如何在R中按小时对变量进行百分位数？

Question

I have a task I need to execute in R. I've done it in python (likely not in the most efficient way.) The end goal: A dataframe with columns start_time, agent, percentile. There are ~8200 agents and the business is open from 7:00 through 23:00, this is annotated by integer (7,8,...23). I need to "re-percentile" these agents by hour.

start_time, agent, percentile
7,          1,     1,
7,          2,     0.99,
...
7,          8200,  0,
...
23,         700,   1,
23,         12,    0.99     

Notice that (A) every agent:hour combination will be represented with its normalized score. For reference, this normalization formula is (x-min)/(max-min) .

The data that I currently have looks like this. Table A (metrics.csv)

idx,  agent,          percentile
1,    z_agent[1],     1
2,    z_agent[2],     0.05
3,    z_agent[3],     0.5
...
8200, z_agent[8200],  0.99

Table B (hours.csv)

agent_idx,  start_hour
1           7
2           7
3           7
4           7

python code:

hours = pd.read_csv('hours.csv')
metrics = pd.read_csv('metrics.csv')

ag_rank = {row['agent']:row['percentile'] for idx,row in metrics.iterrows() if 'agent' in row[0]}
raw_scores = [s for s in ag_rank.values()]
raw_min = min(raw_scores)
raw_max = max(raw_scores)

def normed(x,mn,mx):
    return (x-mn)/(mx-mn)

norm_ag_scores = [normed(x,raw_min,raw_max) for x in raw_scores]

c = 0
for k,v in ag_rank.items():
    n = norm_ag_scores[c]
    ag_rank[k] = n
    c += 1

import operator
tups = []
starts = sorted([hr for hr in hours['start_hour'].unique()]) 
for hr in starts:
    agents = [f'z_agent[{a}]' for a in hours[hours['start_hour'] == hr]['agent_idx'].unique()]
    a_set = set(agents)
    b_set = set(ag_rank.keys())
    missing = list(a_set.symmetric_difference(b_set))
    scores = [ag_rank[a] for a in agents if a in ag_rank.keys()]
    hi = max(scores)
    low = min(scores)
    sort = {a:normed(s,low,hi) for a,s in zip(agents,scores)}
    sort = sorted(sort.items(),key=operator.itemgetter(1),reverse=True)
    for a,s in sort:
        tups.append((hr,a,s))
    for m in missing:
        tups.append((hr,m,0))

And the final table, in the form that I need it:

reperc = pd.DataFrame(data=tups,columns=['hour','agent','percentile'])
reperc.head()

>>>
7   z_agent[2853]   1.000000
7   z_agent[6004]   0.855892
7   z_agent[4366]   0.821758
7   z_agent[1742]   0.370188
7   z_agent[21]     0.000000

My questions are (A): How should I accomplish this affect in R? And (B, optional): What/is there a way to accomplish this effect in python? Perhaps a join would help.

Answer 1

Something like this should work. Happy to test/debug if you share reproducible data.

library(dplyr)
metrics %>% 
  left_join(hours, by = c("idx" = "agent_idx")) %>%
  group_by(start_time) %>%
  mutate(
    new_percentile = (percentile - min(percentile)) / (max(percentile) - min(percentile))
  ) %>%
  arrange(start_time, desc(new_percentile))