将元素插入python列表中

Question

I am trying to write a function in python that should take as input 2 arguments as follow f('k', range(4)) and should return the following:

[['k', 0, 1, 2, 3], [0, 'k', 1, 2, 3], [0, 1, 'k', 2, 3], [0, 1, 2, 'k', 3], [0, 1, 2, 3, 'k']]

I have tried the following but something goes wrong

def f(j,ls):
    return [[ls.insert(x,j)]for x in ls]

Does anyone know how to find the solution?

Answer 1

list.insert() returns None because the list is altered in-place. You also don't want to share the same list object, you'll have to create copies of the list:

def f(j, ls):
    output = [ls[:] for _ in xrange(len(ls) + 1)]
    for i, sublist in enumerate(output):
        output[i].insert(i, j)
    return output

You also could use slicing to produce new sublists with the extra element 'inserted' through concatenation; this then gives you one-liner list comprehension:

def f(j, ls):
    return [ls[:i] + [j] + ls[i:] for i in xrange(len(ls) + 1)]

Demo:

>>> def f(j, ls):
...     output = [ls[:] for _ in xrange(len(ls) + 1)]
...     for i, sublist in enumerate(output):
...         output[i].insert(i, j)
...     return output
...
>>> f('k', range(4))
[['k', 0, 1, 2, 3], [0, 'k', 1, 2, 3], [0, 1, 'k', 2, 3], [0, 1, 2, 'k', 3], [0, 1, 2, 3, 'k']]
>>> def f(j, ls):
...     return [ls[:i] + [j] + ls[i:] for i in xrange(len(ls) + 1)]
...
>>> f('k', range(4))
[['k', 0, 1, 2, 3], [0, 'k', 1, 2, 3], [0, 1, 'k', 2, 3], [0, 1, 2, 'k', 3], [0, 1, 2, 3, 'k']]

Answer 2

This should work for you.

def f(j,num):
    lst=[]
    for i in range(num+1):
        innerList=list(range(num))
        innerList[i:i]=j
        lst.append(innerList)

    return lst

Demo:

print f("j",4)

Output:

[['k', 0, 1, 2, 3], [0, 'k', 1, 2, 3], [0, 1, 'k', 2, 3], [0, 1, 2, 'k', 3], [0, 1, 2, 3, 'k']]