为列pandas数据框分配唯一ID

Question

Hello I have the following dataframe

df = 
A      B   
John   Tom
Homer  Bart
Tom    Maggie
Lisa   John 

I would like to assign to each name a unique ID and returns

df = 
A      B         C    D

John   Tom       0    1
Homer  Bart      2    3
Tom    Maggie    1    4 
Lisa   John      5    0

What I have done is the following:

LL1 = pd.concat([df.a,df.b],ignore_index=True)
LL1 = pd.DataFrame(LL1)
LL1.columns=['a']
nameun = pd.unique(LL1.a.ravel())
LLout['c'] = 0
LLout['d'] = 0
NN = list(nameun)
for i in range(1,len(LLout)):
   LLout.c[i] = NN.index(LLout.a[i])
   LLout.d[i] = NN.index(LLout.b[i])

But since I have a very large dataset this process is very slow.

Answer 1

Here's one way. First get the array of unique names:

In [11]: df.values.ravel()
Out[11]: array(['John', 'Tom', 'Homer', 'Bart', 'Tom', 'Maggie', 'Lisa', 'John'], dtype=object)

In [12]: pd.unique(df.values.ravel())
Out[12]: array(['John', 'Tom', 'Homer', 'Bart', 'Maggie', 'Lisa'], dtype=object)

and make this a Series, mapping names to their respective numbers:

In [13]: names = pd.unique(df.values.ravel())

In [14]: names = pd.Series(np.arange(len(names)), names)

In [15]: names
Out[15]:
John      0
Tom       1
Homer     2
Bart      3
Maggie    4
Lisa      5
dtype: int64

Now use applymap and names.get to lookup these numbers:

In [16]: df.applymap(names.get)
Out[16]:
   A  B
0  0  1
1  2  3
2  1  4
3  5  0

and assign it to the correct columns:

In [17]: df[["C", "D"]] = df.applymap(names.get)

In [18]: df
Out[18]:
       A       B  C  D
0   John     Tom  0  1
1  Homer    Bart  2  3
2    Tom  Maggie  1  4
3   Lisa    John  5  0

Note: This assumes that all the values are names to begin with, you may want to restrict this to some columns only:

df[['A', 'B']].values.ravel()
...
df[['A', 'B']].applymap(names.get)

Answer 2

(Note: I'm assuming you don't care about the precise details of the mapping -- which number John becomes, for example -- but only that there is one.)

Method #1: you could use a Categorical object as an intermediary:

>>> ranked = pd.Categorical(df.stack()).codes.reshape(df.shape)
>>> df.join(pd.DataFrame(ranked, columns=["C", "D"]))
       A       B  C  D
0   John     Tom  2  5
1  Homer    Bart  1  0
2    Tom  Maggie  5  4
3   Lisa    John  3  2

It feels like you should be able to treat a Categorical as providing an encoding dictionary somehow (whether directly or by generating a Series) but I can't see a convenient way to do it.

Method #2: you could use rank("dense") , which generates an increasing number for each value in order:

>>> ranked = df.stack().rank("dense").reshape(df.shape).astype(int)-1
>>> df.join(pd.DataFrame(ranked, columns=["C", "D"]))
       A       B  C  D
0   John     Tom  2  5
1  Homer    Bart  1  0
2    Tom  Maggie  5  4
3   Lisa    John  3  2