以表格格式漂亮地打印列表 [英] Pretty printing a list in a tabular format
问题描述
使用Python 2.4，如何以一种良好的表格格式打印列表?
我的列表采用以下格式.
mylist=[(('VAL1', 'VAL2', 'VAL3', 'VAL4', 'VAL5', 'VAL6'), AGGREGATE_VALUE)]
我已经尝试过pprint，但是它不会以表格格式打印结果.
:我希望以以下格式查看输出:
VAL1        VAL2      VAL3     VAL4     VAL5    VAL6 AGGREGATE_VALUE
此表应说明可变的项目长度，并且仍应使用适当的缩进进行打印.
mylist = [ ( ('12', '47', '4', '574862', '58', '7856'), 'AGGREGATE_VALUE1'),
( ('2', '75', '757', '8233', '838', '47775272785'), 'AGGREG2'),
( ('4144', '78', '78965', '778', '78578', '2'), 'AGGREGATE_VALUE3')]
longg = dict.fromkeys((0,1,2,3,4,5,6),0)
for tu,x in mylist:
for i,el in enumerate(tu):
longg[i] = max(longg[i],len(str(el)))
longg[6] = max(longg[6],len(str(x)))
fofo = ' '.join('%'+str(longg[i])+'s' for i in xrange(0,7))
print '\n'.join(fofo % (a,b,c,d,e,f,g) for (a,b,c,d,e,f),g in mylist)
结果
12 47 4 574862 58 7856 AGGREGATE_VALUE1
2 75 757 8233 838 47775272785 AGGREG2
4144 78 78965 778 78578 2 AGGREGATE_VALUE3
不知道这是否满足您的需求
编辑1
使用带模运算符(％)的字符串格式以恒定长度打印，'％6s'右对齐恒定长度6，'％-6s'以6的恒定长度左对齐.
您会在此处
但是没有必要指定一个恒定的长度来在字符串的末尾打印内容，因为在这种情况下，它在某种程度上是自然左对齐的. 然后:
longg = dict.fromkeys((0,1,2,3,4,5,),0)
for tu,x in mylist:
for i,el in enumerate(tu):
longg[i] = max(longg[i],len(str(el)))
fofo = ' '.join('%'+str(longg[i])+'s' for i in xrange(0,6)) + ' %s'
print '\n'.join(fofo % (a,b,c,d,e,f,g) for (a,b,c,d,e,f),g in mylist)
编辑2
mylist = [ ( (12, 47, 4, 574862, 58, 7856), 'AGGREGATE_VALUE1'),
( (2, 75, 757, 8233, 838, 47775272785), 'AGGREG2'),
( (4144, 78, 78965, 778, 78578, 2), 'AGGREGATE_VALUE3')]
longg = dict.fromkeys((0,1,2,3,4,5),0)
for tu,_ in mylist:
longg.update(( i, max(longg[i],len(str(el))) ) for i,el in enumerate(tu))
fofo = ' '.join('%%%ss' % longg[i] for i in xrange(0,6)) + ' %s'
print '\n'.join(fofo % (a,b,c,d,e,f,g) for (a,b,c,d,e,f),g in mylist)
编辑3
mylist = [ ( (12, 47, 4, 574862, 58, 7856), 'AGGREGATE_VALUE1'),
( (2, 75, 757, 8233, 838, 47775272785), 'AGGREG2'),
( (4144, 78, 78965, 778, 78578, 2), 'AGGREGATE_VALUE3')]
header = ('Price1','Price2','reference','XYD','code','resp','AGGREG values')
longg = dict(zip((0,1,2,3,4,5,6),(len(str(x)) for x in header)))
for tu,x in mylist:
longg.update(( i, max(longg[i],len(str(el))) ) for i,el in enumerate(tu))
longg[6] = max(longg[6],len(str(x)))
fofo = ' | '.join('%%-%ss' % longg[i] for i in xrange(0,7))
print '\n'.join((fofo % header,
'-|-'.join( longg[i]*'-' for i in xrange(7)),
'\n'.join(fofo % (a,b,c,d,e,f,g) for (a,b,c,d,e,f),g in mylist)))
结果
Price1 | Price2 | reference | XYD | code | resp | AGGREG values
-------|--------|-----------|--------|-------|-------------|-----------------
12 | 47 | 4 | 574862 | 58 | 7856 | AGGREGATE_VALUE1
2 | 75 | 757 | 8233 | 838 | 47775272785 | AGGREG2
4144 | 78 | 78965 | 778 | 78578 | 2 | AGGREGATE_VALUE3
请注意，使用Python 2.6中引入的字符串方法 format()，这种格式化会容易得多
Using Python 2.4, how do I print a list in a nice tabular format?
