在Python中根据前4个字母从数组中删除重复项

Question

I have a list of postcodes, eg

DD1 1DB
DD1 5PH
DD10 8JG
DD10 9LJ

What I would like to do is keep the first representative, depending on the first part of the postcode eg

I need to keep:

DD1 1DB
DD10 8JG

I am using pandas and imported the file.csv containing column POSTCODES as:

df = pandas.read_csv('file.csv')
pc = df.POSTCODES

Now I am completely stuck. I managed to get it working using Excel (is this the better option?) but I would like to learn python and decided to see if I can do it in python.

Answer 1

You could use df['POSTCODES'].str[:4] to obtain the first four characters, and use the duplicated method to identify duplicates:

In [89]: df.loc[~df['POSTCODES'].str[:4].duplicated(keep='first')]
Out[89]: 
  POSTCODES
0   DD1 1DB
2  DD10 8JG

Since duplicated(keep='first') marks duplicates as True, the row we wish to keep would be marked False. So to select the False rows with df.loc , the ~ is used to invert the boolean Series .

Answer 2

pc[~pc.POSTCODES.str.split(' ', expand=True)[0].duplicated()]

OR as piRSquared suggests in the comments:

pc[~pc.POSTCODES.str.split().str[0].duplicated()]

Output:

  POSTCODES
0   DD1 1DB
2  DD10 8JG

Answer 3

In[24]: f = '''\
   ...: DD1 1DB
   ...: DD1 5PH
   ...: DD10 8JG
   ...: DD10 9LJ'''.split('\n')
In[25]: d = {}
   ...: for line in f:
   ...:     left, right = line.split()
   ...:     if left not in d:
   ...:         d[left] = right
   ...: 
In[26]: d
Out[26]: {'DD1': '1DB', 'DD10': '8JG'}