我想首先删除重复项来组合两个嵌套列表。
list1 = [[1,2], [1,3], [3,5], [4,1], [9,6]]
list2 = [[1,2], [1,3], [3,5], [6,6], [0,2], [1,7], [7,7]]
results = [[1,2], [1,3], [3,5], [4,1], [9,6], [6,6], [0,2], [1,7], [7,7]]
我的代码：
not_in_list1 = set(list2) - set(list1)
results = list(list1) + list(not_in_list1)
错误：
TypeError: unhashable type: 'list'
是否因为set操作无法在嵌套列表中使用？
由于
答案 0 :(得分：2)
是否因为嵌套列表中不能使用set操作？
是的，这是因为列表是可变的，因此列表可以在创建后更改，这意味着集合中使用的哈希值可以更改。
但是元组是不可变的，所以你可以在集合中使用它们：
<div class="container-fluid" >
<!-- style="height: 90%; width: 60%; float:left;" height="100%" width="49%" align="left" -->
<div class="youtube-video" id="video">
<!-- for live video -->
<iframe src="https://www.facebook.com/plugins/video.php?href=https%3A%2F%2Fwww.facebook.com%2F1277805348996425%2Fvideos%2F<?php echo $liveID; ?>%2F&show_text=0&width=476" width="476" height="476" style="border:none;overflow:hidden" scrolling="no" frameborder="0" allowTransparency="true" allowFullScreen="true"></iframe>
<!-- <iframe class="embed-responsive-item" src="https://www.facebook.com/plugins/video.php?href=https%3A%2F%2Fwww.facebook.com%2F<?php echo $liveID; ?>%2Fvideos%2F1277978488979111%2F&show_text=1&width=560" width="560" height="475" style="border:none;overflow:hidden" scrolling="no" frameborder="0" allowTransparency="true" allowFullScreen="true"></iframe> -->
<!-- <iframe src="https://www.facebook.com/plugins/video.php?href=https%3A%2F%2Fwww.facebook.com%2F1277805348996425%2Fvideos%2F1278782988898661%2F&show_text=1&width=560" width="560" height="475" style="border:none;overflow:hidden" scrolling="no" frameborder="0" allowTransparency="true" allowFullScreen="true"></iframe> -->
<!-- for embbedded facebook video (test purposes) -->
<!-- <iframe src="https://www.facebook.com/plugins/video.php?href=https%3A%2F%2Fwww.facebook.com%2FDota2BestYolo%2Fvideos%2F<?php echo $liveID; ?>%2F&show_text=1&width=560" width="560" height="451" style="border:none;overflow:hidden" scrolling="no" frameborder="0" allowTransparency="true" allowFullScreen="true"></iframe> -->
<!-- youtube embed video -->
<!-- <iframe class="embed-responsive-item" src="https://www.youtube.com/embed/live_stream?channel=UCJaiEVEFaen5QC28rJp0fEw"></iframe> -->
</div>
<div class="chat row" >
<div id="messages" class="chat-area"></div>
<?php
if (loggedin()) { ?>
<table>
<tr>
<td>
<textarea style="padding: 10px;" rows="3" cols="50" class="entry row" placeholder="Type your message here..." name="msg" id="txtBox"></textarea>
</td>
</tr>
</table>
<?php } else { ?>
<table>
<tr>
<td style="width: 400px;">
<textarea style="" id="message" rows="3" cols="50" class="entry row" placeholder="Type your message here..." name="msg"></textarea>
<!-- Modal -->
<div class="modal fade" id="loginModal" tabindex="-1" role="dialog" aria-labelledby="exampleModalLabel" aria-hidden="true">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<h5 class="modal-title" id="exampleModalLabel">Please Login</h5>
<button type="button" class="close" data-dismiss="modal" aria-label="Close">
<span aria-hidden="true">×</span>
</button>
</div>
<div class="modal-body">
<p>You must first login before you can join the conversation.</p>
</div>
<div class="modal-footer">
<input type="button" class="btn btn-primary" value="Login" onclick="location.href='login.html'"/>
<button type="button" class="btn btn-danger" data-dismiss="modal">Cancel</button>
</div>
</div>
</div>
</div>
</td>
</tr>
</table>
<?php } ?>
</div>
</div><!-- end of container -->
将它们转换为元组：
list1 = [[1,2], [1,3], [3,5], [4,1], [9,6]]
list2 = [[1,2], [1,3], [3,5], [6,6], [0,2], [1,7], [7,7]]
tuple1 = [tuple(l) for l in list1]
tuple2 = [tuple(l) for l in list2]
not_in_tuples = set(tuple2) - set(tuple1)
的结果：
not_in_tuples并在 {(0, 2), (1, 7), (6, 6), (7, 7)}
中将它们组合回您想要的内容：
results产生：
results = list1 + list(map(list, not_in_tuples))
编辑
如果有兴趣在将两个列表添加到一起后保留两个列表的顺序：
[[1, 2], [1, 3], [3, 5], [4, 1], [9, 6], [0, 2], [1, 7], [7, 7], [6, 6]]
产生：
list1 = [[1,2], [1,3], [3,5], [4,1], [9,6]]
list2 = [[1,2], [1,3], [3,5], [6,6], [0,2], [1,7], [7,7]]
intersection = set(map(tuple, list1)).intersection(set(map(tuple, list2)))
result = list1 + [list(t) for t in map(tuple, list2) if t not in intersection]
答案 1 :(得分：1)
您需要将子列表[mutable]转换为元组[immutable]并获取集合
async componentWillMount() {
let products = await this.props
.searchLocations(
this.props.navigation.state.params.product_name,
this.props.navigation.state.params.searchRadius
)
.then(
() =>
Promise.all(this.props.locations.map(async(location) =>
({[location.id]: await this.searchProducts(
this.props.navigation.state.params.product_name,
location.storeid
)}))
)
);
console.log(products);
}
输出：
set([tuple(i) for i in list1+list2])
答案 2 :(得分：1)
另一种方法是：
>>> list1 = [[1,2], [1,3], [3,5], [4,1], [9,6]]
>>> list2 = [[1,2], [1,3], [3,5], [6,6], [0,2], [1,7], [7,7]]
>>> k = list1+list2 #We combine both the lists
>>> z = [] #Declare an empty list
>>>for i in k: #Loop through every element of the combined list
if i in z: #If the element is already in the final list
pass #Do nothing
else: #If the element in the combined list is not not there in the final list
z.append(i) #Append that element to the final list
>>>print z
>>>[[1, 2], [1, 3], [3, 5], [4, 1], [9, 6], [6, 6], [0, 2], [1, 7], [7, 7]]