Explanation of SO answer for Emperor Wafi

sorted_dict.py

sortedrows = { 
  # k: v  were replaced with key and value, respectively to make it easier to follow
  key: value
  
  # enumerate through the sorted 
  for key, value in sorted(
    rownums.items(), 
    key=lambda item: item[0]
  )
}

# okay we have to realize that this is actually just a for loop
# We're going through all of the items after they've been sorted and assigning them as key: value in the object

# first let's take a look at the sorted() function because we're iterating through whatever it returns

# rownums.items() returns an array of tuples
# imagine the following:
rownums = { "a": 3, "b": 2, "c": 3 }

# if we call rownums.items() we get the following:
[('a', 3), ('b', 2), ('c', 1)]

# So now let's look at the second parameter to the sorted() function:
# the 'key' parameter takes in a function
# For every item in the first paramter (our unsorted array) (our unsorted array is an array of tuples)
# It is an array of tuples because that's what rownums.items() returned

# the lambda keyword is used to define functions
lambda item: item[0]
# is the same as
def key_function(item):
  return item[0]
# 
# What does the function return?
# Essentially a 'key' is like a comparison score we are giving that item 
# An item with a key of 1 is placed higher in the dict than an item with a key of 2
# An item with a key of 'a' is placed before an item with a key of 'b'
# Two items with the same key are treated as having the same sort value

# okay so now let's put that together

sorted(
   rownums.items(), 

   # since the item is a tuple, item[0] is the key and item[1] is the value
   key=lambda item: item[0]
 )
# We are returning item[0] as the 'key' so we are essentially asking it to order it based on keys

# after the sorted function finishes it returns an array that looks like this:
[('a', 3), ('b', 2), ('c', 1)]

# We then iterate through the array in our for loop to recreate the dictionary object

sortedrows = { 
  key: value for key, value in [('a', 3), ('b', 2), ('c', 1)]
}