CODEWITHSUNDEEP
pythonlistfor-loop
fanronghong
fanronghong

Reputation: 55

for loop within a list

I want to get a ordered list with ascending order, but leave out the first element in the original list. But the first for-loop only runs 4 times. After 4 times loop in first for-loop: 

lst=[6, 5, 4]

I think the first for-loop should keep running, but it finishes. I don't know why?

Source code below:

#coding: utf-8
lst = [0, 1, 6, 2, 5, 3, 4]
new_lst = []  # the ordered list, [0, 1, 2, 3, 4, 5, 6]
p0 = lst[0]  # the first element can be arbitrary
lst.remove(p0)
new_lst.append(p0)
for temp in lst:
    lst.remove(temp)
    while 1:
        for pt in lst:
            if temp < pt:
                continue
            else:
                lst.append(temp)
                temp = pt
                lst.remove(pt)
                break
        else:
            new_lst.append(temp)
            break

print("the ordered list:", new_lst)

At last, I only get:

('the ordered list:', [0, 1, 2, 3])

Upvotes: 0

Views: 107

Answers (3)

kaifi
kaifi

Reputation: 136

If you can use sorted, that's the easiest way to do it. If not, this is my solution:

lst = [0, 1, 6, 2, 5, 3, 4]
new = lst[1:]
for num in range(len(new)-1, 0, -1):
    for i in range(num):
        if new[i] > new[i+1]:
            new[i], new[i+1] = new[i+1], new[i]
new.insert(0, lst[0])
print(new)

Upvotes: 0

rnso
rnso

Reputation: 24535

To ignore first element in sorting: 

newlst = [lst[0]]+sorted(lst[1:])

First element lst[0] has to be enclosed in [ ] to get a list which can be extended using + with rest of the list (lst[1:]).

Upvotes: 1

Michael Ilie
Michael Ilie

Reputation: 459

try new_lst = sorted(lst). This sorts a list by integer value (and ascii value if characters are present). This is a much nicer solution for your problem and is way more pythonic.

Upvotes: 0

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