1

我希望有更好的方法来做到这一点。直接上代码：

print "-I- %-6s%-6s%-6s%-6s%-6s%-6s%-6s%-6s%-8s" % \
      ("A",B","C","D","E","F","G","H","% Done")
print "-I- %-6s%-6s%-6s%-6s%-6s%-6s%-6s%-6s%-8s" % \
      ("-"*5 ,"-"*5 ,"-"*5 ,"-"*5 ,"-"*5 ,"-"*5 ,"-"*5,"-"*5,"-"*8)

理想情况下，我想做这样的事情：

hdrs = ["A",B","C","D","E","F","G","H","% Done"]
<print statement that uses len(hdrs[i]+2) for the column width>
<print statement that uses len(hdrs[i]+2) for the column width and len(hdrs[i]+1 for the number of dashes>

输出将如下所示：

A     B     C
----- ----- -----

这种方法将比我目前的方法更具可扩展性。我使用 join 和 map 尝试了各种方法，但我一直无法找到可行的解决方案。任何帮助将不胜感激。

编辑：

我刚刚让这部分工作：

print " ".join("-"*(len(x)+1) for x in hdrs)

上一行代码按照我在原始帖子中要求的方式打印破折号，但我想知道是否有更清洁的方式。我仍然不知道如何打印字符串。

4

4 回答 4

5

如果您只是想完成它而不是将其作为练习，请使用prettytable. 此示例来自链接教程：

x = PrettyTable(["City name", "Area", "Population", "Annual Rainfall"])
x.align["City name"] = "l" # Left align city names
x.padding_width = 1 # One space between column edges and contents (default)
x.add_row(["Adelaide",1295, 1158259, 600.5])
x.add_row(["Brisbane",5905, 1857594, 1146.4])
x.add_row(["Darwin", 112, 120900, 1714.7])
x.add_row(["Hobart", 1357, 205556, 619.5])
x.add_row(["Sydney", 2058, 4336374, 1214.8])
x.add_row(["Melbourne", 1566, 3806092, 646.9])
x.add_row(["Perth", 5386, 1554769, 869.4])
print x

输出：

+-----------+------+------------+-----------------+
| City name | Area | Population | Annual Rainfall |
+-----------+------+------------+-----------------+
| Adelaide  | 1295 |  1158259   |      600.5      |
| Brisbane  | 5905 |  1857594   |      1146.4     |
| Darwin    | 112  |   120900   |      1714.7     |
| Hobart    | 1357 |   205556   |      619.5      |
| Sydney    | 2058 |  4336374   |      1214.8     |
| Melbourne | 1566 |  3806092   |      646.9      |
| Perth     | 5386 |  1554769   |      869.4      |
+-----------+------+------------+-----------------+
于 2013-03-03T16:42:45.700 回答
3

这个怎么样：

hdrs = ("A","B","C","D","E","F","G","H","% Done")
fmt_string = ''.join("%%-%is" % (len(h)+2) for h in hdrs)
print(fmt_string % hdrs)
print(fmt_string % tuple("-"*(len(h)+1) for h in hdrs))

我使用了描述的列大小而不是示例中的列大小。

于 2013-03-03T16:41:49.743 回答
2

您可以像这样构造格式字符串：

format = "".join(["%-"+str(len(h)+2)+"s" for h in hdrs])

然后用它来打印你的列表，例如：

l = range(hdrs) # example data to print, the number of items is the same as hdrs
print format % tuple(l)
于 2013-03-03T16:42:05.260 回答
1

试试这个，在这里你可以指定每列的宽度（不要让字符串大于宽度） 

widths = [6,6,6,6,6,6,6,6,8]
hdrs = ["A","B","C","D","E","F","G","H","% Done"]

data = [hdrs[i].ljust(widths[i]) for i in range(len(hdrs))]
widths = ['-'*i for i in widths]

print '%s '*len(data) % tuple(data)
print ' '.join(widths)

输出

A      B      C      D      E      F      G      H      % Done 
------ ------ ------ ------ ------ ------ ------ ------ --------
于 2013-03-03T16:46:41.813 回答