I am trying to create a list of tuples where the tuple contents are the number 9 and the number before it in the list.
我试图创建一个元组列表,其中元组内容是数字9和列表中的数字前面的数字。
Input List:
myList = [1, 8, 9, 2, 4, 9, 6, 7, 9, 8]
Desired Output:
sets = [(8, 9), (4, 9), (7, 9)]
Code:
sets = [list(zip(myList[i:i], myList[-1:])) for i in myList if i==9]
Current Result:
[[], [], []]
6 个解决方案
#1
39
Cleaner Pythonic approach:
更清洁的Pythonic方法:
>>> [(x,y) for x,y in zip(myList, myList[1:]) if y == 9]
[(8, 9), (4, 9), (7, 9)]
What is the code above doing:
上面的代码是做什么的:
-
zip(some_list, some_list[1:])would generate a list of pairs of adjacent elements. - Now with that tuple, filter on the condition that the second element is equal to
9. You're done :)
zip(some_list,some_list [1:])将生成一对相邻元素的列表。
现在有了这个元组,过滤第二个元素等于9的条件。你做完了:)
#2
17
Part of your issue is that myList[i:i] will always return an empty list. The end of a slice is exclusive, so when you do a_list[0:0] you're trying to take the elements of a_list that exist between index 0 and index 0.
你的部分问题是myList [i:i]将始终返回一个空列表。切片的结尾是独占的,所以当你执行a_list [0:0]时,你试图获取索引0和索引0之间存在的a_list元素。
You're on the right track, but you want to zip the list with itself.
你是在正确的轨道上,但你想要自己压缩列表。
[(x, y) for x, y in zip(myList, myList[1:]) if y==9]
#3
7
You were pretty close, I'll show you an alternative way that might be more intuitive if you're just starting out:
你非常接近,如果你刚刚开始,我会告诉你另一种可能更直观的方式:
sets = [(myList[i-1], myList[i]) for i in range(len(myList)) if myList[i] == 9]
Get the index in the range of the list lenght, and if the value at the position i is equal to 9, grab the adjacent elements.
获取列表长度范围内的索引,如果位置i的值等于9,则抓取相邻元素。
The result is:
结果是:
sets
[(8, 9), (4, 9), (7, 9)]
This is less efficient than the other approaches but I decided to un-delete it to show you a different way of doing it. You can make it go a bit faster by using enumerate() instead:
这比其他方法效率低,但我决定取消删除它以向您展示不同的方法。你可以使用enumerate()代替它:
sets = [(myList[i-1], j) for i, j in enumerate(myList) if j == 9]
Take note that in the edge case where myList[0] = 9 the behavior of the comprehension without zip and the behavior of the comprehension with zip is different.
请注意,在myList [0] = 9的边缘情况下,没有zip的理解行为和zip的理解行为是不同的。
Specifically, if myList = [9, 1, 8, 9, 2, 4, 9, 6, 7, 9, 8] then:
具体来说,如果myList = [9,1,8,9,2,4,9,6,7,9,8]那么:
[(myList[i-1], myList[i]) for i in range(len(myList)) if myList[i] == 9]
# results in: [(8, 9), (8, 9), (4, 9), (7, 9)]
while:
[(x, y) for x, y in zip(myList, myList[1:]) if y==9]
# results in: [(8, 9), (4, 9), (7, 9)]
It is up to you to decide which of these fits your criteria, I'm just pointing out that they don't behave the same in all cases.
由你决定哪一个符合你的标准,我只是指出它们在所有情况下的行为都不一样。
#4
5
You can also do it without slicing by creating iterators:
您也可以通过创建迭代器来完成它而不进行切片:
l = myList = [1,8,9,2,4,9,6,7,9,8]
it1, it2 = iter(l), iter(l)
# consume first element from it2 -> leaving 8,9,2,4,9,6,7,9,8
next(it2, "")
# then pair up, (1,8), (8,9) ...
print([(i, j) for i,j in zip(it1, it2) if j == 9])
Or use the pairwise recipe to create your pairs
或者使用成对配方来创建配对
from itertools import tee, izip
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return izip(a, b)
If using python3, just import tee and use the regular zip.
如果使用python3,只需导入tee并使用常规zip。
#5
5
It is really surprising that no one has added a functional approach.
令人惊讶的是,没有人添加功能方法。
Another alternative answer is using a filter. This builtin function returns an iterator (list in Python2) consisting of all the elements present in the list that return True for a particular function
另一个替代答案是使用过滤器。这个内置函数返回一个迭代器(Python2中的列表),该迭代器包含列表中存在的所有元素,这些元素为特定函数返回True
>>> myList = [1,8,9,2,4,9,6,7,9,8]
>>> list(filter(lambda x:x[1]==9,zip(myList, myList[1:])))
[(8, 9), (4, 9), (7, 9)]
It is to be noted that the list call is needed only in python3+. The difference between the functional approach and list comprehensions is discussed in detail in this post.
