Pandas Pivot MultiIndex高效
[英]Pandas Pivot MultiIndex efficiently
問題描述
我在Pandas處理大約300 MB的財務數據，這與拍賣中的限價訂單相對應。 它是多維數據，看起來像這樣：
bid ask
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity price quantity
2014-05-13 08:47:16.180000 102.298 1000000 102.297 1500000 102.296 6500000 102.295 8000000 102.294 3000000 102.293 24300000 102.292 6000000 102.291 1000000 102.290 1000000 102.289 2500000 102.288 11000000 102.287 4000000 102.286 10100000 102.284 5000000 102.280 1500000 102.276 3000000 102.275 8100000 102.265 9500000 NaN NaN NaN NaN 102.302 2000000 102.303 6100000 102.304 14700000 102.305 3500000 102.307 9800000 102.308 15500000 102.310 5000000 102.312 7000000 102.313 1000000 102.315 8000000 102.316 4500000 102.320 4000000 102.321 1000000 102.324 4000000 102.325 9500000 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
2014-05-13 08:47:17.003000 102.298 1000000 102.297 2500000 102.296 6500000 102.295 7000000 102.294 3000000 102.293 24300000 102.292 6000000 102.291 1000000 102.290 1000000 102.289 2500000 102.288 11000000 102.287 4000000 102.286 10100000 102.284 5000000 102.280 1500000 102.276 3000000 102.275 8100000 102.265 9500000 NaN NaN NaN NaN 102.302 2000000 102.303 5100000 102.304 14700000 102.305 4500000 102.307 9800000 102.308 15500000 102.310 5000000 102.312 7000000 102.313 1000000 102.315 8000000 102.316 4500000 102.320 4000000 102.321 1000000 102.324 4000000 102.325 9500000 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
2014-05-13 08:47:17.005000 102.298 3000000 102.297 3500000 102.296 6000000 102.295 9300000 102.294 4000000 102.293 17500000 102.292 2000000 102.291 4000000 102.290 1000000 102.289 2500000 102.288 6000000 102.287 4000000 102.286 10100000 102.284 5000000 102.280 1500000 102.276 3000000 102.275 8100000 102.265 9500000 NaN NaN NaN NaN 102.302 2000000 102.303 5100000 102.304 14700000 102.305 4500000 102.307 9000000 102.308 16300000 102.310 5000000 102.312 7000000 102.313 1000000 102.315 8000000 102.316 4500000 102.320 4000000 102.321 1000000 102.324 4000000 102.325 9500000 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
2014-05-13 08:47:17.006000 102.299 1000000 102.298 3000000 102.297 6500000 102.296 5000000 102.295 5300000 102.294 4000000 102.293 15500000 102.292 2000000 102.291 4000000 102.290 1000000 102.289 2500000 102.288 6000000 102.287 4000000 102.286 10100000 102.284 5000000 102.280 1500000 102.276 3000000 102.275 8100000 102.265 9500000 NaN NaN 102.302 2000000 102.303 5100000 102.304 11700000 102.305 7500000 102.307 9000000 102.308 11300000 102.309 5000000 102.310 5000000 102.312 7000000 102.313 1000000 102.315 8000000 102.316 4500000 102.320 4000000 102.321 1000000 102.324 4000000 102.325 9500000 NaN NaN NaN NaN NaN NaN NaN NaN
2014-05-13 08:47:17.007000 102.299 1000000 102.298 3000000 102.297 8500000 102.296 4000000 102.295 4300000 102.294 5000000 102.293 14500000 102.292 2000000 102.291 4000000 102.290 1000000 102.289 2500000 102.288 6000000 102.287 4000000 102.286 10100000 102.284 5000000 102.280 1500000 102.276 3000000 102.275 8100000 102.265 9500000 NaN NaN 102.302 2000000 102.303 4100000 102.304 13700000 102.305 7500000 102.307 8000000 102.308 12300000 102.309 5000000 102.310 5000000 102.312 7000000 102.313 1000000 102.315 8000000 102.316 4500000 102.320 4000000 102.321 1000000 102.324 4000000 102.325 9500000 NaN NaN NaN NaN NaN NaN NaN NaN
