,0,0,0,0,6,6,5,5]
>>> [new_arr[i] if old_arr[i] else 0 for i in range(len(new_arr))]
[9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4]

---

列表在Python中是可修改的序列，因此您可以就地更改它们，这对于大型数据集是有意义的。只需使用两个嵌套循环即可完成此操作：

for i, l in enumerate(new_arr):
    for j in range(len(l)):
        if old_arr[i][j] == 0:
            l[j] = 0

---

列表在Python中是可修改的序列，因此您可以就地更改它们，这对于大型数据集是有意义的。只需使用两个嵌套循环即可完成此操作：

for i, l in enumerate(new_arr):
    for j in range(len(l)):
        if old_arr[i][j] == 0:
            l[j] = 0

例如：
old_arr=[[8,8,8,0,0,0,6,6,5,5]，[1,2,3,4,5,6,7,8]]
新的arr=[[9,9,6,7,3,6,5,0,6,4,3,4]，[8,3,4,5,6,78,8,9]]
自定义比较（旧的、新的）
最后新的arr=[[9,9,6,7,0,0,0,0,6,4,3,4]，[8,3,4,5,6,78,8,9]

例如：
old_arr=[[8,8,8,0,0,0,6,6,5,5]，[1,2,3,4,5,6,7,8]]
新的arr=[[9,9,6,7,3,6,5,0,6,4,3,4]，[8,3,4,5,6,78,8,9]]
自定义比较（旧的、新的）
最后，new_arr=[[9,9,6,7,0,0,0,0,6,4,3,4]，[8,3,4,5,6,78,8,9]

一个简单的方法是这样的：

old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5], [8,8,7,0,0,0,0,0,6,6,5,5]]
new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4], [9,0,6,7,4,6,5,0,6,4,3,4]]

new_arr = [[x if old[i] else 0 for i, x in enumerate(new)] for old, new in zip(old_arr, new_arr)]

print(new_arr)

first=[[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5]]
second = [[9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4]]

for first_1,second_1 in zip(first,second):
    for index,value in enumerate(first_1):
        if value==0:
            try:
                if first_1[index-1]==0 and first_1[index+1]==0:
                    second_1[index]=0
                    second_1[index-1]=0
                    second_1[index+1]=0
            except IndexError:
                pass
print(second)

输出：

[[9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]]

---

一个简单的方法如下：

old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5], [8,8,7,0,0,0,0,0,6,6,5,5]]
new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4], [9,0,6,7,4,6,5,0,6,4,3,4]]

new_arr = [[x if old[i] else 0 for i, x in enumerate(new)] for old, new in zip(old_arr, new_arr)]

print(new_arr)

first=[[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5]]
second = [[9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4]]

for first_1,second_1 in zip(first,second):
    for index,value in enumerate(first_1):
        if value==0:
            try:
                if first_1[index-1]==0 and first_1[index+1]==0:
                    second_1[index]=0
                    second_1[index-1]=0
                    second_1[index+1]=0
            except IndexError:
                pass
print(second)

输出：

[[9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]]

---

---

它们在python中没有数组列表吗？数组列表仅在java中使用。您所说的“数组列表”是指“列表”吗？还是“名单”？因为你不是指“数组列表”。@khelwood我指的是多维数组，如示例所示，你是指numpy数组吗？或列表列表？对于突出显示的单元格

[8,8,8,8,0，**0**，0,0,6,6,5,5]

，只有相邻单元格的

3

是

0

？除非我看得不对。它们在python中没有数组列表？数组列表仅在java中使用。您所说的“数组列表”是指“列表”吗？还是“名单”？因为你不是指“数组列表”。@khelwood我指的是多维数组，如示例所示，你是指numpy数组吗？或列表列表？对于突出显示的单元格

[8,8,8,8,0，**0**，0,0,6,6,5,5]

，只有相邻单元格的

3

是

0

？除非我看得不对。给定一个类型Error range（）预期的整数结束参数，得到list@SergeBallesta但是如何将更新保存回new_arr？@Tak：更新直接在

new_arr

中完成。您的问题是什么？提供了一个类型错误范围（）整数结束参数list@SergeBallesta但是如何将更新保存回new_arr？@Tak：更新直接在

new_arr

中完成。你的问题是什么？只有当它周围的4个相邻单元的值也为0时，才可能将其更新为0吗？它们是表示行和列的多维数组（列表），所以它们就像一个大表。因此，在这张图中，只有当旧_arr中的对应单元格为0，并且它的4个相邻单元格为0（照片中的白色单元格）时，新_arr中的蓝色单元格才会更新为零。Tak完全可能，我会更新我的答案。你能详细描述一下“四个相邻的细胞”吗？零的两边是2吗？还是零的两边各有4个？我不明白这一点。我检查所有8个相邻单元（有时6个或7个，取决于单元位置）是否为零，并对它们进行计数，如果计数为4个或更多，则将单元值设置为0@Tak对不起，我有点忙，忘了回复。您是否可以发布此行为的示例输入和预期输出？这将使它更容易。问题更新，请让我知道，如果我需要澄清任何更多。谢谢，只有当它周围的4个相邻单元格的值也为0时，才可能将其更新为0吗？它们是表示行和列的多维数组（列表），所以它们就像一个大表。所以