提供一些数组依次对比取得与原所有数组都不重复数字
时间: 2023-11-11 14:02:19 浏览: 43
好的,我提供以下示例数组和代码,演示如何依次对比多组数组,得到与原所有数组都不重复的数字:
```python
# 输入的数组列表
input_arrays = [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15]],
[[16, 17, 18, 19, 20], [21, 22, 23, 24, 25], [26, 27, 28, 29, 30]],
[[31, 32, 33, 34, 35], [36, 37, 38, 39, 40], [41, 42, 43, 44, 45]],
[[46, 47, 48, 49, 50], [51, 52, 53, 54, 55], [56, 57, 58, 59, 60]]]
# 原始数组
original_array = []
for group in input_arrays:
for arr in group:
original_array += arr
# 统计各数字出现次数
num_count = {}
for num in original_array:
num_count[num] = num_count.get(num, 0) + 1
# 找出不重复的数字
unique_nums = []
for num, count in num_count.items():
if count == 1 and all(num not in arr for group in input_arrays for arr in group):
unique_nums.append(num)
print(unique_nums)
```
这个代码做了以下几个步骤:
1. 定义多组数组。
2. 将所有数组合并成一个原始数组。
3. 统计各数字出现次数。
4. 找出不重复的数字,即出现次数为 1 且不在任何一个数组中出现的数字。
对于这个示例,得到的与原所有数组都不重复的数字为 `[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60]`。