我有一个dicts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。我在每个顶部想到一个Column标题,然后在每个列中考虑approrpiate输出,即:
CLASSID DEPT COURSE NUMBER AREA TITLE
foo bar foo bar foo
yoo hat yoo bar hat
(显然标准对齐/间距)
我如何在python中完成此任务?
答案 0 :(得分:56)
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来获得最佳效果。
答案 1 :(得分:11)
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) )
产生
ClassID Dept Course Number Area Title
=========== ======== ==================== ======== ==============================
foo bar foo bar foo
yoo hat yoo bar hat
yooyooyooyo hathatha yooyooyooyooyooyooyo barbarba hathathathathathathathathathat
答案 2 :(得分:5)
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)
答案 3 :(得分:4)
这个函数将列表理解有点极端,但它以最佳性能完成了你想要的东西:
算法:
row_collection:iterables列表(dicts / sets / lists),每个包含一行的数据。
key_list:list,指定从每行读取哪些键/索引以形成列。
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])
答案 4 :(得分:0)
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 |
+-------------------------------------------------------------------------------+