confusing python urlencode order

okay, s开发者_如何学Pythono according to http://docs.python.org/library/urllib.html

"The order of parameters in the encoded string will match the order of parameter tuples in the sequence."

except when I try to run this code:

import urllib
values ={'one':'one',
         'two':'two',
         'three':'three',
         'four':'four',
         'five':'five',
         'six':'six',
         'seven':'seven'}
data=urllib.urlencode(values)
print data

outputs as ...

seven=seven&six=six&three=three&two=two&four=four&five=five&one=one

7,6,3,2,4,5,1?

That doesn't look like the order of my tuples.

---

Dictionaries are inherently unordered because of the way they are implemented. If you want them to be ordered, you should use a list of tuples instead (or a tuple of lists, or a tuple of tuples, or a list of lists...):

values = [ ('one', 'one'), ('two', 'two') ... ]

---

Just in case someones arrives here like me searching for a way to get deterministic results from urlencode, to encode the values alphabetically you can do it like this:

from urllib.parse import urlencode
values ={'one':'one',
         'two':'two',
         'three':'three',
         'four':'four',
         'five':'five',
         'six':'six',
         'seven':'seven'}
sorted_values = sorted(values.items(), key=lambda val: val[0])
data=urlencode(sorted_values)
print(data)
#> 'five=five&four=four&one=one&seven=seven&six=six&three=three&two=two'

---

Why not using OrderedDict? Your code would then look like this:

from collections import OrderedDict
from urllib.parse import urlencode

d = OrderedDict()
d['one'] = 'one'
d['two'] = 'two'
d['three'] = 'three'
d['four'] = 'four'
...

data=urlencode(d)
print(data)
# one=one&two=two&three=three&four=four

This way the order of your dictionary would be preserved