（注意當你到達20級時，第1級會發生變化。對於表格的長格式抱歉...）
我需要做一些數據透視操作來處理數據。 例如，不是有0,1,2,3 ......（隊列中訂單的相對位置），而是有102.297,102.296，......即訂單的價格作為指數。 他是這種行動的一個例子：
x.stack([0,0]).reset_index(drop=True,level=2).set_index("price",append=True).unstack([1,2]).fillna(0).diff().stack([1,1])
收益：
quantity
side price
2014-05-13 08:47:17.003000 ask 102.300 0
102.301 0
102.302 0
102.303 -1000000
102.304 0
這可以通過stack/unstack/reset_index的組合來實現，但它看起來效率非常低。 我沒有查看代碼，但我猜測表的副本是在每個stack / unstack ，導致我的8GB系統耗盡內存並開始點擊頁面文件。 在這種情況下我也不認為我可以使用pivot ，因為所需的列在多索引中
有關如何加快速度的任何建議嗎？
這是一個示例輸入csv文件，根據評論：
side,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,bid,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask,ask
level,0,0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,14,15,15,16,16,17,17,18,18,19,19,0,0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,14,15,15,16,16,17,17,18,18,19,19
value,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity,price,quantity
2014-05-13 08:47:16.18,102.298,1000000.0,102.297,1500000.0,102.296,6500000.0,102.295,8000000.0,102.294,3000000.0,102.293,2.43E7,102.292,6000000.0,102.291,1000000.0,102.29,1000000.0,102.289,2500000.0,102.288,1.1E7,102.287,4000000.0,102.286,1.01E7,102.284,5000000.0,102.28,1500000.0,102.276,3000000.0,102.275,8100000.0,102.265,9500000.0,N/A,N/A,N/A,N/A,102.302,2000000.0,102.303,6100000.0,102.304,1.47E7,102.305,3500000.0,102.307,9800000.0,102.308,1.55E7,102.31,5000000.0,102.312,7000000.0,102.313,1000000.0,102.315,8000000.0,102.316,4500000.0,102.32,4000000.0,102.321,1000000.0,102.324,4000000.0,102.325,9500000.0,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A
2014-05-13 08:47:17.003,102.298,1000000.0,102.297,2500000.0,102.296,6500000.0,102.295,7000000.0,102.294,3000000.0,102.293,2.43E7,102.292,6000000.0,102.291,1000000.0,102.29,1000000.0,102.289,2500000.0,102.288,1.1E7,102.287,4000000.0,102.286,1.01E7,102.284,5000000.0,102.28,1500000.0,102.276,3000000.0,102.275,8100000.0,102.265,9500000.0,N/A,N/A,N/A,N/A,102.302,2000000.0,102.303,5100000.0,102.304,1.47E7,102.305,4500000.0,102.307,9800000.0,102.308,1.55E7,102.31,5000000.0,102.312,7000000.0,102.313,1000000.0,102.315,8000000.0,102.316,4500000.0,102.32,4000000.0,102.321,1000000.0,102.324,4000000.0,102.325,9500000.0,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A
2014-05-13 08:47:17.005,102.298,3000000.0,102.297,3500000.0,102.296,6000000.0,102.295,9300000.0,102.294,4000000.0,102.293,1.75E7,102.292,2000000.0,102.291,4000000.0,102.29,1000000.0,102.289,2500000.0,102.288,6000000.0,102.287,4000000.0,102.286,1.01E7,102.284,5000000.0,102.28,1500000.0,102.276,3000000.0,102.275,8100000.0,102.265,9500000.0,N/A,N/A,N/A,N/A,102.302,2000000.0,102.303,5100000.0,102.304,1.47E7,102.305,4500000.0,102.307,9000000.0,102.308,1.63E7,102.31,5000000.0,102.312,7000000.0,102.313,1000000.0,102.315,8000000.0,102.316,4500000.0,102.32,4000000.0,102.321,1000000.0,102.324,4000000.0,102.325,9500000.0,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A
