熊猫数据框中重复位置的频率

Question

Hi I am working to find out repetitive position of the following data frame:

data = pd.DataFrame()
data ['league'] =['A','A','A','A','A','A','B','B','B']
data ['Team'] = ['X','X','X','Y','Y','Y','Z','Z','Z']
data ['week'] =[1,2,3,1,2,3,1,2,3]
data ['position']= [1,1,2,2,2,1,2,3,4]

I will compare the data for position from previous row, it is it the same, I will assign one. If it is different previous row, I will assign as 1

My expected outcome will be as follow:

It means I will group by (League, Team and week) and work out the frequency. Can anyone advise how to do that in Pandas

Thanks,

Zep

Answer 1

Use diff and abs with fillna :

data['frequency'] = data['position'].diff().abs().fillna(0,downcast='infer')

print(data)
  league Team  week  position  frequency
0      A    X     1         1          0
1      A    X     2         1          0
2      A    X     3         2          1
3      A    Y     1         2          0
4      A    Y     2         2          0
5      A    Y     3         1          1
6      B    Z     1         2          1
7      B    Z     2         3          1
8      B    Z     3         4          1

---

Using groupby gives all zeros, since you are comparing within groups not on whole dataframe.

data.groupby(['league', 'Team', 'week'])['position'].diff().fillna(0,downcast='infer')

0    0
1    0
2    0
3    0
4    0
5    0
6    0
7    0
8    0
Name: position, dtype: int64

Answer 2

Use diff , and compare against 0 :

v = df.position.diff()
v[0] = 0
df['frequency'] = v.ne(0).astype(int)

print(df)
  league Team  week  position  frequency
0      A    X     1         1          0
1      A    X     2         1          0
2      A    X     3         2          1
3      A    Y     1         2          0
4      A    Y     2         2          0
5      A    Y     3         1          1
6      B    Z     1         2          1
7      B    Z     2         3          1
8      B    Z     3         4          1

---

For performance reasons, you should try to avoid a fillna call.

df = pd.concat([df] * 100000, ignore_index=True)

%timeit df['frequency'] = df['position'].diff().abs().fillna(0,downcast='infer')
%%timeit
v = df.position.diff()
v[0] = 0
df['frequency'] = v.ne(0).astype(int)

83.7 ms ± 1.55 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
10.9 ms ± 217 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

---

To extend this answer to work in a groupby , use

v = df.groupby(['league', 'Team', 'week']).position.diff()
v[np.isnan(v)] = 0

df['frequency'] = v.ne(0).astype(int)