<p>抱歉,我没有注意到海报只想删除特定属性。以下是正确的解决方案:</p>
<pre><code>tt=[{k:v for k,v in i.items() if k != 'property3' or v != 'value6'} for i in tt]
</code></pre>
<p>LazyScripter的解决方案将删除键或值或两者匹配的字典元素。如果我理解最初的问题,我认为这不是我想要的。你知道吗</p>
<p>我还比较了两种方案的效率,其中一种方案的速度似乎快了12倍。。。你知道吗</p>
<pre><code>python -m cProfile test.py
</code></pre>
<p>输出:</p>
<pre><code>1780059 function calls (1500059 primitive calls) in 2.759 seconds
ncalls tottime percall cumtime percall filename:lineno(function)
...
1 0.026 0.026 2.759 2.759 test.py:1(<module>)
1 0.036 0.036 0.917 0.917 test.py:19(search)
1 0.035 0.035 0.073 0.073 test.py:33(search2)
...
</code></pre>
<p>对于以下代码:</p>
<pre><code>import copy
tt = [
{
'property1': 'value1',
'property2': 'value2',
'property3': 'value3'
},
{
'property1': 'value4',
'property2': 'value5',
'property3': 'value6'
}
]
d=[]
[d.extend(copy.deepcopy(tt)) for _ in range(10000)]
def search(dict, key=None, val=None):
c_dict = copy.deepcopy(dict)
for i, n in enumerate(dict):
for k, v in n.iteritems():
if key and val and k == key and val == v:
c_dict[i].pop(k)
# elif key and k == key:
# c_dict[i].pop(k)
# elif v and val == v:
# c_dict[i].pop(k)
else:
continue
return c_dict
def search2(d,a,b):
return [{k:v for k,v in i.items() if k != a or v != b} for i in d]
search (d, 'property3', 'value6')
search2(d, 'property3', 'value6')
</code></pre>