EX.Deleting a list item while iterating

暮晨 · 暮晨 · commit 61d727779e01 · 2018-12-05T23:26:15.000+08:00
diff --git a/README.md b/README.md
@@ -57,7 +57,7 @@ So, here we go...
     - [Section: Watch out for the landmines!](#section-watch-out-for-the-landmines)
         - [> Modifying a dictionary while iterating over it/迭代字典时的修改](#-modifying-a-dictionary-while-iterating-over-it迭代字典时的修改)
         - [> Stubborn `del` operator/坚强的 `del` *](#-stubborn-del-operator坚强的-del-)
-        - [> Deleting a list item while iterating](#-deleting-a-list-item-while-iterating)
+        - [> Deleting a list item while iterating/迭代列表时删除元素](#-deleting-a-list-item-while-iterating迭代列表时删除元素)
         - [> Loop variables leaking out!](#-loop-variables-leaking-out)
         - [> Beware of default mutable arguments!](#-beware-of-default-mutable-arguments)
         - [> Catching the Exceptions](#-catching-the-exceptions)
@@ -1420,7 +1420,7 @@ Deleted!
 
 ---
 
-### > Deleting a list item while iterating
+### > Deleting a list item while iterating/迭代列表时删除元素
 
 ```py
 list_1 = [1, 2, 3, 4]
@@ -1453,30 +1453,30 @@ for idx, item in enumerate(list_4):
 [2, 4]
 ```
 
-Can you guess why the output is `[2, 4]`?
+你能猜到为什么输出是 `[2, 4]` 吗?
 
-#### 💡 Explanation:
+#### 💡 说明:
 
-* It's never a good idea to change the object you're iterating over. The correct way to do so is to iterate over a copy of the object instead, and `list_3[:]` does just that.
+* 在迭代时修改对象是一个很愚蠢的主意. 正确的做法是迭代对象的副本, `list_3[:]` 就是这么做的.
 
      ```py
      >>> some_list = [1, 2, 3, 4]
      >>> id(some_list)
      139798789457608
-     >>> id(some_list[:]) # Notice that python creates new object for sliced list.
+     >>> id(some_list[:]) # 注意python为切片列表创建了新对象.
      139798779601192
      ```
 
-**Difference between `del`, `remove`, and `pop`:**
-* `del var_name` just removes the binding of the `var_name` from the local or global namespace (That's why the `list_1` is unaffected).
-* `remove` removes the first matching value, not a specific index, raises `ValueError` if the value is not found.
-* `pop` removes the element at a specific index and returns it, raises `IndexError` if an invalid index is specified.
+**`del`, `remove` 和 `pop` 的不同:**
+* `del var_name` 只是从本地或全局命名空间中删除了 `var_name` (这就是为什么 `list_1` 没有受到影响).
+* `remove` 会删除第一个匹配到的指定值, 而不是特定的索引, 如果找不到值则抛出 `ValueError` 异常.
+* `pop` 则会删除指定索引处的元素并返回它, 如果指定了无效的索引则抛出 `IndexError` 异常.
 
-**Why the output is `[2, 4]`?**
-- The list iteration is done index by index, and when we remove `1` from `list_2` or `list_4`, the contents of the lists are now `[2, 3, 4]`. The remaining elements are shifted down, i.e., `2` is at index 0, and `3` is at index 1. Since the next iteration is going to look at index 1 (which is the `3`), the `2` gets skipped entirely. A similar thing will happen with every alternate element in the list sequence.
+**为什么输出是 `[2, 4]`?**
+- 列表迭代是按索引进行的, 所以当我们从 `list_2` 或 `list_4` 中删除 `1` 时, 列表的内容就变成了 `[2, 3, 4]`. 剩余元素会依次位移, 也就是说, `2` 的索引会变为 0, `3` 会变为 1. 由于下一次迭代将获取索引为 1 的元素 (即 `3`), 因此 `2` 将被彻底的跳过. 类似的情况会交替发生在列表中的每个元素上.
 
-* Refer to this StackOverflow [thread](https://stackoverflow.com/questions/45946228/what-happens-when-you-try-to-delete-a-list-element-while-iterating-over-it) explaining the example
-* See also this nice StackOverflow [thread](https://stackoverflow.com/questions/45877614/how-to-change-all-the-dictionary-keys-in-a-for-loop-with-d-items) for a similar example related to dictionaries in Python.
+* 参考这个StackOverflow的[回答](https://stackoverflow.com/questions/45946228/what-happens-when-you-try-to-delete-a-list-element-while-iterating-over-it)来解释这个例子
+* 关于Python中字典的类似例子, 可以参考这个Stackoverflow的[回答](https://stackoverflow.com/questions/45877614/how-to-change-all-the-dictionary-keys-in-a-for-loop-with-d-items).
 
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