Deb*_*har 10 python merge cython pandas numba
正如其他帖子所暗示的那样,我特别没有Pands Merge的性能问题,但我有一个类,其中有很多方法,它们在数据集上进行了大量的合并.
该班有大约10个小组和大约15个合并.虽然groupby相当快,但是对于类的总执行时间为1.5秒,在这15次合并调用中大约需要0.7秒.
我想加快那些合并调用的性能.因为我将有大约4000次迭代,因此在单次迭代中总共节省0.5秒将导致整体性能降低大约30分钟,这将是很好的.
我应该尝试任何建议吗？我试过:Cython Numba,Numba比较慢.
谢谢
编辑1:添加示例代码片段:我的合并语句:
tmpDf = pd.merge(self.data, t1, on='APPT_NBR', how='left')
tmp = tmpDf
tmpDf = pd.merge(tmp, t2, on='APPT_NBR', how='left')
tmp = tmpDf
tmpDf = pd.merge(tmp, t3, on='APPT_NBR', how='left')
tmp = tmpDf
tmpDf = pd.merge(tmp, t4, on='APPT_NBR', how='left')
tmp = tmpDf
tmpDf = pd.merge(tmp, t5, on='APPT_NBR', how='left')
并且,通过实现连接,我合并了以下声明:
dat = self.data.set_index('APPT_NBR')
t1.set_index('APPT_NBR', inplace=True)
t2.set_index('APPT_NBR', inplace=True)
t3.set_index('APPT_NBR', inplace=True)
t4.set_index('APPT_NBR', inplace=True)
t5.set_index('APPT_NBR', inplace=True)
tmpDf = dat.join(t1, how='left')
tmpDf = tmpDf.join(t2, how='left')
tmpDf = tmpDf.join(t3, how='left')
tmpDf = tmpDf.join(t4, how='left')
tmpDf = tmpDf.join(t5, how='left')
tmpDf.reset_index(inplace=True)
注意,所有都是名为的函数的一部分:def merge_earlier_created_values(self):
并且,当我通过以下方式从profilehooks做时间调用时:
@timedcall(immediate=True)
def merge_earlier_created_values(self):
我得到以下结果:
该方法的分析结果给出:
@profile(immediate=True)
def merge_earlier_created_values(self):
使用Merge进行功能分析如下:
*** PROFILER RESULTS ***
merge_earlier_created_values (E:\Projects\Predictive Inbound Cartoon Estimation-MLO\Python\CodeToSubmit\helpers\get_prev_data_by_date.py:122)
function called 1 times
71665 function calls (70588 primitive calls) in 0.524 seconds
Ordered by: cumulative time, internal time, call count
List reduced from 563 to 40 due to restriction <40>
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.012 0.012 0.524 0.524 get_prev_data_by_date.py:122(merge_earlier_created_values)
14 0.000 0.000 0.285 0.020 generic.py:1901(_update_inplace)
14 0.000 0.000 0.285 0.020 generic.py:1402(_maybe_update_cacher)
19 0.000 0.000 0.284 0.015 generic.py:1492(_check_setitem_copy)
7 0.283 0.040 0.283 0.040 {built-in method gc.collect}
15 0.000 0.000 0.181 0.012 generic.py:1842(drop)
10 0.000 0.000 0.153 0.015 merge.py:26(merge)
10 0.000 0.000 0.140 0.014 merge.py:201(get_result)
8/4 0.000 0.000 0.126 0.031 decorators.py:65(wrapper)
4 0.000 0.000 0.126 0.031 frame.py:3028(drop_duplicates)
1 0.000 0.000 0.102 0.102 get_prev_data_by_date.py:264(recreate_previous_cartons)
1 0.000 0.000 0.101 0.101 get_prev_data_by_date.py:231(recreate_previous_appt_scheduled_date)
1 0.000 0.000 0.098 0.098 get_prev_data_by_date.py:360(recreate_previous_freight_type)
10 0.000 0.000 0.092 0.009 internals.py:4455(concatenate_block_managers)
10 0.001 0.000 0.088 0.009 internals.py:4471(<listcomp>)
120 0.001 0.000 0.084 0.001 internals.py:4559(concatenate_join_units)
266 0.004 0.000 0.067 0.000 common.py:733(take_nd)
120 0.000 0.000 0.061 0.001 internals.py:4569(<listcomp>)
120 0.003 0.000 0.061 0.001 internals.py:4814(get_reindexed_values)
1 0.000 0.000 0.059 0.059 get_prev_data_by_date.py:295(recreate_previous_appt_status)
10 0.000 0.000 0.038 0.004 merge.py:322(_get_join_info)
10 0.001 0.000 0.036 0.004 merge.py:516(_get_join_indexers)
25 0.001 0.000 0.024 0.001 merge.py:687(_factorize_keys)
74 0.023 0.000 0.023 0.000 {pandas.algos.take_2d_axis1_object_object}
50 0.022 0.000 0.022 0.000 {method 'factorize' of 'pandas.hashtable.Int64Factorizer' objects}
120 0.003 0.000 0.022 0.000 internals.py:4479(get_empty_dtype_and_na)
88 0.000 0.000 0.021 0.000 frame.py:1969(__getitem__)
