如何在python中将整数和字符串列表更改为整数和浮点数列表?

时间:2011-07-22 15:58:25

标签: python string list

我有很多列表的长列表:

[0,['0.2000','0.2000','3.0000','0.5000']]

如何使'...'浮动并保持整数(0)整数?

我尝试过numpy,但它对我不起作用     numpy.array(list,numpy.float)

我真的不介意整数是否也是浮点数。

5 个答案:

答案 0 :(得分:3)

我会做这样的事情:

newlist = [[element[0], [float(e) for e in element[1]] for element in oldlist]

element[0]是保持原样的整数,element[1]是在内部列表推导中转换为浮点数的字符串列表。

答案 1 :(得分:1)

我需要查看更多数据,以确定这是否合适,但可能会接近:

l1 = [0, ['0.2000', '0.2000', '3.0000', '0.5000']]
l2 = [x if type(x) is int else map(float, x) for x in l1]

同样,需要了解有关您的实际数据的更多信息,以确定是否有效。以上回报:

[0, [0.20000000000000001, 0.20000000000000001, 3.0, 0.5]]

答案 2 :(得分:0)

以下是你如何用numpy做到的:

In [96]: l1=[0, ['0.2000', '0.2000', '3.0000', '0.5000']]

In [97]: l2=np.array(tuple(l1),dtype=[('f1','int'),('f2','float',4)])

In [98]: l2
Out[98]: 
array((0, [0.20000000000000001, 0.20000000000000001, 3.0, 0.5]), 
      dtype=[('f1', '<i4'), ('f2', '<f8', 4)])

如果您的实际列表看起来更像这样:

In [99]: l3=[0, ['0.2000', '0.2000', '3.0000', '0.5000'], 1, ['.1','.2','.3','0.4']]

然后你可以使用石斑鱼习语zip(*[iter(l3)]*2))将每2个元素分组为一个元组:

In [104]: zip(*[iter(l3)]*2)
Out[104]: [(0, ['0.2000', '0.2000', '3.0000', '0.5000']), (1, ['.1', '.2', '.3', '0.4'])]

并将其传递给np.array

In [105]: l4=np.array(zip(*[iter(l3)]*2),dtype=[('f1','int'),('f2','float',4)])

In [106]: l4
Out[106]: 
array([(0, [0.20000000000000001, 0.20000000000000001, 3.0, 0.5]),
       (1, [0.10000000000000001, 0.20000000000000001, 0.29999999999999999, 0.40000000000000002])], 
      dtype=[('f1', '<i4'), ('f2', '<f8', 4)])

答案 3 :(得分:0)

简单地

li = [0, ['0.2000', '0.2000', '3.0000', '0.5000']]
print li

li[1] = map(float,li[1])
print li

编辑1

如果 xyze 是:

xyze = [[0, ['0.2000', '0.2000', '3.000' , '0.5000']],
        [9, ['0.1450', '0.8880', '3.000' , '0.4780']],
        [4, ['5.0025', '7.2000', '12.00' , '6.5013']]]

DO

print '\n'.join(map(str,xyze))
print

for el in xyze:
    el[1] = map(float,el[1])

print '\n'.join(map(str,xyze))

结果

[0, ['0.2000', '0.2000', '3.000', '0.5000']]
[9, ['0.1450', '0.8880', '3.000', '0.4780']]
[4, ['5.0025', '7.2000', '12.00', '6.5013']]

[0, [0.2, 0.2, 3.0, 0.5]]
[9, [0.145, 0.888, 3.0, 0.478]]
[4, [5.0025, 7.2, 12.0, 6.5013]]

答案 4 :(得分:0)

如果你想平整嵌套列表......

import types

def process(l):
    for item in l:
        if isinstance(item, types.ListType):
            for item in process(item):
                yield float(item)
        else:
            yield float(item)

>>> l = [0, 1, 2, [3, 4, '5.000'], ['6.01', 7, [8, 9]]]
>>> list( process(l) )
[0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.01, 7.0, 8.0, 9.0]