My list is in the below format.
mylist=[(('VAL1', 'VAL2', 'VAL3', 'VAL4', 'VAL5', 'VAL6'), AGGREGATE_VALUE)]
I have tried pprint, but it does not print the result in a tabular format.
EDIT : I would like to see the output in the below format:
VAL1 VAL2 VAL3 VAL4 VAL5 VAL6 AGGREGATE_VALUE
This table, should account for variable item lengths and still print with proper indentation.
mylist = [ ( ('12', '47', '4', '574862', '58', '7856'), 'AGGREGATE_VALUE1'),
( ('2', '75', '757', '8233', '838', '47775272785'), 'AGGREG2'),
( ('4144', '78', '78965', '778', '78578', '2'), 'AGGREGATE_VALUE3')]
longg = dict.fromkeys((0,1,2,3,4,5,6),0)
for tu,x in mylist:
for i,el in enumerate(tu):
longg[i] = max(longg[i],len(str(el)))
longg[6] = max(longg[6],len(str(x)))
fofo = ' '.join('%'+str(longg[i])+'s' for i in xrange(0,7))
print '\n'.join(fofo % (a,b,c,d,e,f,g) for (a,b,c,d,e,f),g in mylist)
result
12 47 4 574862 58 7856 AGGREGATE_VALUE1
2 75 757 8233 838 47775272785 AGGREG2
4144 78 78965 778 78578 2 AGGREGATE_VALUE3
Don't know if this fills your need
EDIT 1
Using string formatting with modulo operator (%) to print in a constant length, '%6s' right-justifies in a constant length of 6, and '%-6s' left-justifies in a constant length of 6.
You'll find precisions here
But there is no sense to specify a constant length to print something at the end of a string, because it's somewhat naturally-left-justified in this case. Then :
longg = dict.fromkeys((0,1,2,3,4,5,),0)
for tu,x in mylist:
for i,el in enumerate(tu):
longg[i] = max(longg[i],len(str(el)))
fofo = ' '.join('%'+str(longg[i])+'s' for i in xrange(0,6)) + ' %s'
print '\n'.join(fofo % (a,b,c,d,e,f,g) for (a,b,c,d,e,f),g in mylist)
EDIT 2
mylist = [ ( (12, 47, 4, 574862, 58, 7856), 'AGGREGATE_VALUE1'),
( (2, 75, 757, 8233, 838, 47775272785), 'AGGREG2'),
( (4144, 78, 78965, 778, 78578, 2), 'AGGREGATE_VALUE3')]
longg = dict.fromkeys((0,1,2,3,4,5),0)
for tu,_ in mylist:
longg.update(( i, max(longg[i],len(str(el))) ) for i,el in enumerate(tu))
fofo = ' '.join('%%%ss' % longg[i] for i in xrange(0,6)) + ' %s'
print '\n'.join(fofo % (a,b,c,d,e,f,g) for (a,b,c,d,e,f),g in mylist)
EDIT 3
mylist = [ ( (12, 47, 4, 574862, 58, 7856), 'AGGREGATE_VALUE1'),
( (2, 75, 757, 8233, 838, 47775272785), 'AGGREG2'),
( (4144, 78, 78965, 778, 78578, 2), 'AGGREGATE_VALUE3')]
header = ('Price1','Price2','reference','XYD','code','resp','AGGREG values')
longg = dict(zip((0,1,2,3,4,5,6),(len(str(x)) for x in header)))
for tu,x in mylist:
longg.update(( i, max(longg[i],len(str(el))) ) for i,el in enumerate(tu))
longg[6] = max(longg[6],len(str(x)))
fofo = ' | '.join('%%-%ss' % longg[i] for i in xrange(0,7))
print '\n'.join((fofo % header,
'-|-'.join( longg[i]*'-' for i in xrange(7)),
'\n'.join(fofo % (a,b,c,d,e,f,g) for (a,b,c,d,e,f),g in mylist)))
result
Price1 | Price2 | reference | XYD | code | resp | AGGREG values
-------|--------|-----------|--------|-------|-------------|-----------------
12 | 47 | 4 | 574862 | 58 | 7856 | AGGREGATE_VALUE1
2 | 75 | 757 | 8233 | 838 | 47775272785 | AGGREG2
4144 | 78 | 78965 | 778 | 78578 | 2 | AGGREGATE_VALUE3
Note that this kind of formatting would be much easier with the string's method format() introduced in Python 2.6
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