Pandas：如何根据唯一列值分配随机数

Question

I have a large dataset with the columns 'group' and 'postcode'. An example of the df is given below:

group   postcode
group_1 WC2E 8BU
group_1 WC2E 8BU
group_1 WC2E 8BU
group_2 WC2E 8BU
group_2 WC2E 8BU
group_2 WC2E 8BU
group_2 WC1A 1DD
group_2 WC1A 1DD
group_2 WC1A 1DD
group_2 WC1A 1DD
1488087 WC1A 1DD
1488087 WC1A 1DD

I am trying to create a new column called 'random_val' to assign a random uniform number to each matching postcode in a unique group, for rows where there are no digits in the 'group' column. My code is shown below:

df.loc[~df['group'].astype(str).str.isdigit(), 'random_val'] = df['postcode'].map(dict(zip(df['postcode'].unique(), np.random.uniform(0, 1, size=len(self.data['postcode'].unique())))))

Currently, this code assigns a unique random number to a unique postcode, regardless of the group it is in:

group   postcode    random_val
group_1 WC2E 8BU    0.210917735
group_1 WC2E 8BU    0.210917735
group_1 WC2E 8BU    0.210917735
group_2 WC2E 8BU    0.210917735
group_2 WC2E 8BU    0.210917735
group_2 WC2E 8BU    0.210917735
group_2 WC1A 1DD    0.55733542
group_2 WC1A 1DD    0.55733542
group_2 WC1A 1DD    0.55733542
group_2 WC1A 1DD    0.55733542
1488087 WC1A 1DD    
1488087 WC1A 1DD

However, I would like the random number to be unique to the postcode and the group:

group   postcode    random_val
group_1 WC2E 8BU    0.210917735
group_1 WC2E 8BU    0.210917735
group_1 WC2E 8BU    0.210917735
group_2 WC2E 8BU    0.494920676
group_2 WC2E 8BU    0.494920676
group_2 WC2E 8BU    0.494920676
group_2 WC1A 1DD    0.55733542
group_2 WC1A 1DD    0.55733542
group_2 WC1A 1DD    0.55733542
group_2 WC1A 1DD    0.55733542
1488087 WC1A 1DD    
1488087 WC1A 1DD    

Struggling to figure out how to do this. Any help appreciated. Thanks

Answer 1

You can take advantage of pandas alignment here.

df.set_index('group',inplace=True)
unique_idx = df.index[~df.index.str.isdigit()].unique()
s = pd.Series(np.random.uniform(0,1,len(unique_idx)) , index =unique_idx)
df['random_value'] = s
df.reset_index()

      group  postcode  random_value
0   group_1  WC2E 8BU      0.232501
1   group_1  WC2E 8BU      0.232501
2   group_1  WC2E 8BU      0.232501
3   group_2  WC2E 8BU      0.242696
4   group_2  WC2E 8BU      0.242696
5   group_2  WC2E 8BU      0.242696
6   group_2  WC1A 1DD      0.242696
7   group_2  WC1A 1DD      0.242696
8   group_2  WC1A 1DD      0.242696
9   group_2  WC1A 1DD      0.242696
10  1488087  WC1A 1DD           NaN
11  1488087  WC1A 1DD           NaN

Answer 2

Here's a solution:

def random_val(x):
    return pd.Series([np.random.uniform(0, 1)] * x.size)

df["dummy"] = 1

df["random_val"] = df.groupby(["group", "postcode"])["dummy"].transform(random_val)
df.loc[df['group'].astype(str).str.isdigit(), "random_val"] = None

The result is:

      group  postcode  dummy  random_val
0   group_1  WC2E 8BU      1    0.781711
1   group_1  WC2E 8BU      1    0.781711
2   group_1  WC2E 8BU      1    0.781711
3   group_2  WC2E 8BU      1    0.107743
4   group_2  WC2E 8BU      1    0.107743
5   group_2  WC2E 8BU      1    0.107743
6   group_2  WC1A 1DD      1    0.103295
7   group_2  WC1A 1DD      1    0.103295
8   group_2  WC1A 1DD      1    0.103295
9   group_2  WC1A 1DD      1    0.103295
10  1488087  WC1A 1DD      1         NaN
11  1488087  WC1A 1DD      1         NaN

Answer 3

Hashing the two columns might be the simplest solution:

df['hash'] = pd.Series((hash(tuple(row)) for _, row in df.iterrows()))

    group   postcode    hash
0   group_1 WC2E 8BU    -8918045538474016779
1   group_1 WC2E 8BU    -8918045538474016779
2   group_1 WC2E 8BU    -8918045538474016779
3   group_2 WC2E 8BU    -6943464964421442707
4   group_2 WC2E 8BU    -6943464964421442707
5   group_2 WC2E 8BU    -6943464964421442707
6   group_2 WC1A 1DD    -357652478068898330
7   group_2 WC1A 1DD    -357652478068898330
8   group_2 WC1A 1DD    -357652478068898330
9   group_2 WC1A 1DD    -357652478068898330
10  1488087 WC1A 1DD    1701757393872926575
11  1488087 WC1A 1DD    1701757393872926575