1 0.000 0.000 0.019 0.019 get_prev_data_by_date.py:328(recreate_previous_location_numbers)
39 0.000 0.000 0.018 0.000 internals.py:3495(reindex_indexer)
537 0.017 0.000 0.017 0.000 {built-in method numpy.core.multiarray.empty}
15 0.000 0.000 0.017 0.001 ops.py:725(wrapper)
15 0.000 0.000 0.015 0.001 frame.py:2011(_getitem_array)
24 0.000 0.000 0.014 0.001 internals.py:3625(take)
10 0.000 0.000 0.014 0.001 merge.py:157(__init__)
10 0.000 0.000 0.014 0.001 merge.py:382(_get_merge_keys)
15 0.008 0.001 0.013 0.001 ops.py:662(na_op)
234 0.000 0.000 0.013 0.000 common.py:158(isnull)
234 0.001 0.000 0.013 0.000 common.py:179(_isnull_new)
15 0.000 0.000 0.012 0.001 generic.py:1609(take)
20 0.000 0.000 0.012 0.001 generic.py:2191(reindex)
使用Joins进行分析如下:
65079 function calls (63990 primitive calls) in 0.550 seconds
Ordered by: cumulative time, internal time, call count
List reduced from 592 to 40 due to restriction <40>
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.016 0.016 0.550 0.550 get_prev_data_by_date.py:122(merge_earlier_created_values)
14 0.000 0.000 0.295 0.021 generic.py:1901(_update_inplace)
14 0.000 0.000 0.295 0.021 generic.py:1402(_maybe_update_cacher)
19 0.000 0.000 0.294 0.015 generic.py:1492(_check_setitem_copy)
7 0.293 0.042 0.293 0.042 {built-in method gc.collect}
10 0.000 0.000 0.173 0.017 generic.py:1842(drop)
10 0.000 0.000 0.139 0.014 merge.py:26(merge)
8/4 0.000 0.000 0.138 0.034 decorators.py:65(wrapper)
4 0.000 0.000 0.138 0.034 frame.py:3028(drop_duplicates)
10 0.000 0.000 0.132 0.013 merge.py:201(get_result)
5 0.000 0.000 0.122 0.024 frame.py:4324(join)
5 0.000 0.000 0.122 0.024 frame.py:4371(_join_compat)
1 0.000 0.000 0.111 0.111 get_prev_data_by_date.py:264(recreate_previous_cartons)
1 0.000 0.000 0.103 0.103 get_prev_data_by_date.py:231(recreate_previous_appt_scheduled_date)
1 0.000 0.000 0.099 0.099 get_prev_data_by_date.py:360(recreate_previous_freight_type)
10 0.000 0.000 0.093 0.009 internals.py:4455(concatenate_block_managers)
10 0.001 0.000 0.089 0.009 internals.py:4471(<listcomp>)
100 0.001 0.000 0.085 0.001 internals.py:4559(concatenate_join_units)
205 0.003 0.000 0.068 0.000 common.py:733(take_nd)
100 0.000 0.000 0.060 0.001 internals.py:4569(<listcomp>)
100 0.001 0.000 0.060 0.001 internals.py:4814(get_reindexed_values)
1 0.000 0.000 0.056 0.056 get_prev_data_by_date.py:295(recreate_previous_appt_status)
10 0.000 0.000 0.033 0.003 merge.py:322(_get_join_info)
52 0.031 0.001 0.031 0.001 {pandas.algos.take_2d_axis1_object_object}
5 0.000 0.000 0.030 0.006 base.py:2329(join)
37 0.001 0.000 0.027 0.001 internals.py:2754(apply)
6 0.000 0.000 0.024 0.004 frame.py:2763(set_index)
7 0.000 0.000 0.023 0.003 merge.py:516(_get_join_indexers)
2 0.000 0.000 0.022 0.011 base.py:2483(_join_non_unique)
7 0.000 0.000 0.021 0.003 generic.py:2950(copy)
7 0.000 0.000 0.021 0.003 internals.py:3046(copy)
84 0.000 0.000 0.020 0.000 frame.py:1969(__getitem__)
19 0.001 0.000 0.019 0.001 merge.py:687(_factorize_keys)
100 0.002 0.000 0.019 0.000 internals.py:4479(get_empty_dtype_and_na)
1 0.000 0.000 0.018 0.018 get_prev_data_by_date.py:328(recreate_previous_location_numbers)
15 0.000 0.000 0.017 0.001 ops.py:725(wrapper)
34 0.001 0.000 0.017 0.000 internals.py:3495(reindex_indexer)
83 0.004 0.000 0.016 0.000 internals.py:3211(_consolidate_inplace)
68 0.015 0.000 0.015 0.000 {method 'copy' of 'numpy.ndarray' objects}
15 0.000 0.000 0.015 0.001 frame.py:2011(_getitem_array)
正如您所看到的,合并比连接更快,虽然它是小值,但是超过4000次迭代,这个小值变成了一个巨大的数字,只需几分钟.