需要注意的是,只有在python3 +中才需要列表调用。本文详细讨论了功能方法和列表推导之间的区别。
#6
1
My solution is similar to one of Jim's advanced with zero-index check
我的解决方案类似于Jim先进的零索引检查
myList = [9, 1, 8, 9, 2, 4, 9, 6, 7, 9, 8]
[(myList[i-1], x) for i, x in enumerate(myList) if x==9 and i!=0]
# [(8, 9), (4, 9), (7, 9)]
#1
39
Cleaner Pythonic approach:
更清洁的Pythonic方法:
>>> [(x,y) for x,y in zip(myList, myList[1:]) if y == 9]
[(8, 9), (4, 9), (7, 9)]
What is the code above doing:
上面的代码是做什么的:
-
zip(some_list, some_list[1:])would generate a list of pairs of adjacent elements. - Now with that tuple, filter on the condition that the second element is equal to
9. You're done :)
zip(some_list,some_list [1:])将生成一对相邻元素的列表。
现在有了这个元组,过滤第二个元素等于9的条件。你做完了:)
#2
17
Part of your issue is that myList[i:i] will always return an empty list. The end of a slice is exclusive, so when you do a_list[0:0] you're trying to take the elements of a_list that exist between index 0 and index 0.
你的部分问题是myList [i:i]将始终返回一个空列表。切片的结尾是独占的,所以当你执行a_list [0:0]时,你试图获取索引0和索引0之间存在的a_list元素。
You're on the right track, but you want to zip the list with itself.
你是在正确的轨道上,但你想要自己压缩列表。
[(x, y) for x, y in zip(myList, myList[1:]) if y==9]
#3
7
You were pretty close, I'll show you an alternative way that might be more intuitive if you're just starting out:
你非常接近,如果你刚刚开始,我会告诉你另一种可能更直观的方式:
sets = [(myList[i-1], myList[i]) for i in range(len(myList)) if myList[i] == 9]
Get the index in the range of the list lenght, and if the value at the position i is equal to 9, grab the adjacent elements.
获取列表长度范围内的索引,如果位置i的值等于9,则抓取相邻元素。
The result is:
结果是:
sets
[(8, 9), (4, 9), (7, 9)]
This is less efficient than the other approaches but I decided to un-delete it to show you a different way of doing it. You can make it go a bit faster by using enumerate() instead:
这比其他方法效率低,但我决定取消删除它以向您展示不同的方法。你可以使用enumerate()代替它:
sets = [(myList[i-1], j) for i, j in enumerate(myList) if j == 9]
Take note that in the edge case where myList[0] = 9 the behavior of the comprehension without zip and the behavior of the comprehension with zip is different.
请注意,在myList [0] = 9的边缘情况下,没有zip的理解行为和zip的理解行为是不同的。
Specifically, if myList = [9, 1, 8, 9, 2, 4, 9, 6, 7, 9, 8] then:
具体来说,如果myList = [9,1,8,9,2,4,9,6,7,9,8]那么:
[(myList[i-1], myList[i]) for i in range(len(myList)) if myList[i] == 9]
# results in: [(8, 9), (8, 9), (4, 9), (7, 9)]
while:
[(x, y) for x, y in zip(myList, myList[1:]) if y==9]
# results in: [(8, 9), (4, 9), (7, 9)]
It is up to you to decide which of these fits your criteria, I'm just pointing out that they don't behave the same in all cases.
由你决定哪一个符合你的标准,我只是指出它们在所有情况下的行为都不一样。
#4
5
You can also do it without slicing by creating iterators:
您也可以通过创建迭代器来完成它而不进行切片:
l = myList = [1,8,9,2,4,9,6,7,9,8]
it1, it2 = iter(l), iter(l)
# consume first element from it2 -> leaving 8,9,2,4,9,6,7,9,8
next(it2, "")
# then pair up, (1,8), (8,9) ...
print([(i, j) for i,j in zip(it1, it2) if j == 9])
Or use the pairwise recipe to create your pairs
或者使用成对配方来创建配对
from itertools import tee, izip
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return izip(a, b)
If using python3, just import tee and use the regular zip.
如果使用python3,只需导入tee并使用常规zip。
#5
5
It is really surprising that no one has added a functional approach.
令人惊讶的是,没有人添加功能方法。
Another alternative answer is using a filter. This builtin function returns an iterator (list in Python2) consisting of all the elements present in the list that return True for a particular function
另一个替代答案是使用过滤器。这个内置函数返回一个迭代器(Python2中的列表),该迭代器包含列表中存在的所有元素,这些元素为特定函数返回True
>>> myList = [1,8,9,2,4,9,6,7,9,8]
>>> list(filter(lambda x:x[1]==9,zip(myList, myList[1:])))
[(8, 9), (4, 9), (7, 9)]
It is to be noted that the list call is needed only in python3+. The difference between the functional approach and list comprehensions is discussed in detail in this post.
需要注意的是,只有在python3 +中才需要列表调用。本文详细讨论了功能方法和列表推导之间的区别。
#6
1
My solution is similar to one of Jim's advanced with zero-index check
我的解决方案类似于Jim先进的零索引检查
myList = [9, 1, 8, 9, 2, 4, 9, 6, 7, 9, 8]
[(myList[i-1], x) for i, x in enumerate(myList) if x==9 and i!=0]
# [(8, 9), (4, 9), (7, 9)]