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>
输出看起来像这样:
A B C
----- ----- -----
此方法比我当前的方法更具可扩展性。我尝试过使用 join 和 map 的各种方法，但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)
前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简洁的方法。我仍然不知道如何打印字符串。
请您参考如下方法:
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 |
+-----------+------+------------+-----------------+
