another way to sort like l.sort(key=lambda x:(x[0][0], -x[1][0]))
sajuptpm
l = [((30,50),(70)), ((50,20),(20))]
for i in range(10):
k = ((i+30,i+50),(i+70))#suppose creating new tuple in each iteration
using some random value and in sert it into list.
flag=True
for i, v in enumerate(l):
if v >= k:
l.insert(i,k)
flag = False
break
if flag:
l.append(k)
This code will give a list of tuples sorted in asc order.
I need to change this code to sort this list by k[0][0] ascending and
k[0][1] descending.
What are the modifications needed in this code.
I dont want to user sort() method of list.
i need to implement l.sort(key=lambda x:(x[0][0], -x[1][0])) in
another way .I want to know what the modification needed in the 'if'
check to sort this list of tuples in k[0][0] ascending and k[0][1]
descending.
Peter Otten
> i need to implement l.sort(key=lambda x:(x[0][0], -x[1][0])) in
> another way .I want to know what the modification needed in the 'if'
> check to sort this list of tuples in k[0][0] ascending and k[0][1]
> descending.
It seems you are not getting any closer to your goal. Perhaps it would help
if you could explain that goal clearly rather than describing the means you
are employing to achieve it.
> I have a list of tuples.
>
> l = [((30,50),(70)), ((50,20),(20))]
By the way, (42) is not a tuple, it's an integer. To turn it into a 1-tuple
you have to add a ',':
>>> (42)
42
>>> (42,)
(42,)
>>> 42,
(42,)
Peter
sajuptpm
l = [((30,50),(70,)), ((50,20),(20,))]
for i in range(10):
k = ((i+30,i+50),(i+70))#suppose creating new tuple in each
iteration
using some random value and in sert it into list.
flag=True
for i, v in enumerate(l):
if v >= k:
l.insert(i,k)
flag = False
break
if flag:
l.append(k)
This code will give a list of tuples sorted in asc order.
I need to change this code to sort this list by k[0][0] ascending and
k[0][1] descending.
What are the modifications needed in this code.
I dont want to user sort() method of list.
i need to implement l.sort(key=lambda x:(x[0][0], -x[1][0])) in
Peter Otten
As a thought experiment assume that your audience had never heard of tuples
or even Python. How would you then explain your goal?
Peter
Terry Reedy
> I have a list of tuples.
>
> l = [((30,50),(70)), ((50,20),(20))]
>
> for i in range(10):
> k = ((i+30,i+50),(i+70))
The (i+70) parens do nothing, as already explained to (20)
Your set of test data are not very good as they do not test all the
insertion possibilities. The value of the third number is not relevant
to the problem and might as well be 0 for testing purposes.
#suppose creating new tuple in each iteration
> using some random value and in sert it into list.
>
> flag=True
> for i, v in enumerate(l):
You are reusing 'i' as index variable when it is already in use as index
variable for the outer loop. This is a confusing and a bad idea in
general, even if it happens to work.
> if v>= k:
> l.insert(i,k)
> flag = False
> break
This block should be indented under the inner for loop
> if flag:
> l.append(k)
flay can be avoided by using else clause of for statement
> This code will give a list of tuples sorted in asc order.
> I need to change this code to sort this list by k[0][0] ascending and
> k[0][1] descending.
> What are the modifications needed in this code.
Change the v>=k test to be the one you want. Here is my code:
l = [((30,51),0), ((30,49),0), ((32,20),0)]
for i in range(5):
k = ((i+29,i+51), 0)
for i, v in enumerate(l):
v00 = v[0][0]
k00 = k[0][0]
if v00 > k00 or v00==k00 and v[0][1] <= k[0][1]:
l.insert(i,k)
break
else:
l.append(k)
print(l)
# prints
[((29, 51), 0), ((30, 52), 0), ((30, 51), 0), ((30, 49), 0),
((31, 53), 0), ((32, 54), 0), ((32, 20), 0), ((33, 55), 0)]
> I dont want to user sort() method of list.
Why not? Even for the example above, .sort is faster than doing the
insert sort in Python code! Indent the above with def f1():. Then
def f2():
l = [((30,51),0), ((30,49),0), ((32,20),0)]
for i in range(5):
l.append(((i+29,i+51), 0))
l.sort(key=lambda x:(x[0][0], -x[0][1]))
import timeit
print(timeit.timeit(f1,'from __main__ import f1,f2',number=100000))
print(timeit.timeit(f2,'from __main__ import f1,f2',number=100000))
#prints
2.51296240165
1.63514413145
> i need to implement l.sort(key=lambda x:(x[0][0], -x[1][0])) in
That should be -x[0][1]
It is a good idea to test code before posting.
Any reason for the 'need' other than a wrong idea about relative speed?
--
Terry Jan Reedy
sajuptpm
---------------------
l1 = []
l2 = [
((3,8),(1,2)),
((1,3),(1,7)),
((7,0),(1,8)),
((4,2),(1,2)),
((2,9),(9,1))
]
I need to take each item from l2 and insert into l1 with first
element(column)(3,1,7,4,2) sorted in ascending order and second
element(column)(8,3,0,2,9) sorted in descending order.
#SORTING
for k in l2:
flag=True
for i, v in enumerate(l1):
if v <= k:
l1.insert(i,k)
flag = False
break
if flag:
l1.append(k)
for a in l1:
print a
output
-------
((7, 0), (1, 8))
((4, 2), (1, 2))
((3, 8), (1, 2))
((2, 9), (9, 1))
((1, 3), (1, 7))
This will not give l1 with first element(column)(3,1,7,4,2) sorted in
ascending order and second element(column)(8,3,0,2,9) sorted in
descending order.
-------------- I added a -ve signe to all first elements
l2 = [
((-3,8),(1,2)),
((-1,3),(1,7)),
((-7,0),(1,8)),
((-4,2),(1,2)),
((-2,9),(9,1))
]
#SORTING
for k in l2:
flag=True
for i, v in enumerate(l1):
if v <= k:
l1.insert(i,k)
flag = False
break
if flag:
l1.append(k)
for a in l1:
print a
output
-------
((-1, 3), (1, 7))
((-2, 9), (9, 1))
((-3, 8), (1, 2))
((-4, 2), (1, 2))
((-7, 0), (1, 8))
Now output is similar to first elements asc and second elements
desc.But the problem is the -ve sign, i dont need that.
Have any other method to do it??
Peter Otten
> Now output is similar to first elements asc and second elements
> desc.But the problem is the -ve sign, i dont need that.
>
> Have any other method to do it??
Sort twice:
>>> for item in items:
... print item
...
('adams', 'anne')
('miller', 'arnold')
('miller', 'bill')
('adams', 'berta')
('adams', 'charlotte')
('miller', 'thomas')
>>> items.sort(key=lambda x: x[1], reverse=True)
>>> items.sort(key=lambda x: x[0])
>>> for item in items:
... print item
...
('adams', 'charlotte')
('adams', 'berta')
('adams', 'anne')
('miller', 'thomas')
('miller', 'bill')
('miller', 'arnold')
See? First column ascending, second column descending within groups of rows
with equal first column.
Peter
Terry Reedy
yet, try the gmane.org archives.
--
Terry Jan Reedy