在Pandas中將列拆分為兩個新列

Question

我有一個具有4列的數據框，其值如下所示：

value_1
over 1 - 42 -> take this ; over 3 - 4
over 3 - 26 -> take this ; over 3 - 45 
over 5 - 25 -> take this ; over 2 - 80 

而且我需要通過用';'分隔每列中的兩列。 當我嘗試這個：

s = df['value_1'].apply(lambda x: x.split(';'))
df['value_left'] = s.apply(lambda x: x[0])
df['value_right'] = s.apply(lambda x: x[1])

或這個：

f['new_value1'] = df['value_1']
df['value_1_right'] = None
df['value_1_right'].update(df.new_band_bandw_1.apply(lambda x: x.str.split(';')[1] 
                      if len(x.str.split()) == 2 else None))

我遇到了同樣的error: List index out of range 。 問題出在哪里，這些值是否在某種列表中？

任何一種解決方案都歡迎。 謝謝

Answer 1

您需要split ：

df[['value_left','value_right']] = df['value_1'].str.split(';', expand=True)
print (df)

                                  value_1                 value_left  \
0   over 1 - 42 -> take this ; over 3 - 4  over 1 - 42 -> take this    
1  over 3 - 26 -> take this ; over 3 - 45  over 3 - 26 -> take this    
2  over 5 - 25 -> take this ; over 2 - 80  over 5 - 25 -> take this    

    value_right  
0    over 3 - 4  
1   over 3 - 45  
2   over 2 - 80  

樣品多次; -可以指定哪個; 用於拆分：

df = pd.DataFrame({
    'value_1': ['a;r;e','b;r','c;g;t;e']
})
print (df)

   value_1
0    a;r;e
1      b;r
2  c;g;t;e

df[['value_left','value_right']] = df['value_1'].str.split(';', expand=True, n=1)
print (df)
   value_1 value_left value_right
0    a;r;e          a         r;e
1      b;r          b           r
2  c;g;t;e          c       g;t;e

Answer 2

字符串partition另一種方法，即

df[['val1','val2']] = df[0].str.partition(';').iloc[:,0::2]

                                     0                       val1          val2
0    over 1 - 42 -> take this ; over 3 - 4  over 1 - 42 -> take this     over 3 - 4  
1   over 3 - 26 -> take this ; over 3 - 45  over 3 - 26 -> take this     over 3 - 45
2  over 5 - 25 -> take this ; over; 2 - 80  over 5 - 25 -> take this     over; 2 - 80