Python 如果出现错误，则执行其他操作-字符串拆分

我希望在上面的数据框中拆分字符串，如下所示

df = pd.DataFrame(['BERGEPAINT20FEB550PE', 'BANKNIFTY2020631300CE', 'BANKNIFTY2020631300PE'], columns=list('A'))
df['StrikePrice'] = df.A.str.split('(\d+)').apply(lambda x: x[3])
df['CallPut'] = df.A.str[-2:]
print(df.head())

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但是如果得到一个错误

请使用regex'或'表达式对给定的数据执行此操作。按5位数或2位数拆分：

BERGEPAINT20FEB550PE -> BERGEPAINT, 550, PE

BANKNIFTY2020631300CE -> BANKNIFTY, 31300, CE

BANKNIFTY2020631300PE -> BANKNIFTY, 31300, PE

输出：

df = pd.DataFrame(['BERGEPAINT20FEB550PE', 'BANKNIFTY2020631300CE', 'BANKNIFTY2020631300PE'], columns=list('A'))
df['StrikePrice'] = df.A.str.split('(\d{5}|\d{2})').str[-2]
df['CallPut'] = df.A.str[-2:]
df['Name'] = df.A.str.split('(\d+)').str[0]
print(df.head())

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假设您不想要的部件20FEB、20206、20206都从20FEB开始，并且由5个字符组成，那么您可以使用：

                       A StrikePrice CallPut        Name
0   BERGEPAINT20FEB550PE          55      PE  BERGEPAINT
1  BANKNIFTY2020631300CE       31300      CE   BANKNIFTY
2  BANKNIFTY2020631300PE       31300      PE   BANKNIFTY

输出：

df = pd.DataFrame(['BERGEPAINT20FEB550PE', 'BANKNIFTY2020631300CE', 'BANKNIFTY2020631300PE'], columns=list('A'))
df['Toto'] = df.A.apply(lambda x: x[:x.index("20")])
df['StrikePrice'] = df.A.apply(lambda x: x[x.index("20")+5:-2])
df['CallPut'] = df.A.str[-2:]
print(df)

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也许这就是你想要的：

                       A        Toto StrikePrice CallPut
0   BERGEPAINT20FEB550PE  BERGEPAINT         550      PE
1  BANKNIFTY2020631300CE   BANKNIFTY       31300      CE
2  BANKNIFTY2020631300PE   BANKNIFTY       31300      PE

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您在这样的任意数字处进行拆分，并尝试通过x[3]访问第四个元素。但是BANKNIFTY2020631300CE中没有4个元素，因为它像[BANKNIFTY，2020631300，CE]一样分裂

s = df['A'].str.split('(\d+)').apply(lambda x: [x[0], x[-2][-5:], x[-1]])
s.apply(lambda x: pd.Series(x)).rename(columns={0: 'A', 1: 'StrikePrice', 2: 'CallPut'})

            A StrikePrice CallPut
0  BERGEPAINT         550      PE
1   BANKNIFTY       31300      CE
2   BANKNIFTY       31300      PE