使用python将[a b c]转换为[a，b，c]

Question

我想在python3中将[a b c]这样的列表转换为[a,b,c]

我的名单：

[john jr 0, Akbar AK 1001, doctor rd 9999, Mohammedans med 1000, pat mat 200, cajole Jul 21]

我要把这个列表的int加起来！我该怎么做？你知道吗

Answer 1

我英语不是很好，但我想这是你所期望的

L = 'john jr 0, Akbar AK 1001, doctor rd 9999, Mohammedans med 1000, pat mat 200, 

cajole Jul 21'
L = L.split(',')
Sum_L = 0
for i in range(0,len(L)):
  L[i] = (L[i].strip()).split(' ')
  Sum_L = Sum_L + int(L[i][2])
print(Sum_L)

Answer 2

首先，这不起作用。。。你知道吗

a = [john jr 0, Akbar AK 1001, doctor rd 9999, Mohammedans med 1000, pat mat 200, cajole Jul 21]

因为它不是python的有效语法。 所以我想你是说

a = "[john jr 0, Akbar AK 1001, doctor rd 9999, Mohammedans med 1000, pat mat 200, cajole Jul 21]"

我们可以在分离器上把它分成几部分。。。 默认分隔符为空格“” 现在我们的代码是

s = "[john jr 0, Akbar AK 1001, doctor rd 9999, Mohammedans med 1000, pat mat 200, cajole Jul 21]"
s = s.replace("[", "").replace("]", "")
s = s.replace(",", " ")
a = s.split()
print(s)
print(a)

我们会被打印出来

[john jr 0, Akbar AK 1001, doctor rd 9999, Mohammedans med 1000, pat mat 200, cajole Jul 21]
['[john', 'jr', '0,', 'Akbar', 'AK', '1001,', 'doctor', 'rd', '9999,', 'Mohammedans', 'med', '1000,', 'pat', 'mat', '200,', 'cajole', 'Jul', '21]']

有点难读，你可能不想要的领先和尾随括号，所以我们剥离这些。逗号会导致数字解析错误，所以我们用空格替换它们。因为空格是分隔符，我们想要任何像12，hi这样没有空格的部分变成12hi

接下来，让我们找出所有的数字并把它们加起来。我们只需尝试将每个字符串解析为一个数字，看看它是否成功。你知道吗

s = "[john jr 0, Akbar AK 1001, doctor rd 9999, Mohammedans med 1000, pat mat 200, cajole Jul 21]"
s = s.replace("[", "").replace("]", "")
s = s.replace(",", " ")
a = s.split()
print(s)
print(a)
mysum = 0.0
for w in a:
    print("Word is %s" % (w,))
    try:
        n = float(w)
        print("Found number %f" % (n,))
        mysum += n
    except:
        # Ignore any exceptions
        pass
print("Sum is %f" %(mysum,))
pass

我们得到了

john jr 0  Akbar AK 1001  doctor rd 9999  Mohammedans med 1000  pat mat 200  cajole Jul 21
['john', 'jr', '0', 'Akbar', 'AK', '1001', 'doctor', 'rd', '9999', 'Mohammedans', 'med', '1000', 'pat', 'mat', '200', 'cajole', 'Jul', '21']
Word is john
Word is jr
Word is 0
Found number 0.000000
Word is Akbar
Word is AK
Word is 1001
Found number 1001.000000
Word is doctor
Word is rd
Word is 9999
Found number 9999.000000
Word is Mohammedans
Word is med
Word is 1000
Found number 1000.000000
Word is pat
Word is mat
Word is 200
Found number 200.000000
Word is cajole
Word is Jul
Word is 21
Found number 21.000000
Sum is 12221.000000

所以现在我们有了符合你所说的工作代码，它找到了数字并将它们相加。 但是这是可维护的、高效的和功能性的等等。 当我遇到上述问题时，我会使用一些我自己做的方法，使之更容易。 他们来了。你知道吗

def tryParseInt(value=None, default_value=0):
    """
    Try to parse a string value to an int.  Returns the value and True
    e.g.  tryParseInt("42", 7) returns (42, True)
    tryParseInt("abcdef", 7) returns (7, False)
    See twit_test.py
    """
    try:
        return int(value), True
    except (ValueError, TypeError):
        return default_value, False


def tryParseFloat(value=None, default_value=0.0):
    """
    Try to parse a string value to an float.  Returns the value and True
    e.g.  tryParseInt("42.42", 7.3) returns (42.42, True)
    tryParseInt("abcdef", 7.3) returns (7.3, False)
    See twit_test.py
    """
    try:
        return float(value), True
    except (ValueError, TypeError):
        return default_value, False


def split_strip(s: str, sep=' '):
    """ 
    Split s into parts on delimiter, then strip the sub strings and remove any blanks.
    Never returns None.
    Returns an array of the sub strings.  The returned array my be empty.
    See twit_test.py for examples.
    """
    if s is None:
        return []
    parts = s.split(sep=sep)
    ret = []
    if parts is not None:
        for ss in parts:
            sss = ss.strip()
            if len(sss) > 0:
                ret.append(sss)
    return ret

以上帮助程序在GitHib项目https://github.com/RMKeene/twit上可用