从列值中拆分并获取字符串的一部分，并从pandas python中创建新列

Question

I have a strings like this as a value of one column in my df.

ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232
ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232

How to get new column with part of this columns. Part that I need is

74
89

Answer 1

string.split() allows you to explode a string into a list of parts according to a separator (here / and - ).

s = 'ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232'
print s.split('/')[2].split('-')[1]
# 74

Use pandas.apply() to apply it to your column

df['b'] = df['a'].apply(lambda s:s.split('/')[2].split('-')[1])
print (df)

output

                                              a   b
0  ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232  74
1  ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232  89

nb: Use @A-Za-z 's solution, it's faster than mine.

Answer 2

If this is the df

    val
0   ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232
1   ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232

You can use str.extract

df['num_val'] = df.val.str.extract('LNFFF-(\d+)/', expand = False)

You get

    val                                             num_val
0   ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232   74
1   ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232   89

Answer 3

假设您的数据框名为df且列col：

 df['sub_col'] = pd.Series([s[21:23] for s in df['col'].values], index=df.index)

Answer 4

It seems you need str.extract :

df = pd.DataFrame({'a': ['ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232',
                         'ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232']})  
print (df)
                                               a
0  ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232
1  ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232

df['new'] = df['a'].str.extract('LNFFF-(\d+)', expand=False)
#if necessary convert to ints
df['new'] = df['new'].astype(int)
print (df)
                                               a new
0  ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232  74
1  ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232  89

Solution with splitting by split and selecting by indexing with str :

df['new'] = df['a'].str.split('/').str[2].str.extract('(\d+)', expand=False)
print (df)
                                               a new
0  ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232  74
1  ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232  89

---

df['new'] = df['a'].str.split('/').str[2].str.split('-').str[1]
print (df)
                                               a new
0  ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232  74
1  ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232  89