tinylambda
diff --git a/‎cookbook/c04/p11_iterate_simultaneous.py‎
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diff --git a/‎source/c04/p11_iterate_over_multiple_sequences_simultaneously.rst‎
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@@ -0,0 +1,35 @@
+#!/usr/bin/env python
+# -*- encoding: utf-8 -*-
+"""
+Topic: 同时迭代多个序列
+Desc : 
+"""
+from itertools import zip_longest
+
+
+def iterate_simul():
+    xpts = [1, 5, 4, 2, 10, 7]
+    ypts = [101, 78, 37, 15, 62, 99]
+    for x, y in zip(xpts, ypts):
+        print(x, y)
+
+    a = [1, 2, 3]
+    b = ['w', 'x', 'y', 'z']
+    for i in zip(a,b):
+        print(i)  # 默认是按最短长度
+    for i in zip_longest(a,b):
+        print(i)
+    for i in zip_longest(a, b, fillvalue=0):
+        print(i)
+
+    headers = ['name', 'shares', 'price']
+    values = ['ACME', 100, 490.1]
+    s = dict(zip(headers,values))
+
+    for name, val in zip(headers, values):
+        print(name, '=', val)
+
+
+if __name__ == '__main__':
+    iterate_simul()
+
@@ -5,14 +5,117 @@
 ----------
 问题
 ----------
-todo...
+你想同时迭代多个序列，每次分别从一个序列中同时选取取某些元素。
+
+|
 
 ----------
 解决方案
 ----------
-todo...
+为了同时迭代多个序列，使用zip()函数。比如：
+
+.. code-block:: python
+
+    >>> xpts = [1, 5, 4, 2, 10, 7]
+    >>> ypts = [101, 78, 37, 15, 62, 99]
+    >>> for x, y in zip(xpts, ypts):
+    ...     print(x,y)
+    ...
+    1 101
+    5 78
+    4 37
+    2 15
+    10 62
+    7 99
+    >>>
+
+zip(a, b)会生成一个可返回元组(x, y)的迭代器，其中x来自a，y来自b。
+一旦其中某个序列到底结尾，迭代宣告结束。
+因此迭代长度跟参数中最短序列长度一致。
+
+.. code-block:: python
+
+    >>> a = [1, 2, 3]
+    >>> b = ['w', 'x', 'y', 'z']
+    >>> for i in zip(a,b):
+    ...     print(i)
+    ...
+    (1, 'w')
+    (2, 'x')
+    (3, 'y')
+    >>>
+
+如果这个不是你想要的效果，那么还可以使用itertools.zip_longest()函数来代替。比如：
+
+.. code-block:: python
+
+    >>> from itertools import zip_longest
+    >>> for i in zip_longest(a,b):
+    ...     print(i)
+    ...
+    (1, 'w')
+    (2, 'x')
+    (3, 'y')
+    (None, 'z')
+
+    >>> for i in zip_longest(a, b, fillvalue=0):
+    ...     print(i)
+    ...
+    (1, 'w')
+    (2, 'x')
+    (3, 'y')
+    (0, 'z')
+    >>>
+
+|
 
 ----------
 讨论
 ----------
-todo...
+当你想成对处理数据的时候zip()函数是很有用的。
+比如，假设你头列表和一个值列表，就像下面这样：
+
+.. code-block:: python
+
+    headers = ['name', 'shares', 'price']
+    values = ['ACME', 100, 490.1]
+
+使用zip()可以让你将它们打包并生成一个字典：
+
+.. code-block:: python
+
+    s = dict(zip(headers,values))
+
+或者你也可以像下面这样产生输出：
+
+.. code-block:: python
+
+    for name, val in zip(headers, values):
+        print(name, '=', val)
+
+虽然不常见，但是zip()可以被传入多于两个的序列。
+这时候所生成的结果元组中元素个数跟输入序列个数一样。比如;
+
+.. code-block:: python
+
+    >>> a = [1, 2, 3]
+    >>> b = [10, 11, 12]
+    >>> c = ['x','y','z']
+    >>> for i in zip(a, b, c):
+    ...     print(i)
+    ...
+    (1, 10, 'x')
+    (2, 11, 'y')
+    (3, 12, 'z')
+    >>>
+
+最后强调一点就是，zip()会创建一个迭代器来作为结果返回。
+如果你需要将结对的值存储在列表中，要使用list()函数。比如：
+
+.. code-block:: python
+
+    >>> zip(a, b)
+    <zip object at 0x1007001b8>
+    >>> list(zip(a, b))
+    [(1, 10), (2, 11), (3, 12)]
+    >>>