给出了一个包含3列的数据框:['User', 'Feature', 'Value']
。我需要为每个用户获得TOP-N功能。
我尝试了几种选择,但它们似乎都太慢了。实际上,pandas最大的方法似乎比手工制作的方法要慢,如以下代码所示:
import pandas as pd
import numpy as np
import time
np.random.seed(0)
max_rows = 1000000
n_users = 10000
max_entries = 100
df = pd.DataFrame()
df['Value'] = [i * 0.1 for i in range(max_rows)]
df['User'] = ['u%i'%i for i in np.random.randint(0, n_users, size=max_rows)]
df['Feature'] = ['f%i'%i for i in range(max_rows)]
# PART 1
t0 = time.time()
dfg = df.groupby(['User', 'Feature'], sort=False, as_index=True).sum()
t1 = time.time()
print "first groupby takes: ", t1-t0, "for nrows=", max_rows
# PART 2
#Option1
dfg2 = dfg.groupby(dfg.index.get_level_values(0),
group_keys=False,
sort=False).apply(lambda x: x.sort_values('Value', ascending=False).\
iloc[:max_entries])
t2 = time.time()
print "second groupby takes:", t2-t1
# PART 2
#Option2
dfg = dfg.reset_index(level='Feature')
dfg3 = dfg.groupby(dfg.index.get_level_values(0), sort=False, group_keys=False).\
apply(lambda x: x.nlargest(max_entries, 'Value'))
t3 = time.time()
print "Third groupby takes", t3-t2
示例输出:
n [29]: first groupby takes: 0.914537191391 for nrows= 1000000
second groupby takes: 17.5580070019
Third groupby takes 59.3013348579
有没有办法使其更快?我应该避免像熊猫这样的问题吗?
为什么最大的东西这么慢?我期望它会更快。
答案 0 :(得分:4)
您可以使用head
Your solution 1
%timeit dfg.groupby(dfg.index.get_level_values(0),group_keys=False,sort=False).apply(lambda x: x.sort_values('Value', ascending=False).iloc[:max_entries])
1 loop, best of 3: 9.1 s per loop
My solution
%timeit dfg.sort_values('Value',ascending=False).groupby(level=0).head(max_entries)
1 loop, best of 3: 201 ms per loop