Assume there is a list [1,9,7,3,6]. I want to produce a new list which is sorted and smaller than one of the integers, for example the integer is 7, so the new list should be list:
[1,3,6]
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7 Answers
You can use list-comprehension to do that:
my_list = [1,9,7,3,6]
result = sorted([x for x in my_list if x < 7])
Comments
oldList = [1,9,7,3,6]
myInt = 7
newList = sorted([x for x in oldList if x < myInt])
This is a Python "list comprehension" - it looks at each element in oldList and adds it to newList if and only if it's smaller than the integer you've chosen. Additionally, the sorted() method is native in Python.
Here's a good resource for list comprehensions for the future: http://www.pythonforbeginners.com/basics/list-comprehensions-in-python
Hope that helps!
1 Comment
Burhan Khalid
You don't need a list comprehension.
new_list = sorted(x for x in old_list if x < my_int)Lists have a sort method:
old_list = [1,9,7,3,6]
new_list = [x for x in old_list if x < 7]
new_list.sort()
Comments
You can use a list comprehension:
sorted([i for i in [1,9,7,3,6] if i < 7])
You can also use a generator:
sorted(i for i in [1,9,7,3,6] if i < 7)
Note: The list comprehension version executes ~2x faster.
1 Comment
Qlaus
Note that it is only faster for small lists. For big ones the generator is slightly faster. Also interestingly using filter() instead is always slower.
Try this:
num = #some number
new_list = sorted(old_list)[:sorted(old_list).index(num)]
OR Alternative
num = 7
somelist = [3,6,7,1,2,8,9,4,5,12]
somelist.sort()
somelist[:somelist.index(num)]
OUTPUT:
[0, 1, 2, 3, 4, 5, 6]
1 Comment
Qlaus
Note that this is quite inefficient: You sort the full list (instead of the filtered one), you sort it twice and then you traverse it to find the index to truncate it. This of course only matters for much bigger lists.
aa = filter(lambda x: x < 7, [1,9,7,3,6])
aa.sort()
print aa
OUTPUT:
[1, 3, 6]