S [,] forxins: ifx0s. remove (x) Why is the result [, 3] instead? S = [0, 0, 5, 3]
For x in s:
If x = 0
S. remove (x)
Why is the result [, 3] instead? Reply content: initial s [0] = 0, s [1] = 0, s [2] = 5, s [3] = 3
When the for command is executed for the first time,
X = s [0], that is, x = 0 and then remove this element,
S is changed to [0, 5, 3], that is, s [0] = 0, s [1] = 5, s [2] = 3
For the second execution,
X = s [1], that is, x = 5
Because x is not 0, s remains unchanged.
The reason for this result is that list s has changed during the operation.
A reasonable way is to place elements not 0 in a new list.
print [x for x in s if x != 0]
During the loop, try not to add or delete the list. Do not change the array in the loop. I do not want to say anything important three times. Locke is right. It's not easy. Let's share my opinion.
In for x in s: s is the list of your iterations, and in iteration, you use s. remove (x), this line of code modifies the original list s, so that the second execution of "for x in s:", Here s and the first execution
Not the same listThe first time s = [0, 0, 5, 3], the second time s = [0, 5, 3], but the subscript they use is the same group of subscript, that is, if the first subscript is 0 and the second subscript is 1, then the list s corresponding to the two times shows that the first s [0] is 0, the second s [1] is 2 (corresponding to the second s)
Therefore, this easy mistake is that the list of iterations has changed, so in python iteration,
Do not modify the list(Here is a mistake you will make in the future --
List variability and multi-Reference, Which will change the original list when modifying other lists)
Then, except for Locke's
List push(You must know the best, most distinctive, and most X-Day bombing skills in python). Other solutions:
1.) create a new list result = [] and save the result. append (x) value that meets the conditions. In this way, the original list is not modified.
2 .) use a = s, iterate a, for x in a:, and then use s. remove (x) modify the list s. Congratulations, you are wrong, that is, the multi-reference problem mentioned above. Please refer to this document to understand it.
If you do not understand it, share a sentence"
The book has been read for hundreds of times."@ Locke's explanation is correct, but the method can be improved.
This is more suitable for the Python filter:
#!/usr/bin/env python3def check_number(x): if x == 0: return False else: return Trues = [0,0,5,3]print(filter(check_number,s))
s = [0,0,5,3]print(filter(lambda x: x != 0, s))
S = [I for I in s if I! = 0] # cell phone code word, take a look
There are good examples in the official document, which is in the list section of version 2.7.
When looping, the original column a [] is not used, but a [:] is used.
This method is the easiest and requires no additional code.
In addition, @'s problem is very good, and @ Locke's explanation is correct. Because the for loop does not obtain the length of a single column first, the efficiency is too low. For is implemented through the iter iterator, so it is a dynamic iteration by subscript until the end.
Life is short, I use python. s is no longer the original s. If x is indeed the original x, it will be executed four times in a loop. The fourth time should be because the array is out of bounds. Is it true that Mao does not report an error? How does python check the mechanism?
