我有一个文件,其中有一些看起来像这样的行:
[16, 1, 4, 15][0, 4, 5, 14][8, 9, 10, 3][2, 11, 12, 6][0, 1, 10, 11][1, 19, 12, 14][19, 3, 13, 15][9, 17, 14, 15][9, 2, 18, 17][8, 2, 13, 7][4, 2, 19, 12][16, 18, 3, 4][10, 3, 5, 15][16, 9, 18, 6][1, 19, 5, 7][0, 12, 6, 7][0, 17, 11, 13][16, 8, 18, 7][8, 17, 11, 13][10, 6, 5, 14]
我希望阅读它们,并列出每行的列表。
我已尝试split()
功能,但无效。
我尝试的是:
file = open(filename, 'r')
string.split(',')
print(string[3])
但它返回,
,而不是[2, 11, 12, 6]
任何猜测?提前谢谢!
答案 0 :(得分:3)
您可以使用json
和列表理解
import json
line = [16, 1, 4, 15][0, 4, 5, 14][8, 9, 10, 3][2, 11, 12, 6][0, 1, 10, 11][1, 19, 12, 14][19, 3, 13, 15][9, 17, 14, 15][9, 2, 18, 17][8, 2, 13, 7][4, 2, 19, 12][16, 18, 3, 4][10, 3, 5, 15][16, 9, 18, 6][1, 19, 5, 7][0, 12, 6, 7][0, 17, 11, 13][16, 8, 18, 7][8, 17, 11, 13][10, 6, 5, 14]
lst = [json.loads(sublist+']') for sublist in line.split(']') if sublist]
#[[16, 1, 4, 15],
# [0, 4, 5, 14],
# [8, 9, 10, 3],
# [2, 11, 12, 6],
# [0, 1, 10, 11],
# [1, 19, 12, 14],
# [19, 3, 13, 15],
# [9, 17, 14, 15],
# [9, 2, 18, 17],
# [8, 2, 13, 7],
# [4, 2, 19, 12],
# [16, 18, 3, 4],
# [10, 3, 5, 15],
# [16, 9, 18, 6],
# [1, 19, 5, 7],
# [0, 12, 6, 7],
# [0, 17, 11, 13],
# [16, 8, 18, 7],
# [8, 17, 11, 13],
# [10, 6, 5, 14]]
在此代码中,我根据']'分割了这一行,这为我提供了一个字符串列表,如'[16, 1, 4, 15'
,'[0, 4, 5, 14'
,...然后为每个字符串这些字符串,我添加结束括号并使用json
来解释它并将其转换为列表。
答案 1 :(得分:2)
在另一个网站上,我得到了一个有效的答案(我不知道这是否是一个很好的方法,但它适用于我)。
我文件中的典型行如下所示:
[16, 1, 4, 15][0, 4, 5, 14][8, 9, 10, 3][2, 11, 12, 6][0, 1, 10, 11][1, 19, 12, 14][19, 3, 13, 15][9, 17, 14, 15][9, 2, 18, 17][8, 2, 13, 7][4, 2, 19, 12][16, 18, 3, 4][10, 3, 5, 15][16, 9, 18, 6][1, 19, 5, 7][0, 12, 6, 7][0, 17, 11, 13][16, 8, 18, 7][8, 17, 11, 13][10, 6, 5, 14]
是字符串,不是列表。我想从那个str中列出一个列表。
我的代码现在看起来像这样:
line=file.readline() # stores the str line from the file
line = '[' + line + ']'
line = line.replace('][', '],[')
line = ast.literal_eval(line)
现在我可以访问(大)列表中的每个列表,以及每个列表中的每个值。
答案 2 :(得分:0)
如果string
是您在上面给出的行,则使用列表推导的单行解决方案是:
[[int(s) for s in t.split(',')] for t in string.strip()[1:-1].split('][')]
喜欢这样:
>>> string = '[16, 1, 4, 15][0, 4, 5, 14][8, 9, 10, 3][2, 11, 12, 6][0, 1, 10, 11][1, 19, 12, 14][19, 3, 13, 15][9, 17, 14, 15][9, 2, 18, 17][8, 2, 13, 7][4, 2, 19, 12][16, 18, 3, 4][10, 3, 5, 15][16, 9, 18, 6][1, 19, 5, 7][0, 12, 6, 7][0, 17, 11, 13][16, 8, 18, 7][8, 17, 11, 13][10, 6, 5, 14]\n'
>>> [[int(s) for s in t.split(',')] for t in string.strip()[1:-1].split('][')]
[[16, 1, 4, 15], [0, 4, 5, 14], [8, 9, 10, 3], [2, 11, 12, 6], [0, 1, 10, 11], [1, 19, 12, 14], [19, 3, 13, 15], [9, 17, 14, 15], [9, 2, 18, 17], [8, 2, 13, 7], [4, 2, 19, 12], [16, 18, 3, 4], [10, 3, 5, 15], [16, 9, 18, 6], [1, 19, 5, 7], [0, 12, 6, 7], [0, 17, 11, 13], [16, 8, 18, 7], [8, 17, 11, 13], [10, 6, 5, 14]]
这最后显然是整数列表的列表,而不是字符串,如以下输出所示:
>>> [sum(nums) for nums in [[int(s) for s in t.split(',')] for t in string.strip()[1:-1].split('][')]]
[36, 23, 30, 31, 22, 46, 50, 55, 46, 30, 37, 41, 33, 49, 32, 25, 41, 49, 49, 35]
答案 3 :(得分:0)
s='[16, 1, 4, 15][0, 4, 5, 14][8, 9, 10, 3][2, 11, 12, 6][0, 1, 10, 11][1, 19, 12, 14][19, 3, 13, 15][9, 17, 14, 15][9, 2, 18, 17][8, 2, 13, 7][4, 2, 19, 12][16, 18, 3, 4][10, 3, 5, 15][16, 9, 18, 6][1, 19, 5, 7][0, 12, 6, 7][0, 17, 11, 13][16, 8, 18, 7][8, 17, 11, 13][10, 6, 5, 14]'
[l.split(',') for l in s[1:-1].split('][')]
答案 4 :(得分:0)
如果您不想使用额外的模块,可以使用列表推导和字符串split
和strip
方法来执行此操作:
[[int(s.strip()) for s in sublist.split(',')] for sublist in line[1:-1].split('][')]
#[[16, 1, 4, 15],
# [0, 4, 5, 14],
# [8, 9, 10, 3],
# [2, 11, 12, 6],
# [0, 1, 10, 11],
# [1, 19, 12, 14],
# [19, 3, 13, 15],
# [9, 17, 14, 15],
# [9, 2, 18, 17],
# [8, 2, 13, 7],
# [4, 2, 19, 12],
# [16, 18, 3, 4],
# [10, 3, 5, 15],
# [16, 9, 18, 6],
# [1, 19, 5, 7],
# [0, 12, 6, 7],
# [0, 17, 11, 13],
# [16, 8, 18, 7],
# [8, 17, 11, 13],
# [10, 6, 5, 14]]
答案 5 :(得分:0)
您正在阅读的字符串几乎准备就绪:
修改你的字符串,就像我从你的问题中看到的那样,你在变量string
中修改,然后用json.loads
或ast.literal_eval
import json # or ast
parse = json.loads
# or
# parse = ast.literal_eval
new_string = parse("".join(["[", string.replace("][", "],["), "]"])