Pandas Dataframe 得到两列之间的最小差异

Question

Dataframe

d = {'Resource': ['A','A','A','B','B','B'], 'User': ['1','2','3','4','5','6'], 'earliestSlot': [1,2,3,5,4,6], 'latestSlot': [1.2,2.5,3.9,6,5,6.1]}
pd.DataFrame(data=d)

I want to aggregate the data in a way that calculates the minimum difference between latestSlot and earliestSlot of different users of the same resource. Basically I want to calculate the minimum idle time of each resource before the next user accesses it.

Target Data

Resource	MinHeadway
A	0.5
B	0

I have the following code but I am sure there is a faster method.

´´´

def get_min_headway(resource_id):
    latestSlots = d[d['Resource'] == resource_id].latestSlot
    earliestSlots = d[d['Resource'] == resource_id].earliestSlot
    min_headway = float('inf')
    for time in latestSlots:
        headways = earliestSlots - time
        for headway in headways:
            if headway >= 0:
                if headway < min_headway:
                    min_headway = headway
    return min_headway

d['min_headway'] = d['Resource'].apply(get_min_headway)

´´´

Answer 1

It's not exactly clear what your desired outcome is, but by minimum idle time , I assumed you wanted the minimum difference between "latestSlot" of the previous user and the "earliestSlot" of the next user (since an unused time is an idle time ). So in that case, you can use the following.

We sort by "earliestSlot"; then groupby "Resource" and do exactly what's explained above.

out = (df.sort_values(by='earliestSlot')
       .groupby('Resource')
       .apply(lambda x: (x['earliestSlot']-x['latestSlot'].shift()).min())
       .reset_index()
       .rename(columns={0:'MinHeadway'}))

Output:

  Resource  MinHeadway
0        A         0.5
1        B         0.0

The same result could also be obtained without applying a lambda:

tmp = df.sort_values(by='earliestSlot')
out = (tmp.groupby('Resource')['latestSlot'].shift()
       .rsub(tmp['earliestSlot'])
       .groupby(x['Resource']).min()
       .reset_index()
       .rename(columns={0:'MinHeadway'}))