2014-05-13 08:47:17.006,102.299,1000000.0,102.298,3000000.0,102.297,6500000.0,102.296,5000000.0,102.295,5300000.0,102.294,4000000.0,102.293,1.55E7,102.292,2000000.0,102.291,4000000.0,102.29,1000000.0,102.289,2500000.0,102.288,6000000.0,102.287,4000000.0,102.286,1.01E7,102.284,5000000.0,102.28,1500000.0,102.276,3000000.0,102.275,8100000.0,102.265,9500000.0,N/A,N/A,102.302,2000000.0,102.303,5100000.0,102.304,1.17E7,102.305,7500000.0,102.307,9000000.0,102.308,1.13E7,102.309,5000000.0,102.31,5000000.0,102.312,7000000.0,102.313,1000000.0,102.315,8000000.0,102.316,4500000.0,102.32,4000000.0,102.321,1000000.0,102.324,4000000.0,102.325,9500000.0,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A
2014-05-13 08:47:17.007,102.299,1000000.0,102.298,3000000.0,102.297,8500000.0,102.296,4000000.0,102.295,4300000.0,102.294,5000000.0,102.293,1.45E7,102.292,2000000.0,102.291,4000000.0,102.29,1000000.0,102.289,2500000.0,102.288,6000000.0,102.287,4000000.0,102.286,1.01E7,102.284,5000000.0,102.28,1500000.0,102.276,3000000.0,102.275,8100000.0,102.265,9500000.0,N/A,N/A,102.302,2000000.0,102.303,4100000.0,102.304,1.37E7,102.305,7500000.0,102.307,8000000.0,102.308,1.23E7,102.309,5000000.0,102.31,5000000.0,102.312,7000000.0,102.313,1000000.0,102.315,8000000.0,102.316,4500000.0,102.32,4000000.0,102.321,1000000.0,102.324,4000000.0,102.325,9500000.0,N/A,N/A,N/A,N/A,N/A,N/A,N/A,N/A
1 個解決方案
解決方案1
1 2014-05-15 11:45:44
Unstack實質上創建了一個索引x列的枚舉，因此當你有很多列和行時它可以創建一個巨大的內存空間。
這是一個解決方案，速度較慢，但應該具有更低的峰值內存使用率（我認為）。 它提供了一個略小的總空間，因為你可能在原件中有一些零條目不在這里（但你總是可以重新索引和填充來修復它）。
定義此功能，這可能會針對這種情況進行優化（已經在級別上進行分組）
In [79]: def f(x):
try:
y = x.stack([0,0]).reset_index(drop=True,level=2).set_index("price",append=True).unstack([1,2]).fillna(0).diff().stack([1,1])
return y[y!=0].dropna()
except:
return None
....:
按列上的“級別”分組並應用f; 不要直接使用apply，而只是將結果作為行連接（這是'unstacking'部分）。
然而，這會產生重復（在價格水平上），因此需要聚合它們。
In [76]: concat([ f(grp) for g, grp in df.groupby(level='level',axis=1) ]).groupby(level=[0,1,2]).sum().sortlevel()
Out[76]:
value quantity
side price
2014-05-13 08:47:17.003 ask 102.303 -1000000
102.305 1000000
bid 102.295 -1000000
102.297 1000000
2014-05-13 08:47:17.005 ask 102.307 -800000
102.308 800000
bid 102.288 -5000000
102.291 3000000
102.292 -4000000
102.293 -6800000
102.294 1000000
102.295 2300000
102.296 -500000
102.297 1000000
102.298 2000000
2014-05-13 08:47:17.006 ask 102.304 -3000000
102.305 3000000
102.308 -5000000
102.309 5000000
102.310 0
102.312 0
102.313 0
102.315 0
102.316 0
102.320 0
102.321 0
102.324 0
102.325 0
bid 102.265 -9500000
102.275 0
102.276 0
102.280 0
102.284 0
102.286 0
102.287 0
102.288 0
102.289 0
102.290 0
102.291 0
102.292 0
102.293 -2000000
102.294 0
102.295 -4000000
102.296 -1000000
102.297 3000000
102.298 0
102.299 1000000
2014-05-13 08:47:17.007 ask 102.303 -1000000
102.304 2000000
102.307 -1000000
102.308 1000000
bid 102.293 -1000000
102.294 1000000
102.295 -1000000
102.296 -1000000
102.297 2000000
計時（我認為優化f會使這個更快）
In [77]: %timeit concat([ f(grp) for g, grp in df.groupby(level='level',axis=1) ]).groupby(level=[0,1,2]).sum().sortlevel()
1 loops, best of 3: 319 ms per loop
In [78]: %memit concat([ f(grp) for g, grp in df.groupby(level='level',axis=1) ]).groupby(level=[0,1,2]).sum().sortlevel()
maximum of 1: 67.515625 MB per loop
原始方法
In [7]: %timeit df.stack([0,0]).reset_index(drop=True,level=2).set_index("price",append=True).unstack([1,2]).fillna(0).diff().stack([1,1])
10 loops, best of 3: 56.4 ms per loop
In [8]: %memit df.stack([0,0]).reset_index(drop=True,level=2).set_index("price",append=True).unstack([1,2]).fillna(0).diff().stack([1,1])
maximum of 1: 61.187500 MB per loop
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