将Sven Marnach的答案更新到Python 3.4中:

row_format ="{:>15}" * (len(teams_list) + 1)
print(row_format.format("", *teams_list))
for team, row in zip(teams_list, data):
    print(row_format.format(team, *row))

当我这样做时，我喜欢对表格的格式细节有一定的控制。特别是，我希望标题单元格与主体单元格具有不同的格式，并且表列的宽度仅为每个列所需的宽度。以下是我的解决方案:

def format_matrix(header, matrix,
                  top_format, left_format, cell_format, row_delim, col_delim):
    table = [[''] + header] + [[name] + row for name, row in zip(header, matrix)]
    table_format = [['{:^{}}'] + len(header) * [top_format]] \
                 + len(matrix) * [[left_format] + len(header) * [cell_format]]
    col_widths = [max(
                      len(format.format(cell, 0))
                      for format, cell in zip(col_format, col))
                  for col_format, col in zip(zip(*table_format), zip(*table))]
    return row_delim.join(
               col_delim.join(
                   format.format(cell, width)
                   for format, cell, width in zip(row_format, row, col_widths))
               for row_format, row in zip(table_format, table))

print format_matrix(['Man Utd', 'Man City', 'T Hotspur', 'Really Long Column'],
                    [[1, 2, 1, -1], [0, 1, 0, 5], [2, 4, 2, 2], [0, 1, 0, 6]],
                    '{:^{}}', '{:<{}}', '{:>{}.3f}', '\n', ' | ')

输出如下:

                   | Man Utd | Man City | T Hotspur | Really Long Column
Man Utd            |   1.000 |    2.000 |     1.000 |             -1.000
Man City           |   0.000 |    1.000 |     0.000 |              5.000
T Hotspur          |   2.000 |    4.000 |     2.000 |              2.000
Really Long Column |   0.000 |    1.000 |     0.000 |              6.000

我发现这只是为了寻找一种输出简单列的方法。如果你只是需要简单的列，那么你可以使用这个:

print("Titlex\tTitley\tTitlez")
for x, y, z in data:
    print(x, "\t", y, "\t", z)

编辑:我试图尽可能简单，因此手动做了一些事情，而不是使用团队列表。概括一下OP的实际问题:

#Column headers
print("", end="\t")
for team in teams_list:
    print(" ", team, end="")
print()
# rows
for team, row in enumerate(data):
    teamlabel = teams_list[team]
    while len(teamlabel) < 9:
        teamlabel = " " + teamlabel
    print(teamlabel, end="\t")
    for entry in row:
        print(entry, end="\t")
    print()

Ouputs:

          Man Utd  Man City  T Hotspur
  Man Utd       1       2       1   
 Man City       0       1       0   
T Hotspur       2       4       2   

但这似乎不再比其他答案简单，好处可能是它不需要更多的进口。但是@campkeith的答案已经满足了这一点，并且更加健壮，因为它可以处理更广泛的标签长度。

一些特别的代码:

row_format ="{:>15}" * (len(teams_list) + 1)
print(row_format.format("", *teams_list))
for team, row in zip(teams_list, data):
    print(row_format.format(team, *row))

这依赖于str.format()和格式规范迷你语言。

Python实际上让这变得非常简单。

类似的

for i in range(10):
    print '%-12i%-12i' % (10 ** i, 20 ** i)

会有输出

1           1           
10          20          
100         400         
1000        8000        
10000       160000      
100000      3200000     
1000000     64000000    
10000000    1280000000  
100000000   25600000000
1000000000  512000000000

字符串中的%本质上是一个转义字符，它后面的字符告诉python数据应该是什么样的格式。字符串外部和后面的%告诉python您打算使用前面的字符串作为格式字符串，并且下面的数据应该放入指定的格式中。

在本例中，我使用了两次“%-12i”。分解每一部分:

'-' (left align)
'12' (how much space to be given to this part of the output)
'i' (we are printing an integer)

来自文档:https://docs.python.org/2/library/stdtypes.html#string-formatting

Pure Python 3

def print_table(data, cols, wide):
    '''Prints formatted data on columns of given width.'''
    n, r = divmod(len(data), cols)
    pat = '{{:{}}}'.format(wide)
    line = '\n'.join(pat * cols for _ in range(n))
    last_line = pat * r
    print(line.format(*data))
    print(last_line.format(*data[n*cols:]))

data = [str(i) for i in range(27)]
print_table(data, 6, 12)

将打印

0           1           2           3           4           5           
6           7           8           9           10          11          
12          13          14          15          16          17          
18          19          20          21          22          23          
24          25          26