如何在forLOOP中使用2種不同的方法拆分串列？

如果我無法正確提出問題，我很抱歉。但這是我的代碼

data1 = ['TOOK22JAN1515100HG','BOGGOK22MAR1742200HG']
data2 = ['TOOK2231515100HG','BOGGOK2221643200GH']

for i in data1:
  splt_1 = re.split(r'(TOOK|BOGGOK)([0-9]{2})(JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)([0-9]{2})([0-9]{5})(HG|GH)', i)
  print('data1:', splt_1)

for I in data2:
  splt_2 = re.split(r'(TOOK|BOGGOK)([0-9]{5})([0-9]{5})(HG|GH)', i)
  print('data2:', splt_2)

輸出結果

data1: ['', 'TOOK', '22', 'JAN', '15', '15100', 'HG', '']
data1: ['', 'BOGGOK', '22', 'MAR', '17', '42200', 'HG', '']

data2: ['', 'TOOK', '22315', '15100', 'HG', '']
data2: ['', 'BOGGOK', '22216', '43200', 'GH', '']

我想做的事？

如果

data = ['TOOK22JAN1515100HG','BOGGOK22MAR1742200HG', 'TOOK2231515100HG','BOGGOK2221643200GH']

我希望能夠使用 2 方法回圈和拆分資料串列

re.split(r'(TOOK|BOGGOK)([0-9]{5})([0-9]{5})(HG|GH)', i) or
re.split(r'(TOOK|BOGGOK)([0-9]{2})(JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)([0-9]

PS：輸出結果可以是相同或相似的格式

我試過這個代碼

data5 = ['TOOK22JAN1515100HG','BOGGOK22MAR1742200HG','TOOK2231515100HG','BOGGOK2221643200GH']

for i in data5:
  dk = re.split(r'(TOOK|BOGGOK)([0-9]{2})(JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)([0-9]{2})([0-9]{5})(HG|GH)|(TOOK|BOGGOK)([0-9]{5})([0-9]{5})(HG|GH)', i)
  print(dk)

結果

['', 'TOOK', '22', 'JAN', '15', '15100', 'HG', None, None, None, None, '']
['', 'BOGGOK', '22', 'MAR', '17', '42200', 'HG', None, None, None, None, '']
['', None, None, None, None, None, None, 'TOOK', '22315', '15100', 'HG', '']
['', None, None, None, None, None, None, 'BOGGOK', '22216', '43200', 'GH', '']

我想要的結果

['', 'TOOK', '22', 'JAN', '15', '15100', 'HG', '']
['', 'BOGGOK', '22', 'MAR', '17', '42200', 'HG', '']
['', 'TOOK', '22315', '15100', 'HG', '']
['', 'BOGGOK', '22216', '43200', 'GH', '']

or

['TOOK', '22', 'JAN', '15', '15100', 'HG']
['BOGGOK', '22', 'MAR', '17', '42200', 'HG']
['TOOK', '22315', '15100', 'HG']
['BOGGOK', '22216', '43200', 'GH']

感謝您抽出寶貴時間回答我的問題.. 非常感謝。

uj5u.com熱心網友回復：

第一個解決方案：

data5 = ['TOOK22JAN1515100HG','BOGGOK22MAR1742200HG','TOOK2231515100HG','BOGGOK2221643200GH']
for i in range(len(data5)):
    data5[i] = [item for item in (re.split(r'(TOOK|BOGGOK)([0-9]{2})([A-Z]{3})([0-9]{2})([0-9]{5})(HG|GH)|(TOOK|BOGGOK)([0-9]{5})([0-9]{5})(HG|GH)', data5[i])) if (item is not None) and len(item)>0]
print(data5)

第二種解決方案：

def resultant_string_list(data5):
    return [([item for item in (re.split(r'(TOOK|BOGGOK)([0-9]{2})([A-Z]{3})([0-9]{2})([0-9]{5})(HG|GH)|(TOOK|BOGGOK)([0-9]{5})([0-9]{5})(HG|GH)', data5[i])) if (item is not None) and len(item)>0]) for i in range(len(data5))]

data5 = ['TOOK22JAN1515100HG','BOGGOK22MAR1742200HG','TOOK2231515100HG','BOGGOK2221643200GH']
print(resultant_string_list(data5))

上述代碼的輸出：

[['TOOK', '22', 'JAN', '15', '15100', 'HG'], ['BOGGOK', '22', 'MAR', '17', '42200', 'HG'], ['TOOK', '22315', '15100', 'HG'], ['BOGGOK', '22216', '43200', 'GH']]

我在撰寫代碼時牢記了使用的空間，因此您當前的變數將替換為結果串列。 示例：

Before: data5[0] = "TOOK22JAN1515100HG"
After: data5[0] = ['TOOK', '22', 'JAN', '15', '15100', 'HG']

uj5u.com熱心網友回復：

您可以洗掉代碼中的所有 None 值：

data5 = ['TOOK22JAN1515100HG','BOGGOK22MAR1742200HG','TOOK2231515100HG','BOGGOK2221643200GH']

for i in data5:
  dk = re.split(r'(TOOK|BOGGOK)([0-9]{2})(JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)([0-9]{2})([0-9]{5})(HG|GH)|(TOOK|BOGGOK)([0-9]{5})([0-9]{5})(HG|GH)', i)
  dk = list(filter(lambda a: a != None, dk)) #This removes all the None values from your list
  print(dk)

uj5u.com熱心網友回復：

我有兩個想法給你：

import re

data5 = ['TOOK22JAN1515100HG','BOGGOK22MAR1742200HG','TOOK2231515100HG','BOGGOK2221643200GH']

# first approach (ugly): keep the current, simple code, then get rid of Nones
for i in data5:
    dk = re.split(r'(TOOK|BOGGOK)([0-9]{2})(JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)([0-9]{2})([0-9]{5})(HG|GH)|(TOOK|BOGGOK)([0-9]{5})([0-9]{5})(HG|GH)', i)
    dk = list([s for s in dk if s != None])
    print(dk)

# second approach: condition to find out which case holds
for i in data5:
    dk = None
    if re.search(r'JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC', i):
        dk = re.split(r'(TOOK|BOGGOK)([0-9]{2})(JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)([0-9]{2})([0-9]{5})(HG|GH)', i)
    else:
        dk = re.split(r'(TOOK|BOGGOK)([0-9]{5})([0-9]{5})(HG|GH)', i)
    print(dk)

uj5u.com熱心網友回復：

嘗試使用 filter 函式從串列中洗掉 None ：

dk = list(filter(lambda x:x!=None, re.split(Your reg expression here), i))

你會得到None原因有兩個捕獲組，如果它們都不匹配，則None進入串列。

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