合并 2 字典并删除重复元素的键的重复和减去值

Question

我有一个 2 文本文件，我在其中存储字典值并读取它。 所以我正在读取 file1.txt 并将值存储在 counter 变量中，对于 file2.txt 我将值存储在 counter1 变量中。 当我打印 counter 和 counter1 时，它会给出以下输出。

counter= {'laptop': 5, 'chair':6 , 'tv': 3}

counter1= {'laptop': 2, 'chair': 4, 'tv': 4,'plant':2}

p=len(计数器) 对于 z 在 (0 ,p) 范围内：

        if (list(counter.values())[v]!=list(counter1.values())[v]):

            print("Objects value on main list %s and Second List %s  "%(list(counter.items())[v],list(counter1.items())[v]))
        else:
             pass
      v=v+1

我想退货：

[{'laptop': 3}, {'chair': 2}, {'tv': 1}, {'plant': 2}]

将 counter 和 counter1 的值相减并打印结果。

Answer 1

使用这个

counter= {'laptop': 5, 'chair':6 , 'tv': 3}
counter1= {'laptop': 2, 'chair': 4, 'tv': 4,'plant':2}
out = []
for key, val in counter.items():
    if key in counter1:
        out.append({key:abs(val-counter1[key])})
    else:
        out.append({key:val})
for key, val in counter1.items():
    if key not in counter:
        out.append({key:val})
print(out)

Answer 2

counter= {'laptop': 5, 'chair':6 , 'tv': 3}
counter1= {'laptop': 2, 'chair': 4, 'tv': 4,'plant':2}
out = [{key : abs(counter1[key] - counter[key])} if ((key in counter) and (key in counter1)) else {key : counter[key]} if (key in counter) else {key:counter1[key]} for key in set(counter.keys() + counter1.keys())]
print out

似乎可以在 1 行中完成，这应该可以。