Question

以这种格式提供字符串数组：

stringList = ['(0.0,[0.0,0.0,0.0])\n' , '(1.0,[1.0,0.0,0.0])\n']

如何转换为：

inputs = [0.0 , 1.0]
outputs = [ [0.0,0.0,0.0] , [1.0,0.0,0.0] ]

我最近的是：

inputs = []
outputs = []

for l in stringList : 
    inputs.append(np.array(float(l.split(',')[0].replace('(' , ''))))

for l in stringList : 
     outputs.append(float(l.split(',')[1].replace('[' , '')))

print(inputs)
print(outputs)

打印哪些：

[array(0.0), array(1.0)]
[0.0, 1.0]

如何转换为：

inputs = [0.0 , 1.0]
outputs = [ [0.0,0.0,0.0] , [1.0,0.0,0.0] ]

Answer 1

您的字符串是元组格式。您可以使用ast.literal_eval将其评估为元组，然后将结果相应地附加到输出：

stringList = ['(0.0,[0.0,0.0,0.0])\n' , '(1.0,[1.0,0.0,0.0])\n']
inputs = []
outputs = []

from ast import literal_eval
for s in stringList:
    t = literal_eval(s)
    inputs.append(t[0])
    outputs.append(t[1])

inputs
# [0.0, 1.0]

outputs
# [[0.0, 0.0, 0.0], [1.0, 0.0, 0.0]]

或者是较短的版本，您可以在zip之后使用literal_eval：

inputs, outputs = zip(*map(literal_eval, stringList))

Answer 2

其他答案不是pythonic，但这是我的尝试：

<强>代码：

stringList = ['(0.0,[0.0,0.0,0.0])\n' , '(1.0,[1.0,0.0,0.0])\n']

inp = []
outp = []

for i in stringList:    
    x = i.split(',[')[0].replace("(", "").replace("\n", "")
    y = i.split(',[')[1].replace("])", "").replace("\n", "").split(',')
    z = [float(i) for i in y]

    inp.append(float(x))
    outp.append(z)

print(inp)
print(outp)

<强>输出：

[0.0, 1.0]
[[0.0, 0.0, 0.0], [1.0, 0.0, 0.0]]

Answer 3

Here's another option, using split and some character removal:

import re

inputs = []
outputs = []

for s in stringList:
    # delete (, ), [, ], \n characters from string
    tmp = re.sub('[\(\)\n\[\]]', '', s) 

    # first float to inputs, others to outputs
    tmp_i, *tmp_o = [float(i) for i in tmp.split(',')]

    inputs.append(tmp_i)
    outputs.append(tmp_o)

EDIT

in case you're still using Python 2, change the assignment of tmp_i and tmp_o:

    tmp_list = [float(i) for i in tmp.split(',')]

    inputs.append(tmp_list[0])
    outputs.append(tmp_list[1:])

Answer 4

你可以使用这个

stringList = ['(0.0,[0.0,0.0,0.0])\n' , '(1.0,[1.0,0.0,0.0])\n']
inp_list = map(lambda x: eval(x)[0], stringList)
op_list = map(lambda x: eval(x)[1], stringList)

输出：

print inp_list # [0.0, 1.0]
print op_list # [[0.0, 0.0, 0.0], [1.0, 0.0, 0.0]]

将元组的字符串数组转换为输入和输出

4 个答案:

EDIT