Question

我有这个：

[s[8] = 5,

 s[4] = 3,

 s[19] = 2,

 s[17] = 8,

 s[16] = 8,

 s[2] = 8,

 s[9] = 7,

 s[1] = 2,

 s[3] = 9,

 s[15] = 7,

 s[11] = 0,

 s[10] = 9,

 s[12] = 3,

 s[18] = 1,

 s[0] = 4,

 s[14] = 5,

 s[7] = 4,

 s[6] = 2,

 s[5] = 7,

 s[13] = 9]

如何将其转换为python数组，我可以在其中for items in x:？

Answer 1

import re

data = """[s[8] = 5,

 s[4] = 3,

 s[19] = 2,

 s[17] = 8,

 s[16] = 8,

 s[2] = 8,

 s[9] = 7,

 s[1] = 2,

 s[3] = 9,

 s[15] = 7,

 s[11] = 0,

 s[10] = 9,

 s[12] = 3,

 s[18] = 1,

 s[0] = 4,

 s[14] = 5,

 s[7] = 4,

 s[6] = 2,

 s[5] = 7,

 s[13] = 9]"""

d = {int(m.group(1)): int(m.group(2)) for m in re.finditer(r"s\[(\d*)\] = (\d*)", data)}
seq = [d.get(x) for x in range(max(d))]
print(seq)
#result: [4, 2, 8, 9, 3, 7, 2, 4, 5, 7, 9, 0, 3, 9, 5, 7, 8, 8, 1]

Answer 2

如果希望数组与输入字符串中的名称相同，则可以使用exec。这不是非常pythonic，但它适用于简单的东西

string = ("[s[8] = 5, s[4] = 3, s[19] = 2,"
   "s[17] = 8, s[16] = 8, s[2] = 8,"
   "s[9] = 7,  s[1] = 2,  s[3] = 9,"
   "s[15] = 7, s[11] = 0, s[10] = 9,"
   "s[12] = 3, s[18] = 1, s[0] = 4,"
   "s[14] = 5, s[7] = 4,  s[6] = 2,"
   "s[5] = 7, s[13] = 9]")

items = [item.rstrip().lstrip() for item in string[1:-1].split(",")]
name = items[0].partition("[")[0]
# Create the array
exec("{} = [None] * {}".format(name, len(items)))
# Populate with the values of the string
for item in items:
    exec(items[0].partition("[")[0] )

这将生成一个名为“s”的数组，如果缺少一个索引，它将被初始化为None

Answer 3

假设这是一个巨大的字符串，如果您希望print s[0]打印4，那么您需要用逗号分隔它，然后遍历每个项目。

inputArray = yourInput[1:-1].replace(' ','').split(',\n\n')
endArray = [0]*20
for item in inputArray:
    endArray[int(item[item.index('[')+1:item.index(']')])]= int(item[item.index('=')+1:])
print endArray

Answer 4

一种简单的方法，虽然效率低于凯文的解决方案（不使用正则表达式），但如下所示（其中some_array是你的字符串）：

sub_list = some_array.split(',')
some_dict = {}

for item in sub_list:
    sanitized_item = item.strip().rstrip().lstrip().replace('=', ':')

    # split item in key val
    k = sanitized_item.split(':')[0].strip()
    v = sanitized_item.split(':')[1].strip()

    if k.startswith('['):
        k = k.replace('[', '')

    if v.endswith(']'):
        v = v.replace(']', '')

    some_dict.update({k: int(v)})

print(some_dict)
print(some_dict['s[9]'])

输出样本：

{'s[5]': 7, 's[16]': 8, 's[0]': 4, 's[9]': 7, 's[2]': 8, 's[3]': 9, 's[10]': 9, 's[15]': 7, 's[6]': 2, 's[7]': 4, 's[14]': 5, 's[19]': 2, 's[17]': 8, 's[4]': 3, 's[12]': 3, 's[11]': 0, 's[13]': 9, 's[18]': 1, 's[1]': 2, 's8]': 5}
7

将字符串类型数组转换为数组

4 个答案: