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[Introduction]: In Python, if we want to modify a set of data in the process of traversing a set of data, there are usually many problems. For example, when performing the above operation on a list, some data will be ignored; when traversing a dictionary , the data cannot be modified. This paper proposes a variety of solutions to these problems.
Introduction
1. About the list
1. Problem description
In Python, if you try to modify a set of data while iterating over it, this is usually fine. E.g:
l = [3, 4, 56, 7, 10, 9, 6, 5]
for i in l:
if not i % 2 == 0:
continue
l.remove(i)
print(l)
复制代码The above code traverses a list of numbers, and directly modifies the list l in order to remove all even numbers. However, the output after running is:
[3, 56, 7, 9, 5]
复制代码Wait a moment! The output doesn't seem right. The final result still contains an even number 56. Why is this number not successfully removed? We can try to print out all the elements traversed by the for loop, running the following code:
l = [3, 4, 56, 7, 10, 9, 6, 5]
for i in l:
print(i)
if not i % 2 == 0:
continue
l.remove(i)
print(l)
复制代码The output of this code is:
3
4
7
10
6
[3, 56, 7, 9, 5]
复制代码As you can see from the output, the for loop doesn't seem to be accessing all the elements in the list. To understand what the for loop does internally, we can simulate it using iter and next. Take a look at the example below, where I use the ipython shell to run the code:
In [1]: l = [3, 4, 56, 7, 10, 9, 6, 5]
In [2]: # 把列表变成一个迭代器
In [3]: it = iter(l)
In [4]: # 使用 next() 方法来模拟 for循环
In [5]: next(it)
Out[5]: 3
In [6]: next(it)
Out[6]: 4
In [7]: # 移除一个迭代器已经访问过的元素
In [8]: l.remove(3)
In [9]: next(it)
Out[9]: 7
In [10]: # 注意此处跳过了56，我们可以再移除一个元素
In [11]: l.remove(4)
In [12]: next(it)
Out[12]: 9
复制代码The above experiment reveals that when you remove an element that an iterator has already visited, on the next iteration, you skip an element on the right and go directly to the next one.
The opposite is still true, that is, after starting the iteration, if you add an element to the beginning of the list, the next iteration may access the element that has already been iterated. This happens in the following code:
In[1]: l = [3, 4, 56, 7, 10, 9, 6, 5]
In[2]: it = iter(l)
In[3]: next(it)
Out[3]: 3
In[4]: next(it)
Out[4]: 4
In[5]: l.insert(0, 44)
In[6]: next(it)
Out[6]: 4
复制代码Note that when 44 is added to the head of the list, 4 is accessed twice.
2. Solutions
In order to solve the above problem, we must ensure that the elements visited by the iterator cannot be removed.
Option One
We can first flip the original list to get a new list, then iterate over the new list and remove the elements that do not meet the conditions in the original list l. The program code is as follows:
l = [3, 4, 56, 7, 10, 9, 6, 5]
# 迭代翻转后的列表
for i in reversed(l):
print(i)
if not i % 2 == 0:
continue
l.remove(i)
print(l)
复制代码The result is as follows:
5
6
9
10
7
56
4
3
[3, 7, 9, 5]
复制代码Note that the iterator now successfully accesses all the elements in the list, and finally outputs a list containing only odd numbers.
Option II
We can also copy the list l before starting the iteration. But when there is too much data in the list l, doing so is obviously more performance-intensive. The program code is as follows:
l = [3, 4, 56, 7, 10, 9, 6, 5]
# 在这里使用 'l.copy()' 来对列表 l 进行浅拷贝
for i in l.copy():
print(i)
if not i % 2 == 0:
continue
l.remove(i)
print(l)
复制代码The output is as follows:
3
4
56
7
10
9
6
5
[3, 7, 9, 5]
复制代码This scheme ensures that the order of iteration is the same as the order in which elements are removed. But since the two operations iterate and remove are on two different lists, it doesn't matter that the order is the same.
2. About the dictionary
1. Problem description
The dictionary cannot be modified while iterating over the dictionary. as follows:
# {0: 0, 1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9}
d = {k: k for k in range(10)}
for k, v in d.items():
if not v % 2 == 0:
continue
d.pop(k)
复制代码This code produces a RuntimeError:
Traceback (most recent call last):
File "F:/Documents/pythonprojects/01practice/app.py", line 7, in <module>
for k, v in d.items():
RuntimeError: dictionary changed size during iteration
复制代码2. Solutions
We can copy all the keys of the dictionary first, and then in the process of iterating the keys, remove the elements that do not meet the conditions. The process is as follows:
# {0: 0, 1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9}
d = {k: k for k in range(10)}
# 这里复制了字典中的所有key值
# 没有复制整个字典
# 同时使用tuple()速度更快
for k in tuple(d.keys()):
if not d[k] % 2 == 0:
continue
d.pop(k)
print(d)
复制代码The output after running the code is as follows:
{1: 1, 3: 3, 5: 5, 7: 7, 9: 9}
复制代码We successfully removed all even key-value pairs from the dictionary!
in conclusion
In this paper, we propose different solutions for the problem that cannot be modified when iterating over a set of data: if we want to modify the list when traversing the list, we can first flip or copy the original list to get A new list, and then in the process of traversing the new list, modify the data in the original list; if we want to modify the dictionary while traversing the dictionary, we can copy all the key values of the dictionary first, and then iterate the key values. , modify the data in the dictionary.
Thanks
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