如何更好地打印列表？

Question

这与How to print a list in Python “nicely”类似，但我希望打印列表时更加美观——不使用括号、撇号和逗号，在列中甚至更好。

foolist = ['exiv2-devel', 'mingw-libs', 'tcltk-demos', 'fcgi', 'netcdf', 
    'pdcurses-devel',     'msvcrt', 'gdal-grass', 'iconv', 'qgis-devel', 
    'qgis1.1', 'php_mapscript']

evenNicerPrint(foolist)

预期结果：

exiv2-devel       msvcrt        
mingw-libs        gdal-grass    
tcltk-demos       iconv         
fcgi              qgis-devel    
netcdf            qgis1.1       
pdcurses-devel    php_mapscript 

谢谢！

Answer 1

简单：

l = ['exiv2-devel', 'mingw-libs', 'tcltk-demos', 'fcgi', 'netcdf', 
    'pdcurses-devel',     'msvcrt', 'gdal-grass', 'iconv', 'qgis-devel', 
    'qgis1.1', 'php_mapscript']

if len(l) % 2 != 0:
    l.append(" ")

split = len(l)/2
l1 = l[0:split]
l2 = l[split:]
for key, value in zip(l1,l2):
    print '%-20s %s' % (key, value)         #python <2.6
    print "{0:<20s} {1}".format(key, value) #python 2.6+

Answer 2

灵感来自吉梅尔的回答，above。

import math

def list_columns(obj, cols=4, columnwise=True, gap=4):
    """
    Print the given list in evenly-spaced columns.

    Parameters
    ----------
    obj : list
        The list to be printed.
    cols : int
        The number of columns in which the list should be printed.
    columnwise : bool, default=True
        If True, the items in the list will be printed column-wise.
        If False the items in the list will be printed row-wise.
    gap : int
        The number of spaces that should separate the longest column
        item/s from the next column. This is the effective spacing
        between columns based on the maximum len() of the list items.
    """

    sobj = [str(item) for item in obj]
    if cols > len(sobj): cols = len(sobj)
    max_len = max([len(item) for item in sobj])
    if columnwise: cols = int(math.ceil(float(len(sobj)) / float(cols)))
    plist = [sobj[i: i+cols] for i in range(0, len(sobj), cols)]
    if columnwise:
        if not len(plist[-1]) == cols:
            plist[-1].extend(['']*(len(sobj) - len(plist[-1])))
        plist = zip(*plist)
    printer = '\n'.join([
        ''.join([c.ljust(max_len + gap) for c in p])
        for p in plist])
    print printer

结果（第二个满足您的要求）：

>>> list_columns(foolist)
exiv2-devel       fcgi              msvcrt            qgis-devel        
mingw-libs        netcdf            gdal-grass        qgis1.1           
tcltk-demos       pdcurses-devel    iconv             php_mapscript     

>>> list_columns(foolist, cols=2)
exiv2-devel       msvcrt            
mingw-libs        gdal-grass        
tcltk-demos       iconv             
fcgi              qgis-devel        
netcdf            qgis1.1           
pdcurses-devel    php_mapscript     

>>> list_columns(foolist, columnwise=False)
exiv2-devel       mingw-libs        tcltk-demos       fcgi              
netcdf            pdcurses-devel    msvcrt            gdal-grass        
iconv             qgis-devel        qgis1.1           php_mapscript     

>>> list_columns(foolist, gap=1)
exiv2-devel    fcgi           msvcrt         qgis-devel     
mingw-libs     netcdf         gdal-grass     qgis1.1        
tcltk-demos    pdcurses-devel iconv          php_mapscript  

Answer 3

这个答案在@Aaron Digulla的答案中使用了相同的方法，使用的是略为python的语法。这可能会使上面的一些答案更容易理解。

>>> for a,b,c in zip(foolist[::3],foolist[1::3],foolist[2::3]):
>>>     print '{:<30}{:<30}{:<}'.format(a,b,c)

exiv2-devel                   mingw-libs                    tcltk-demos
fcgi                          netcdf                        pdcurses-devel
msvcrt                        gdal-grass                    iconv
qgis-devel                    qgis1.1                       php_mapscript

这可以很容易地适应任意数量的列或可变列，这将导致类似@gnibbler的答案。间距可根据屏幕宽度调整。

---

更新：按要求解释。

索引

foolist[::3]选择foolist的每三个元素。foolist[1::3]选择每三个元素，从第二个元素开始（'1'，因为python使用零索引）。

In [2]: bar = [1,2,3,4,5,6,7,8,9]
In [3]: bar[::3]
Out[3]: [1, 4, 7]

压缩

压缩列表（或其他iterable）生成列表元素的元组。例如：

In [5]: zip([1,2,3],['a','b','c'],['x','y','z'])
Out[5]: [(1, 'a', 'x'), (2, 'b', 'y'), (3, 'c', 'z')]

一起

把这些想法放在一起，我们就得到了解决方案：

for a,b,c in zip(foolist[::3],foolist[1::3],foolist[2::3]):

在这里，我们首先生成foolist的三个“切片”，每个切片按第三个元素索引，偏移量为1。它们各自只包含列表的三分之一。现在，当我们压缩这些片段并进行迭代时，每次迭代都会给出foolist的三个元素。

这就是我们想要的：

In [11]: for a,b,c in zip(foolist[::3],foolist[1::3],foolist[2::3]):
   ....:      print a,b,c                           
Out[11]: exiv2-devel mingw-libs tcltk-demos
         fcgi netcdf pdcurses-devel
        [etc]

而不是：

In [12]: for a in foolist: 
   ....:     print a
Out[12]: exiv2-devel
         mingw-libs
         [etc]