[英]remove double quotes from first element of the list
我想從此列表中刪除雙引號:
>>> a
["6/1, 0, 0, ['79', '80', '81', '82']"]
我嘗試將list的第一個元素轉換為list,但它是一個字符串,但比較簡單:
>>> a[0]
"6/1, 0, 0, ['79', '80', '81', '82']"
>>> list(a[0])
['6', '/', '1', ',', ' ', '0', ',', ' ', '0', ',', ' ', '[', "'", '7', '9', "'", ',', ' ', "'", '8', '0', "'", ',', ' ', "'", '8', '1', "'", ',', ' ', "'", '8', '2', "'", ']']
預期產量:
>>> a
[6/1, 0, 0, ['79', '80', '81', '82']]
or
>>> a
['6/1', '0', '0', ['79', '80', '81', '82']]
有人可以幫我解決這個問題嗎?
這會將字符串評估為列表:
a = list(eval(a[0]))
[6, 0, 0, ['79', '80', '81', '82']]
如果您知道端口將是6/1,則可以將其替換為以下字符串:
a = list(eval(a[0]))
a[0] = '6/1'
['6/1', 0, 0, ['79', '80', '81', '82']]
如果您的端口將要更改,但是您知道它將始終是第一個元素,則可以執行以下操作:
a = list(eval("'" + a[0].split(',')[0] + "'" + "," + ','.join(a[0].split(',')[1:])))
['6/1', 0, 0, ['79', '80', '81', '82']]
或更清潔的版本:
l = a[0].split(',')
a = list(eval("'" + l[0] + "' ," + ','.join(l[1:])))
['6/1', 0, 0, ['79', '80', '81', '82']]
x= ["6/1, 0, 0, ['79', '80', '81', '82']" ]
a=x[0].replace('\'','').split(', [')
finallist=a[0].split(',')
finallist.append(a[1].replace(']','').split(','))
print (finallist)
輸出
['6/1', ' 0', ' 0', ['79', ' 80', ' 81', ' 82']]
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