Is there a shorter way to code this challenge?

Question

month = {'January':1, 'February':2, 'March':3, 'April':4, 'May':5, 'June':6, 'July':7, 'August':8, 'September':9, 'October':10, 'November':11, 'December':12}

def usDate(a):
y = a[0]

if y.isdigit():
x = a.split("/")
temp = x[0]
x[0] = x[1]
x[1] = temp
return "/".join(x)
elif y.isalpha():
c = a.replace(",","")
x = c.split(" ")
b = month.get(x[0])
x[0] = str(b)
temp = x[0]
x[0] = x[1]
x[1] = temp
return "/".join(x)

res = usDate(input())

print(res)

Answer 1

Angelmar Dayangco some of the logic can be consolidated since both clauses in the if statement do a swap and then return the same join statement. Furthermore, the swap can be eliminated if you place elements of x in the desired order inside a tuple, and pass the tuple to join.

month = {'January':1, 'February':2, 'March':3, 'April':4, 'May':5, 'June':6, 'July':7, 'August':8, 'September':9, 'October':10, 'November':11, 'December':12}

def usDate(a):
    y = a[0]

    if y.isdigit():
        x = a.split("/")
    elif y.isalpha():
        c = a.replace(",","")
        x = c.split()
        x[0] = str(month.get(x[0]))

    return "/".join((x[1],x[0],x[2]))

res = usDate(input())

print(res)

My own version is a little shorter because I compare only the first 3 characters of the month names.
https://code.sololearn.com/cYLtndwVshDA/?ref=app

Answer 2

Use
x[0], x[1] = x[1], x[0]

instead of
temp = x[0]
x[0] = x[1]
x[1] = temp

Good Luck