如何垂直显示列表？

Question

I have a list of letters and want to be able to display them vertically like so:

a d
b e
c f

def main():
    letters = ["a", "b", "c", "d","e", "f"]
    for i in letters:
       print(i)

this code only display them like this:

a
b
c
d
e

Answer 1

That's because you're printing them in separate lines. Although you haven't given us enough info on how actually you want to print them, I can infer that you want the first half on the first column and the second half on the second colum.

Well, that is not that easy, you need to think ahead a little and realize that if you calculate the half of the list and keep it: h=len(letters)//2 you can iterate with variable i through the first half of the list and print in the same line letters[i] and letters[h+i] correct? Something like this:

def main():
    letters = ["a", "b", "c", "d","e", "f"]
    h = len(letters)//2 # integer division in Python 3
    for i in range(h):
       print(letters[i], letters[h+i])

You can easily generalize it for lists without pair length, but that really depends on what you want to do in that case.

That being said, by using Python you can go further :). Look at this code:

def main():
    letters = ["a", "b", "c", "d","e", "f"]
    for s1,s2 in zip(letters[:len(letters)//2], letters[len(letters)//2:]): #len(letters)/2 will work with every paired length list
       print(s1,s2)

This will output the following in Python 3:

a d
b e
c f

What I just did was form tuples with zip function grouping the two halves of the list.

For the sake of completeness, if someday your list hasn't a pair length, you can use itertools.zip_longest which works more or less like zip but fills with a default value if both iterables aren't of the same size.

Hope this helps!

Answer 2

I'll just throw in another solution:

letters = ["a", "b", "c", "d","e", "f"]

for n,i in enumerate(letters[:3]):
    print(i,letters[n+3])

Also outputs:

a d
b e
c f

Answer 3

Answer Peter Goldsborough has issue

what about this letters array

>>> letters = ["a", "b", "c", "d","e"]
>>> for n,i in enumerate(letters[:3]):
    print(i,letters[n+3])


a d
b e
Traceback (most recent call last):
  File "<pyshell#187>", line 2, in <module>
    print(i,letters[n+3])
IndexError: list index out of range

I added condition for it

>>> for n,i in enumerate(letters[:3]):
    if n + 3 < len(letters):
        print(i,letters[n+3])
    else:
        print(i)


a d
b e
c

The same problem with Paulo Bu's answer.

And here is my I think more simple and universal solution

>>> import math
>>> def main():
    letters = ["a", "b", "c", "d", "e", "f"]
    rows = 3
    columns = int(math.ceil(len(letters) / rows))
    for i in range(min(rows, len(letters))):
        for j in range(columns):
            next_column_i = i + rows * j
            if next_column_i < len(letters):
                print(letters[next_column_i], end = ' ')
        print()


>>> main()
a d 
b e 
c f 
>>> 

I can change rows count set it 2 and easy to get this if I need

>>> main()
a c e 
b d f 
>>> 

Answer 4

If the goal is to specify the number of COLUMNS in which to display a collections, here's my code

# Python 3

from math import ceil

def pt(x, nbcol):
    print ('nbcol ==',nbcol)
    y = int(ceil(len(x)/float(nbcol)))
    wcol = len(x)//y
    if wcol==nbcol or (wcol+1==nbcol and 0 < len(x) - (wcol*y) <= y):
        print ('\n'.join('\t'.join(str(el) for el in x[i::y])
                         for i in range(y)) , '\n' )
    else:
        print ("I can't do it\n")


li = ['ab','R','uio','b',4578,'h','yu','mlp','AZY12','78']
for nbcol in range(1,9):
    pt(li,nbcol)
print ('===============================')
for nbcol in range(1,9):
    pt("abcdef",nbcol)
print ('===============================')
for nbcol in range(1,9):
    pt('abcdefghijk',nbcol)

example

nbcol == 1
ab
R
uio
b
4578
h
yu
mlp
AZY12
78 

nbcol == 2
ab  h
R   yu
uio mlp
b   AZY12
4578    78 

nbcol == 3
ab  4578    AZY12
R   h   78
uio yu
b   mlp 

nbcol == 4
ab  b   yu  78
R   4578    mlp
uio h   AZY12 

nbcol == 5
ab  uio 4578    yu  AZY12
R   b   h   mlp 78 

nbcol == 6
I can't do it

nbcol == 7
I can't do it

nbcol == 8
I can't do it

===============================
nbcol == 1
a
b
c
d
e
f 

nbcol == 2
a   d
b   e
c   f 

nbcol == 3
a   c   e
b   d   f 

nbcol == 4
I can't do it

nbcol == 5
I can't do it

nbcol == 6
a   b   c   d   e   f 

nbcol == 7
I can't do it

nbcol == 8
I can't do it

===============================
nbcol == 1
a
b
c
d
e
f
g
h
i
j
k 

nbcol == 2
a   g
b   h
c   i
d   j
e   k
f 

nbcol == 3
a   e   i
b   f   j
c   g   k
d   h 

nbcol == 4
a   d   g   j
b   e   h   k
c   f   i 

nbcol == 5
I can't do it

nbcol == 6
a   c   e   g   i   k
b   d   f   h   j 

nbcol == 7
I can't do it

nbcol == 8
I can't do it

Answer 5

Without context for what you are doing it is difficult to provide an answer.

If your data need to be in pairs then perhaps you should create a differently structured object. A simple example would be a list of tuples.

def main():
    letters = [("a", "d"), ("b", "e"), ("c", "f")]
    for pair in letters:
       print("%s %s" % pair)

Answer 6

For a general solution, you can do something along these lines:

from itertools import zip_longest

lets = list('abcdefghijklmnop')

def table(it, rows):
    return zip_longest(*[it[i:i+rows] for i in range(0, len(it), rows)], 
                            fillvalue=' ')

for t in table(lets, 6):
    print(*t) 

Prints:

a g m
b h n
c i o
d j p
e k 
f l 

Since we are using zip_longest ( izip_longest for Python 2), you can supply a fillvalue and the odd ending value will not be truncated.

If you want to change the columns, change the number of rows:

for t in table(lets, 2):
    print(*t)  

Prints:

a c e g i k m o
b d f h j l n p

Of course, it is easy math to figure out the relationship between cols and rows if you have a list X items long.

So How does this work?

By definition, a table is a matrix. So first, create a matrix that is rows long:

>>> lets
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p']
>>> rows=6
>>> [lets[i:i+rows] for i in range(0, len(lets), rows)]
[['a', 'b', 'c', 'd', 'e', 'f'], ['g', 'h', 'i', 'j', 'k', 'l'], ['m', 'n', 'o', 'p']]

Then invert that matrix:

>>> for t in zip_longest(*[lets[i:i+rows] for i in range(0, len(lets), rows)]):
...     print(t)
... 
('a', 'g', 'm')
('b', 'h', 'n')
('c', 'i', 'o')
('d', 'j', 'p')
('e', 'k', None)
('f', 'l', None)

And go from there...

Answer 7

一个简单的方法是： D=["A","B","C"] print('\\n'.join(D))

Answer 8

>>> letters = ["a", "b", "c", "d", "e", "f", "g"]
>>> odd_even = zip(letters[::2], letters[1::2] + len(letters) % 2 * [""])
>>> pairs = [" ".join([odd, even]) for odd, even in odd_even]
a b
c d
e f
g