I am given the following list

mylist=[25,42,58,31,20,11,42]

which I separated them into a new list

newlist=[[25,42,58,31],[20,11,42]]

Then I need to print it as follow. How do I code it in a way where all the index will be printed?

First sequence is : 0,1,2,3
Second sequence is : 4,5,6

CodePudding user response：

The following is one implementation using itertools.count:

import itertools

newlist=[[25,42,58,31],[20,11,42]]

cnt = itertools.count()
for i, sublst in enumerate(newlist):
    print(f"Sequence {i}:", [next(cnt) for _ in sublst])

# Sequence 0: [0, 1, 2, 3]
# Sequence 1: [4, 5, 6]

I personally find this method concise.

CodePudding user response：

Assuming this only depends of newlist (i.e newlist is a valid split of mylist):

newlist=[[25,42,58,31],[20,11,42]]
start = 0
for i, l in enumerate(newlist):
    seq = range(start, start len(l))
    start  = len (l)
    print(f'sequence {i} is: {",".join(map(str,seq))}')

Output:

sequence 0 is: 0,1,2,3
sequence 1 is: 4,5,6

CodePudding user response：

What you have here is a list of lists that you could interate through and count the results.

addition = 0
for idx, item in enumerate(newlist):
    if idx != 0:
       addition = len(newlist[idx-1])
    print(f"Sequence of item {idx 1}: {[i addition for i in range(len(item))]}"

CodePudding user response：

You can use the inflect package to convert 1, 2, 3 to 1st, 2nd, 3rd...

Quick prototype:

import inflect
p = inflect.engine()

for i in range(len(newlist)):
    print(f"{p.ordinal(i)} sequence is {','. join(map(str , seq))}}")