如何在 Python 中打印“漂亮”的字符串输出

Question

我有一个字典列表，其中包含来自 sql 查询的字段 classid、dept、coursenum、area 和 title。我想以人类可读的格式输出值。我在想每个顶部的列标题，然后在每列中适当的输出，即：

CLASSID     DEPT     COURSE NUMBER        AREA     TITLE
foo         bar      foo                  bar      foo
yoo         hat      yoo                  bar      hat

（显然是标准对齐/间距）

我将如何在 python 中完成此操作？

Answer 1

Standard Python string formatting 可能就足够了。

# assume that your data rows are tuples
template = "{0:8}|{1:10}|{2:15}|{3:7}|{4:10}" # column widths: 8, 10, 15, 7, 10
print template.format("CLASSID", "DEPT", "COURSE NUMBER", "AREA", "TITLE") # header
for rec in your_data_source: 
  print template.format(*rec)

或者

# assume that your data rows are dicts
template = "{CLASSID:8}|{DEPT:10}|{C_NUM:15}|{AREA:7}|{TITLE:10}" # same, but named
print template.format( # header
  CLASSID="CLASSID", DEPT="DEPT", C_NUM="COURSE NUMBER", 
  AREA="AREA", TITLE="TITLE"
) 
for rec in your_data_source: 
  print template.format(**rec)

使用对齐、填充和精确格式说明符以获得最佳结果。

Answer 2

class TablePrinter(object):
    "Print a list of dicts as a table"
    def __init__(self, fmt, sep=' ', ul=None):
        """        
        @param fmt: list of tuple(heading, key, width)
                        heading: str, column label
                        key: dictionary key to value to print
                        width: int, column width in chars
        @param sep: string, separation between columns
        @param ul: string, character to underline column label, or None for no underlining
        """
        super(TablePrinter,self).__init__()
        self.fmt   = str(sep).join('{lb}{0}:{1}{rb}'.format(key, width, lb='{', rb='}') for heading,key,width in fmt)
        self.head  = {key:heading for heading,key,width in fmt}
        self.ul    = {key:str(ul)*width for heading,key,width in fmt} if ul else None
        self.width = {key:width for heading,key,width in fmt}

    def row(self, data):
        return self.fmt.format(**{ k:str(data.get(k,''))[:w] for k,w in self.width.iteritems() })

    def __call__(self, dataList):
        _r = self.row
        res = [_r(data) for data in dataList]
        res.insert(0, _r(self.head))
        if self.ul:
            res.insert(1, _r(self.ul))
        return '\n'.join(res)

并在使用中：

data = [
    {'classid':'foo', 'dept':'bar', 'coursenum':'foo', 'area':'bar', 'title':'foo'},
    {'classid':'yoo', 'dept':'hat', 'coursenum':'yoo', 'area':'bar', 'title':'hat'},
    {'classid':'yoo'*9, 'dept':'hat'*9, 'coursenum':'yoo'*9, 'area':'bar'*9, 'title':'hathat'*9}
]

fmt = [
    ('ClassID',       'classid',   11),
    ('Dept',          'dept',       8),
    ('Course Number', 'coursenum', 20),
    ('Area',          'area',       8),
    ('Title',         'title',     30)
]

print( TablePrinter(fmt, ul='=')(data) )

生产

ClassID     Dept     Course Number        Area     Title                         
=========== ======== ==================== ======== ==============================
foo         bar      foo                  bar      foo                           
yoo         hat      yoo                  bar      hat                           
yooyooyooyo hathatha yooyooyooyooyooyooyo barbarba hathathathathathathathathathat

Answer 3

如果你想保持简单，你可以简单地将字符串左对齐到一定数量的字符：

print string1.ljust(20) + string2.ljust(20)

Answer 4

此函数将列表理解推向了一个极端，但它以最佳性能完成了您正在寻找的内容：

算法：

1. 在每列中查找最长的字段；即'max(map(len, column_vector))'
2. 对于每个字段（从左到右，从上到下），调用 str.ljust 将其与其所属列的左边界对齐。
3. 使用所需数量的分隔空格连接字段（创建一行）。
4. 用换行符加入行集合。

row_collection：可迭代的列表（dicts/sets/lists），每个包含一行的数据。

key_list：指定从每一行读取哪些键/索引以形成列的列表。

def getPrintTable(row_collection, key_list, field_sep=' '*4):
  return '\n'.join([field_sep.join([str(row[col]).ljust(width)
    for (col, width) in zip(key_list, [max(map(len, column_vector))
      for column_vector in [ [v[k]
        for v in row_collection if k in v]
          for k in key_list ]])])
            for row in row_collection])

Answer 5

对我来说最简单的方法是转换为数组数组：

datas = [
  [ 'ClassID', 'Dept', 'Course Number', 'Area', 'Title' ],
  [ 'foo', 'bar', 'foo', 'bar', 'foo' ],
  [ 'yoo', 'hat', 'yoo', 'bar', 'hat' ],
  [ 'line', 'last', 'fun', 'Lisa', 'Simpson' ]
]

for j, data in enumerate(datas):
  if j == 0:
    # print(list(data))
    max = 0
    for i in datas:
      max_len =  len(''.join(i))
      if max_len > max:
        max = max_len
        max = max + 4 * len(datas[0])
    max = 79
    print(f"max is {max}")
    print('+' + '-' * max + '+')
    v1, v2, v3, v4, v5 = datas[0]
    print(f"|{v1:^15s}|{v2:^15s}|{v3:^15s}|{v4:^15s}|{v5:^15s}|")
    print('+' +  '-' * max + '+')
    continue
  else:
    # print( '+' + '-' * max + '+')
    v1, v2, v3, v4, v5 = data
    print(f"|{v1:^15s}|{v2:^15s}|{v3:^15s}|{v4:^15s}|{v5:^15s}|")
    print('+' +  '-' * max + '+')
    # print(type(data))

你明白了：

+-------------------------------------------------------------------------------+
|    ClassID    |     Dept      | Course Number |     Area      |     Title     |
+-------------------------------------------------------------------------------+
|      foo      |      bar      |      foo      |      bar      |      foo      |
+-------------------------------------------------------------------------------+
|      yoo      |      hat      |      yoo      |      bar      |      hat      |
+-------------------------------------------------------------------------------+
|     line      |     last      |      fun      |     Lisa      |    Simpson    |
+-------------------------------------------------------------------------------+