谢谢
Jul*_*rec 16
我建议您将合并列设置为索引，并使用df1.join(df2)而不是merge，它会快得多。
这是一些示例，包括分析：
In [1]:
import pandas as pd
import numpy as np
df1 = pd.DataFrame(np.arange(1000000), columns=['A'])
df1['B'] = np.random.randint(0,1000,(1000000))
df2 = pd.DataFrame(np.arange(1000000), columns=['A2'])
df2['B2'] = np.random.randint(0,1000,(1000000))
这是 A 和 A2 上的常规左合并：
In [2]: %%timeit
x = df1.merge(df2, how='left', left_on='A', right_on='A2')
1 loop, best of 3: 441 ms per loop
这是相同的，使用 join：
In [3]: %%timeit
x = df1.set_index('A').join(df2.set_index('A2'), how='left')
1 loop, best of 3: 184 ms per loop
现在很明显，如果您可以在循环之前设置索引，则时间方面的收益会大得多：
# Do this before looping
In [4]: %%time
df1.set_index('A', inplace=True)
df2.set_index('A2', inplace=True)
CPU times: user 9.78 ms, sys: 9.31 ms, total: 19.1 ms
Wall time: 16.8 ms
然后在循环中，你会得到在这种情况下快 30 倍的东西：
In [5]: %%timeit
x = df1.join(df2, how='left')
100 loops, best of 3: 14.3 ms per loop
合并列上的set_index确实加快了这一点.以下是@ julien-marrec答案的更现实的版本.
import pandas as pd
import numpy as np
myids=np.random.choice(np.arange(10000000), size=1000000, replace=False)
df1 = pd.DataFrame(myids, columns=['A'])
df1['B'] = np.random.randint(0,1000,(1000000))
df2 = pd.DataFrame(np.random.permutation(myids), columns=['A2'])
df2['B2'] = np.random.randint(0,1000,(1000000))
%%timeit
x = df1.merge(df2, how='left', left_on='A', right_on='A2')
#1 loop, best of 3: 664 ms per loop
%%timeit
x = df1.set_index('A').join(df2.set_index('A2'), how='left')
#1 loop, best of 3: 354 ms per loop
%%time
df1.set_index('A', inplace=True)
df2.set_index('A2', inplace=True)
#Wall time: 16 ms
%%timeit
x = df1.join(df2, how='left')
#10 loops, best of 3: 80.4 ms per loop
当要连接的列在两个表上的整数不是相同时,您仍然可以期望加速8次.
小智 8
我不知道这是否值得一个新的答案，但就个人而言，以下技巧帮助我进一步改进了在大型 DataFrame（数百万行和数百列）上必须执行的连接：
\nimport random\nimport pandas as pd\nimport numpy as np\n\nnbre_items = 100000\n\nids = np.arange(nbre_items)\nrandom.shuffle(ids)\n\ndf1 = pd.DataFrame({"id": ids})\ndf1[\'value\'] = 1\ndf1.set_index("id", inplace=True)\n\nrandom.shuffle(ids)\n\ndf2 = pd.DataFrame({"id": ids})\ndf2[\'value2\'] = 2\ndf2.set_index("id", inplace=True)\n我得到以下结果：
\n%timeit df1.join(df2)\n13.2 ms \xc2\xb1 349 \xc2\xb5s per loop (mean \xc2\xb1 std. dev. of 7 runs, 100 loops each)\n对索引进行排序后（这需要有限的时间）：
\ndf1.sort_index(inplace=True)\ndf2.sort_index(inplace=True)\n%timeit df1.join(df2)\n764 \xc2\xb5s \xc2\xb1 17.7 \xc2\xb5s per loop (mean \xc2\xb1 std. dev. of 7 runs, 1000 loops each)\nfor i in range(0, df2.shape[1], 100):\n df1 = df1.join(df2.iloc[:, i:min(df2.shape[1], (i + 100))], how=\'outer\')\n| 归档时间： |
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