所以我有這個清單：
[
'92\nMetaverse Psychedelics Anonymous Pass\n44.6\n 1185.27%\n---\n0.08\n6.4K\n9.5K',
'100\nCryptoDickbutts\n41.42\n 896.92%\n 105.73%\n0.75\n1.5K\n5.2K',
'20\nArt Blocks Factory\n355.01\n 798.96%\n 122.79%\n---\n20.4K\n79.3K'
]
我需要幫助找到一種在每個“\n”處拆分串列項的好方法。拆分不起作用，因為我假設它沒有將 \n 注冊為字串的一部分......
uj5u.com熱心網友回復：
我不是 100% 確定結果應該是什么，但這是一個想法：
>>> t = [
... '92\nMetaverse Psychedelics Anonymous Pass\n44.6\n 1185.27%\n---\n0.08\n6.4K\n9.5K',
... '100\nCryptoDickbutts\n41.42\n 896.92%\n 105.73%\n0.75\n1.5K\n5.2K',
... '20\nArt Blocks Factory\n355.01\n 798.96%\n 122.79%\n---\n20.4K\n79.3K'
... ]
>>> for i in t:
... print(i.splitlines())
...
['92', 'Metaverse Psychedelics Anonymous Pass', '44.6', ' 1185.27%', '---', '0.08', '6.4K', '9.5K']
['100', 'CryptoDickbutts', '41.42', ' 896.92%', ' 105.73%', '0.75', '1.5K', '5.2K']
['20', 'Art Blocks Factory', '355.01', ' 798.96%', ' 122.79%', '---', '20.4K', '79.3K']
>>> p = [x for i in t for x in i.splitlines()]
>>>
>>> p
['92', 'Metaverse Psychedelics Anonymous Pass', '44.6', ' 1185.27%', '---', '0.08', '6.4K', '9.5K', '100', 'CryptoDickbutts', '41.42', ' 896.92%', ' 105.73%', '0.75', '1.5K', '5.2K', '20', 'Art Blocks Factory', '355.01', ' 798.96%', ' 122.79%', '---', '20.4K', '79.3K']
uj5u.com熱心網友回復：
你可以這樣做
my_list = [
'92\nMetaverse Psychedelics Anonymous Pass\n44.6\n 1185.27%\n---\n0.08\n6.4K\n9.5K',
'100\nCryptoDickbutts\n41.42\n 896.92%\n 105.73%\n0.75\n1.5K\n5.2K',
'20\nArt Blocks Factory\n355.01\n 798.96%\n 122.79%\n---\n20.4K\n79.3K'
]
new_list = []
for string in my_list:
# .extend is like append but it appends all items of the list which in this case is the string split by "\n"
new_list.extend(string.split("\n"))
print(new_list)
它遍歷字串串列并按換行符拆分它們。
uj5u.com熱心網友回復：
您可以使用此代碼
my_list = [
'92\nMetaverse Psychedelics Anonymous Pass\n44.6\n 1185.27%\n---\n0.08\n6.4K\n9.5K',
'100\nCryptoDickbutts\n41.42\n 896.92%\n 105.73%\n0.75\n1.5K\n5.2K',
'20\nArt Blocks Factory\n355.01\n 798.96%\n 122.79%\n---\n20.4K\n79.3K'
]
new_list = []
for item in my_list:
new = item.split('\n')
new_list.append(' '.join(new))
那么我想你會有你需要的
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