print dictionary keys and values in ordered manner

Question

I have a dictionary with keys and values as shown below.

from datetime import datetime
dic = {'pack1':'stage1','pack2':'stage2','pack3':'stage3','pack4':'stage4'}

I want to print the keys and their corresponding values in ordered manner in new line, like

Update at 2018-09-18 09:58:03.263575

**Delivery**           **State** 

'pack1'                 'stage1'

'pack2'                 'stage2'

'pack3'                 'stage3'

'pack4'                 'stage4'

The code I tried is given below

print("Update at %s\nDelivery\t\t\t\t\t\t\tState \n{}\t\t\t{}".format({key for key in dic.keys()}, {val for val in dic.values()}) %datetime.now())

But it give the output as

Update at 2018-09-18 09:58:03.263575

**Delivery**                                          **State** 

set(['pack4', 'pack1', 'pack2', 'pack3'])           set(['stage1', 'stage2', 'stage3', 'stage4'])

The values doesn't correspond to their keys, and the output is given in a single line as a set of lists. How to format the output as my requirement? (There are whitespaces between 'delivery' and State but stack overflow doesn't show them)

Answer 1

Dictionary order is not preserved by default. If you want that for any reason then you need to use OrderedDict instead

Answer 2

It is good to note that since Python 3.7, a dict preserves the order in which data is entered. Prior to that, you can use an OrderedDict .

Although, it seems what you want is to sort the key-value pairs in alphabetical order of keys. In that case, you can use sorted .

dic = {'pack1': 'stage1', 'pack2': 'stage2', 'pack3': 'stage3', 'pack4': 'stage4'}

for k, v in sorted(dic.items()):
    print (k + '\t' + v)

Output

pack1    stage1
pack2    stage2
pack3    stage3
pack4    stage4

From there you can work on the printing format.

Answer 3

You need to loop over your input, you can't do this cleanly in a single format and print operation.

If the keys only need to match their values, just do:

dic = {'pack1':'stage1','pack2':'stage2','pack3':'stage3','pack4':'stage4'}

for k, v in dic.items():  # .viewitems() on Python 2.7
    print("{}\t\t\t{}".format(k, v))

If you need the output ordered, not just matched up, dict don't provide that guarantee before 3.7 (and don't provide even implementation dependent guarantees until 3.6). You'd need a collections.OrderedDict . You could do a presort though:

for k, v in sorted(dic.items()):  # .viewitems() on Python 2.7
    print("{}\t\t\t{}".format(k, v))

so the output ordering is predictable.

Answer 4

Can do something like:

dic = {'pack1':'stage1','pack2':'stage2','pack3':'stage3','pack4':'stage4'}
print('Delivery\t\tState')
for i in dic.items():
   print(('\t'*3).join(i))

Dictionaries are not ordered under python 3.6, so OrderdDict :

from collections import OrderedDict
dic = OrderedDict([('pack1', 'stage1'), ('pack2', 'stage2'), ('pack3', 'stage3'), ('pack4', 'stage4')])
print('Delivery\t\tState')
for i in dic.items():
   print(('\t'*3).join(i))

All Outputs:

Delivery        State
pack1           stage1
pack2           stage2
pack3           stage3
pack4           stage4