Sort Dictionary Alphabetically When The Key Is A String (name)

First, I know there are a LOT of posts on dictionary sorting but I couldn't find one that was exactly for my case - and I am just not understanding the sorted(...lambda) stuff - so

Solution 1:

simple algorithm to sort dictonary keys in alphabetical order, First sort the keys using sorted

sortednames=sorted(dictUsers.keys(), key=lambda x:x.lower())

for each key name retreive the values from the dict

for i in sortednames:
   values=dictUsers[i]
   print("Name= " + i)
   print ("   Age= " + values.age)
   print ("   Address= " + values.address)
   print ("   Phone Number= " + values.phone)

Solution 2:

dictio = { 'Epsilon':5, 'alpha':1, 'Gamma':3, 'beta':2, 'delta':4 }

sortedDict = dict( sorted(dictio.items(), key=lambda x: x[0].lower()) )

for k,v in sortedDict.items():
    print('{}:{}'.format(k,v))

output

 alpha:1
 beta:2
 delta:4
 Epsilon:5
 Gamma:3

Solution 3:

i would do it like:

sorted_dict = {key: value for key, value in sorted(unsorted_dict.items())}

Solution 4:

Just to reiterate @hughdbrown's comment in a separate answer, because I missed it at first and believe it to be the true answer to this question:

You have a dictionary you want to sort alphabetically, say dictio:

dictio = {'Beta': 2, 'alpha': 1}

sorted() will alphabetically sort the keys for you into a list, given that you convert all keys to lowercase (otherwise uppercase entries will be sorted first, then all lowercase):

sortedkeys = sorted(dictio, key=str.lower)

By calling the original dictionary again using that list of sorted keys you then can output it alphabetically:

foriinsortedkeys:
    printdictio[i]

Solution 5:

dict1={"d":1,"a":3,"z":0}

x=sorted(dict1.items())

print(x)

output

[('a', 3), ('d', 1), ('z', 0)]