增加列表的所有元素，除了 Python 中的第一个元素

Question

I have a list T1 . I want to increase all elements of T1 by 1 except the first element. How do I do so?

T1=[0,1,2,3,4,5,6,7]
T2=[i+1 for i in T1]

The current output is

T2= [1, 2, 3, 4, 5, 6, 7, 8]

The expected output is

T2= [0, 2, 3, 4, 5, 6, 7, 8]

Answer 1

Use slicing!

T1 = [0,1,2,3,4,5,6,7]
T2 = [T1[0]] + [i+1 for i in T1[1:]]

Or, with enumerate :

T1 = [0,1,2,3,4,5,6,7]
T2 = [x+(i>0) for i, x in enumerate(T1)]

Output:

[0, 2, 3, 4, 5, 6, 7, 8]

Answer 2

You could separate the first element from the rest and increment the rest by 1 -

first, *rest = t1
t2 = [first] + [(i + 1) for i in rest]

Or you could enumerate your list and use the index -

t2 = [(val + 1) if idx > 0 else val for (idx, val) in enumerate(t1)]

Answer 3

The correct answer would be to slice the T1 list in two, the part that you don't want to modify and the part that you do. Just combine them afterwards:

T1=[0,1,2,3,4,5,6,7]
T2= T1[0:1] + [i+1 for i in T1[1:]]

Answer 4

There are several methods, 2 of them

T2 = [T1[0]] + [ j+1 for j in T1[1:]]

a longer but funnier one

T2 = [x+y for x,y in zip(T1, [0] + [1]*(len(T1)-1))]

or just a for loop with index, so you can decide which index to operate in

T2 = list()
for i,el in enumerate(T1):
    if i == 0:
        T2.append(el)
    else:
        T2.append(el+1)

Have fun

Answer 5

I've tried this:

given:

t1 = [0, 1, 2, 3, 4]

the t2 list may be calculated as following:

t2 = t1[0:1] + [v + 1 for i, v in enumerate(t1) if i != 0]

The first part of the list concatenation is a list: in fact if you try t1[0] you get a TypeError due to the fact that is impossible to concatenate int with list .

Answer 6

I would do it like this:

for i in range(1,len(T1)):   
  T1[i]=T1[i]+1

This would not create a new list, but would change the values in the current list

Answer 7

Relatively simple answer where you use if else condition.

T1=[0,1,2,3,4,5,6,7]
T2 = [x+1 if i>0 else x for i, x in enumerate(T1)]

# Output:
[0, 2, 3, 4, 5, 6, 7, 8]