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I am using list in python program, following is the code

x = [['Port', 'Status']]
x.append({11,'Open'})
x.append({22,'Close'})
x.append({356,'Open'})
x.append({1024,'Close'})
x.append({512,'Open'})
x.append({777,'Close'})
print(x)

the output of above is 

    [['Port', 'Status'], set([11, 'Open']), set(['Close', 22]), set(['Open', 356]), set([1024, 'Close'])
, set([512, 'Open']), set([777, 'Close'])]

Problem is that the output is not in the order It was entered. i.e on some cases string is appearing before integer vice versa on other cases. Please help.

Salman Raza
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  • From [\[Python\]: set definition](https://docs.python.org/2/library/stdtypes.html#set-types-set-frozenset): _A set object is an **unordered** collection of distinct hashable objects_. To keep the order use *list*s (as you do for the header - 1st element), or *tuple*s instead of *set*s. – CristiFati Apr 12 '17 at 10:55

4 Answers4

1

set is an unordered, you can use tuple or list instead:

x = [['Port', 'Status']]
x.append((11,'Open'))
Hackaholic
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0

A set is an unordered collection with no duplicate elements

    for example:
    >> basket = {'apple', 'orange', 'apple', 'pear', 'orange', 'banana'}
    >> print(basket)         # see that duplicates have been removed
    >> {'orange', 'banana', 'pear', 'apple'}

you can see that order has changed.

You are appending {11,'Open'} which is set and can change order at run time. so better use ordered data structure if you want to maintain the order.

for example use like this:
>> x.append((11, 'Open'))
or 
>> x.append([11, 'Open'])
JkShaw
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0

Simply use [] instead of {} as jyotish implied.

dylan_fan
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0

You're using a set as a container for the values, which do not have a fixed order.

Instead, you can use a tuple (e.g. (11, 'Open')) or another list (e.g. [11, 'Open']). But a tuple is probably a better match here.

Arjen
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