Python 如何按小数点位置对齐数字?
我的桌子:Python 如何按小数点位置对齐数字?,python,python-3.x,alignment,numeric,Python,Python 3.x,Alignment,Numeric,我的桌子: 2 4,5 100 400 3 20 10 100 4,2 10.000 500 600 9,2 10.024,5 610 1.100 mylist = [['2', '4,5', '100', '400'], ['3', '20', '10', '100'], ['4,2', '10.000', '500', '600'], ['9,2', '10.024,5', '610', '1.100']]
2 4,5 100 400
3 20 10 100
4,2 10.000 500 600
9,2 10.024,5 610 1.100
mylist = [['2', '4,5', '100', '400'], ['3', '20', '10', '100'], ['4,2', '10.000', '500', '600'], ['9,2', '10.024,5', '610', '1.100']]
我怎样才能像这样排列数字:
2 4,5 100 400
3 20 10 100
4,2 10.000 500 600
9,2 10.024,5 610 1.100
我所做的:
for i in range(0, len(mylist)):
'{:>10} {:>10} {:>10} {:>10}'.format(mylist[i][0], mylist[i][1], mylist[i][2], mylist[i][3])
上述结果如表1所示,但小数未对齐
更新我根据列表中元素的长度调整了Francisco的解决方案
mylist = [['1300', '300', '300'], ['2300,5', '300,5', '300,5'], ['3600000000000', '6000000', '6000000'], ['4200,2345', '200,2345', '200,2345'], ['5301', '301', '301'], ['6200', '200', '200'], ['7300', '300', '300000000'], ['8300,5', '300,500000', '300,5'], ['960000', '600', '6000'], ['10200,2345', '200999,2345', '200,2345'], ['11301', '301', '301'], ['12200', '200', '200'], ['13300', '300', '300'], ['14300,5', '300,5', '300,5']]
column2row = zip(*mylist)
maxbefore = []
maxafter = []
for m in column2row:
maxlengthbefore = 0
maxlengthafter = 0
for e in m:
if ',' in e:
lengthbefore = len(str(e).split(",")[0])
if lengthbefore > maxlengthbefore:
maxlengthbefore = lengthbefore
lengthafter = len(str(e).split(",")[1])
if lengthafter > maxlengthafter:
maxlengthafter = lengthafter
else:
lengthbefore = len(str(e))
if lengthbefore > maxlengthbefore:
maxlengthbefore = lengthbefore
maxbefore.append(maxlengthbefore)
maxafter.append(maxlengthafter + 1)
for m in mylist:
r = 0
for e in m:
if ',' in e:
print(e.rjust(int(maxbefore[r]) + len(e) - e.index(',')).ljust(int(maxbefore[r])+int(maxafter[r])), end=' ')
else:
print(e.rjust(int(maxbefore[r])).ljust(int(maxbefore[r])+int(maxafter[r])), end=' ')
r += 1
print()
输出
1300 300 300
2300,5 300,5 300,5
3600000000000 6000000 6000000
4200,2345 200,2345 200,2345
5301 301 301
6200 200 200
7300 300 300000000
8300,5 300,500000 300,5
960000 600 6000
10200,2345 200999,2345 200,2345
11301 301 301
12200 200 200
13300 300 300
14300,5 300,5 300,5
印刷品:
2 4,5 100 400
3 20 10 100
4,2 10.000 500 600
9,2 10.024,5 610 1.100
基本上,它所做的是检查字符串中是否有逗号
如果它这样做了,它会检查它的位置,将字符串居中到右边,检查有多少个字符,而不计算逗号后面的所有字符,然后将它居中到左边,这次计算逗号后面的位置
如果字符串中没有逗号,它只是将字符串左右居中。您似乎将
,
视为十进制分隔符,但最后一列的最后一行中有一个
。。。我认为
在这里没有得到特殊处理?@mgilson,我使用西欧格式。小数='、'和千='。'Francisco,它在我的系统(python3)上不起作用。很多错误。4,2(无,)10.000(无,)500(无,)600(无,)
很抱歉,我用Python3实现了这一点。我想我的问题没有解决方案?你从来不会在循环中增加r
,你必须更改maxafter.append(maxlengteafter)
tomaxafter.append(maxlengteafter+1)
看起来不错
2 4,5 100 400
3 20 10 100
4,2 10.000 500 600
9,2 10.024,5 610 1.100