How to capitalize only first letter of first word in a list

I want to capital only first letter of first word of list and remaining all words to be in lower case in ascending order of the length of word.

z=['The' ,'lines', 'are', 'printed', 'in', 'reverse', 'order']
z.sort(key=len) 
print(" ".join(x.lower() for x in z))

I want the result like this in ascending order of length.

In the are lines order printed reverse :

z=['The' ,'lines', 'are', 'printed', 'in', 'reverse', 'order']
# i tried like this 
z.sort(key=len)
s=[]
for x in z:
     if (x==0):
        s.append(z[0].capitalize())
     else:
        s.append(z[x].lower())

Actual output that I am trying to get:

In the are lines order printed reverse

Answer

Your code

for x in z:    # ['in', 'The', 'are', 'lines', 'order', 'printed', 'reverse']
     if (x==0):
        s.append(z[0].capitalize())
     else:
        s.append(z[x].lower())

does not work because each x is a word from z – it will never be equal to 0 so all that things added to s are lower case version of your words.

---

You can use enumerate to get the index of all elements in the list while iterating it.

The first element has to use .title() casing, all others .lower():

z= ['The' ,'lines', 'are', 'printed', 'in', 'reverse', 'order']
z.sort(key=len) 

# reformat capitalization based in index, any number != 0 is true-thy
z = [ a.lower() if idx else a.capitalize() for idx, a in enumerate(z) ]

print(" ".join(z))

Output:

In the are lines order printed reverse

---

z = [ a.lower() if idx else a.capitalize() for idx, a in enumerate(z) ] is a list comprehension – as normal loop it would look like:

s = []
for idx, word in enumerate(z):
    if idx:   # any idx > 0 is True-thy
        s.append(word.lower())
    else:
        s.append(word.capitalize())
# print the joined s 

You can learn what python considers True for non-boolean values here: Truth-value-testing