跳过python中范围函数中的值
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Skip over a value in the range function in python
提问by David
What is the pythonic way of looping through a range of numbers and skipping over one value? For example, the range is from 0 to 100 and I would like to skip 50.
循环遍历一系列数字并跳过一个值的pythonic方法是什么？例如，范围是从 0 到 100，我想跳过 50。
Edit: Here's the code that I'm using
编辑：这是我正在使用的代码
for i in range(0, len(list)):
x= listRow(list, i)
for j in range (#0 to len(list) not including x#)
...
采纳答案by njzk2
You can use any of these:
您可以使用以下任何一种：
# Create a range that does not contain 50
for i in [x for x in xrange(100) if x != 50]:
print i
# Create 2 ranges [0,49] and [51, 100] (Python 2)
for i in range(50) + range(51, 100):
print i
# Create a iterator and skip 50
xr = iter(xrange(100))
for i in xr:
print i
if i == 49:
next(xr)
# Simply continue in the loop if the number is 50
for i in range(100):
if i == 50:
continue
print i
回答by Daniel
for i in range(100):
if i == 50:
continue
dosomething
回答by MxLDevs
It depends on what you want to do. For example you could stick in some conditionals like this in your comprehensions:
这取决于你想做什么。例如，你可以在你的理解中坚持一些这样的条件：
# get the squares of each number from 1 to 9, excluding 2
myList = [i**2 for i in range(10) if i != 2]
print(myList)
# --> [0, 1, 9, 16, 25, 36, 49, 64, 81]
回答by Anonemuss
what you could do, is put an if statement around everything inside the loop that you want kept away from the 50. e.g.
您可以做的是在循环内的所有内容周围放置一个 if 语句，您希望远离 50。例如
for i in range(0, len(list)):
if i != 50:
x= listRow(list, i)
for j in range (#0 to len(list) not including x#)
回答by Smittie
In addition to the Python 2 approach here are the equivalents for Python 3:
除了 Python 2 方法之外，这里还有 Python 3 的等效方法：
# Create a range that does not contain 50
for i in [x for x in range(100) if x != 50]:
print(i)
# Create 2 ranges [0,49] and [51, 100]
from itertools import chain
concatenated = chain(range(50), range(51, 100))
for i in concatenated:
print(i)
# Create a iterator and skip 50
xr = iter(range(100))
for i in xr:
print(i)
if i == 49:
next(xr)
# Simply continue in the loop if the number is 50
for i in range(100):
if i == 50:
continue
print(i)
Ranges are lists in Python 2 and iterators in Python 3.
范围是 Python 2 中的列表和 Python 3 中的迭代器。
回答by Acumenus
It is time inefficient to compare each number, needlessly leading to a linear complexity. Having said that, this approach avoids any inequality checks:
比较每个数字的时间效率低下，不必要地导致线性复杂性。话虽如此，这种方法避免了任何不等式检查：
import itertools
m, n = 5, 10
for i in itertools.chain(range(m), range(m + 1, n)):
print(i) # skips m = 5
As an aside, you woudn't want to use (*range(m), *range(m + 1, n))even though it works because it will expand the iterables into a tuple and this is memory inefficient.
顺便说一句，(*range(m), *range(m + 1, n))即使它可以工作，您也不想使用，因为它会将可迭代对象扩展为一个元组，而这是内存效率低下的。
Credit: comment by njzk2, answer by Locke
信用：njzk2 的评论，Locke 的回答
回答by Maria
for i in range(0, 101):
if i != 50:
do sth
else:
pass
