在python中记录每个id的最大系列

Question

I want to to keep one record that has the largest series for each id. So for each id I need one row. I think I need something like

df_new = df.groupby('id')['series'].nlargest(1)

, but that's definitely wrong.

That's how my dataset looks:

id  series s1 s2 s3
1   2      4  9  1
1   8      6  2  2
1   3      9  1  3
2   9      4  1  5
2   2      2  5  5
2   5      1  7  8
3   6      7  2  3
3   2      4  4  1
3   1      3  9  9

This should be the result:

id  series s1 s2 s3
1   8      6  2  2
2   9      4  1  5
3   6      7  2  3

Answer 1

IIUC you want to groupby on 'id' column and get the index label where the 'Series' value is the largest using idxmax() and use this to index back in the orig df:

In [91]:
df.loc[df.groupby('id')['series'].idxmax()]

Out[91]:
   id  series  s1  s2  s3
1   1       8   6   2   2
3   2       9   4   1   5
6   3       6   7   2   3

Answer 2

Another solution with sort_values and aggregate first :

df = df.sort_values(by="series", ascending=False).groupby("id", as_index=False).first()
print (df)
   id  series  s1  s2  s3
0   1       8   6   2   2
1   2       9   4   1   5
2   3       6   7   2   3

Answer 3

Here's one NumPy based solution -

def grouby_max(df):
    arr = df[['id','series']].values
    n = arr.shape[0]-1
    idx = (arr[:,0]*(arr[:,1].max()+1) + arr[:,1]).argsort()
    sidx = np.append(np.nonzero(arr[idx[1:],0] > arr[idx[:-1],0])[0],n)
    return df.iloc[idx[sidx]]

Runtime test -

In [201]: # Setup input
     ...: N = 100 # Number of groups
     ...: data = np.random.randint(11,999999,(10000,5))
     ...: data[:,0] = np.sort(np.random.randint(1,N+1,(data.shape[0])))
     ...: df = pd.DataFrame(data, columns=[['id','series','s1','s2','s3']])
     ...: 

In [202]: %timeit df.loc[df.groupby('id')['series'].idxmax()]
100 loops, best of 3: 15.8 ms per loop #@EdChum's soln

In [203]: %timeit df.sort_values(by="series", ascending=False).groupby("id", as_index=False).first()
100 loops, best of 3: 4.52 ms per loop #@jezrael's soln

In [204]: %timeit grouby_max(df)
100 loops, best of 3: 1.96 ms per loop