识别Ansible字典中值含重复元素的键值对并筛选输出

实现逻辑 #

• 先统计所有元素在全部值列表中的出现总次数
• 遍历原字典筛选键值对：只要当前值列表中存在至少一个元素出现次数大于1（即与其他列表有公共元素），就保留该键值对，否则移除

代码实现（使用Python标准库Counter，简洁易读） #

from collections import Counter

Ip = {'a':['one','two','three'],'b':['one','five','six'],'c':['seven','eight','nine'],'d':['five']}

# 统计所有元素的全局出现次数
all_elements = []
for val_list in Ip.values():
    all_elements.extend(val_list)
elem_counter = Counter(all_elements)

# 筛选符合要求的键值对
Op = {
    key: val_list
    for key, val_list in Ip.items()
    if any(elem_counter[elem] > 1 for elem in val_list)
}

print(Op)

运行输出：

{'a': ['one', 'two', 'three'], 'b': ['one', 'five', 'six'], 'd': ['five']}

无导入依赖版本 #

如果不想导入任何库，也可以手动实现计数逻辑，效果完全一致：

Ip = {'a':['one','two','three'],'b':['one','five','six'],'c':['seven','eight','nine'],'d':['five']}

# 手动统计元素出现次数
elem_count = {}
for val_list in Ip.values():
    for elem in val_list:
        elem_count[elem] = elem_count.get(elem, 0) + 1

# 筛选键值对
Op = {k: v for k, v in Ip.items() if any(elem_count[e] > 1 for e in v)}

内容的提问来源于stack exchange，提问作者Ramarao Kedarasetti