Python中如何在字符前后插入空格并换行？

Question

I need to insert a character before x1 and after x1 and put on a new line

my code:

f='p11p21p31'
y='x1'
k=str(y).join(f[i:i+3] for i in range(0, len(f), 3))
#p=' '.join(k[i:i+3] for i in range(0, len(k), 3))
print(k)

My outout

p11x1p21x1p31

I need:

p11 x1
p21 x1
p31 x1

Answer 1

You were almost there. Using your code:

f='p11p21p31'
y='x1'
k=str(" "+y+"\n").join(f[i:i+3] for i in range(0, len(f), 3))+" "+y
print (k)

Output:

p11 x1
p21 x1
p31 x1

You need to join with a linebreak after your delimiter, and a space preceding that delimiter. You finally add space+delimiter at the end of your string, cause join() on list will not set your delimiter after the last element.

Alternatively:

k= "".join([f[i] if (i+1) % 3 != 0 else f[i]+" "+y+"\n" for i in range(0,len(f))])

In a function:

f='p11p21p31'
y='x1'

def split_fxy(f,x,y):
    return "".join([f[i] if (i+1) % x != 0 else f[i]+" "+y+"\n" for i in range(0,len(f))])

print (split_fxy(f,3,y))

Answer 2

The problem you are facing with join is that it will only add 'x1' between elements you want to join ( see the docs ). To fix it, you will need to add an extra " x1" at the end.

If you are using Python 3.6+ (which is highly recommended) use the following:

f='p11p21p31'
y='x1'
k= f' {y}\n'.join(f[i:i+3] for i in range(0, len(f), 3)) + f' {y}'
print(k)

If you are using a version prior version to 3.6, use:

f='p11p21p31'
y='x1'
k= ' {}\n'.format(y).join(f[i:i+3] for i in range(0, len(f), 3)) + ' ' + y
print(k)

The previous code will work also even in the case you use Python2 which at this point is strongly discouraged.

Output:

p11 x1
p21 x1
p31 x1

Answer 3

Try join the \n .

f = 'p11p21p31'
y = 'x1'
size = 3  # split each 3 chars

k = '\n'.join(f[i:i+size] + ' ' + y for i in range(0, len(f), 3))
print(k)

Answer 4

You could use a f string to make you intent more clear:

f='p11p21p31'
y='x1'
print(''.join([f'{f[i:i+3]} {y}\n' for i in range(0,len(f),3) ]))

Prints:

p11 x1
p21 x1
p31 x1

Answer 5

With a regex based solution

1/ split according to your pattern

2/ filter null item

3/ concat x1 to all item

4/ join with \n

import re

f='p11p21p31'
#      4                3                         2          1
print("\n".join(map(lambda i: i+" x1", list(filter(None, re.split(r'([a-z]*[0-9]+)', f))))))