附录：

Question

对于每对src和dest机场城市，我想返回列a的百分位数，并给出b列的值。

我可以这样手动执行此操作：

示例df只有2对src / dest（我的实际df中有数千个）：

dt  src dest    a   b
0   2016-01-01  YYZ SFO 548.12  279.28
1   2016-01-01  DFW PDX 111.35  -65.50
2   2016-02-01  YYZ SFO 64.84   342.35
3   2016-02-01  DFW PDX 63.81   61.64
4   2016-03-01  YYZ SFO 614.29  262.83

{'a': {0: 548.12,
  1: 111.34999999999999,
  2: 64.840000000000003,
  3: 63.810000000000002,
  4: 614.28999999999996,
  5: -207.49000000000001,
  6: 151.31999999999999,
  7: -56.43,
  8: 611.37,
  9: -296.62,
  10: 6417.5699999999997,
  11: -376.25999999999999,
  12: 465.12,
  13: -821.73000000000002,
  14: 1270.6700000000001,
  15: -1410.0899999999999,
  16: 1312.6600000000001,
  17: -326.25999999999999,
  18: 1683.3699999999999,
  19: -24.440000000000001,
  20: 583.60000000000002,
  21: -5.2400000000000002,
  22: 1122.74,
  23: 195.21000000000001,
  24: 97.040000000000006,
  25: 133.94},
 'b': {0: 279.27999999999997,
  1: -65.5,
  2: 342.35000000000002,
  3: 61.640000000000001,
  4: 262.82999999999998,
  5: 115.89,
  6: 268.63999999999999,
  7: 2.3500000000000001,
  8: 91.849999999999994,
  9: 62.119999999999997,
  10: 778.33000000000004,
  11: -142.78,
  12: 1675.53,
  13: -214.36000000000001,
  14: 983.80999999999995,
  15: -207.62,
  16: 632.13999999999999,
  17: -132.53,
  18: 422.36000000000001,
  19: 13.470000000000001,
  20: 642.73000000000002,
  21: -144.59999999999999,
  22: 213.15000000000001,
  23: -50.200000000000003,
  24: 338.27999999999997,
  25: -129.69},
 'dest': {0: 'SFO',
  1: 'PDX',
  2: 'SFO',
  3: 'PDX',
  4: 'SFO',
  5: 'PDX',
  6: 'SFO',
  7: 'PDX',
  8: 'SFO',
  9: 'PDX',
  10: 'SFO',
  11: 'PDX',
  12: 'SFO',
  13: 'PDX',
  14: 'SFO',
  15: 'PDX',
  16: 'SFO',
  17: 'PDX',
  18: 'SFO',
  19: 'PDX',
  20: 'SFO',
  21: 'PDX',
  22: 'SFO',
  23: 'PDX',
  24: 'SFO',
  25: 'PDX'},
 'dt': {0: Timestamp('2016-01-01 00:00:00'),
  1: Timestamp('2016-01-01 00:00:00'),
  2: Timestamp('2016-02-01 00:00:00'),
  3: Timestamp('2016-02-01 00:00:00'),
  4: Timestamp('2016-03-01 00:00:00'),
  5: Timestamp('2016-03-01 00:00:00'),
  6: Timestamp('2016-04-01 00:00:00'),
  7: Timestamp('2016-04-01 00:00:00'),
  8: Timestamp('2016-05-01 00:00:00'),
  9: Timestamp('2016-05-01 00:00:00'),
  10: Timestamp('2016-06-01 00:00:00'),
  11: Timestamp('2016-06-01 00:00:00'),
  12: Timestamp('2016-07-01 00:00:00'),
  13: Timestamp('2016-07-01 00:00:00'),
  14: Timestamp('2016-08-01 00:00:00'),
  15: Timestamp('2016-08-01 00:00:00'),
  16: Timestamp('2016-09-01 00:00:00'),
  17: Timestamp('2016-09-01 00:00:00'),
  18: Timestamp('2016-10-01 00:00:00'),
  19: Timestamp('2016-10-01 00:00:00'),
  20: Timestamp('2016-11-01 00:00:00'),
  21: Timestamp('2016-11-01 00:00:00'),
  22: Timestamp('2016-12-01 00:00:00'),
  23: Timestamp('2016-12-01 00:00:00'),
  24: Timestamp('2017-01-01 00:00:00'),
  25: Timestamp('2017-01-01 00:00:00')},
 'src': {0: 'YYZ',
  1: 'DFW',
  2: 'YYZ',
  3: 'DFW',
  4: 'YYZ',
  5: 'DFW',
  6: 'YYZ',
  7: 'DFW',
  8: 'YYZ',
  9: 'DFW',
  10: 'YYZ',
  11: 'DFW',
  12: 'YYZ',
  13: 'DFW',
  14: 'YYZ',
  15: 'DFW',
  16: 'YYZ',
  17: 'DFW',
  18: 'YYZ',
  19: 'DFW',
  20: 'YYZ',
  21: 'DFW',
  22: 'YYZ',
  23: 'DFW',
  24: 'YYZ',
  25: 'DFW'}}

我想要每组src和dest对的百分位数。因此，每对只应有1百分位数值。我只想为b执行date = 2017-01-01的百分位数，其中src对于每对dest和a对，每对i.e. src=YYZ and dest=SFT。有意义吗？

我可以手动执行此操作，例如针对特定对from scipy import stats import datetime as dt import pandas as pd p0 = dt.datetime(2017,1,1) # lets slice df for src=YYZ and dest = SFO x = df[(df.src =='YYZ') & (df.dest =='SFO') & (df.dt ==p0)].b.values[0] # given B, what percentile does it fall in for the entire column A for YYZ, SFO stats.percentileofscore(df['a'],x) 61.53846153846154：

vectorize

在上述情况下，我手动对YYZ和SFO对进行了此操作。但是，我的df中有成千上万对。

我如何pandas features使用groupby而不是遍历每一对？

必须有一种方法可以使用apply并在函数上使用src dest percentile 0 YYZ SFO 61.54 1 DFW PDX 23.07 2 XXX YYY blahblah1 3 AAA BBB blahblah2 ...吗？

我想要的df应该类似于：

def b_percentile_a(df,x,y,b):
    z = df[(df['src'] == x ) & (df['dest'] == y)].a
    r = stats.percentileofscore(z,b)
    return r

b_vector_df = df[df.dt == p0]

b_vector_df['p0_a_percentile_b'] = \
    b_vector_df.apply(lambda x: b_percentile_a(df,x.src,x.dest,x.b), axis=1)

更新

我实施了以下内容：

5.16

100对需要55,000秒。我有~50对。所以这需要36分钟。我需要运行此several days次，以便花费body {margin:0}的运行时间。

必须有更快的方法吗？

Answer 1

获得了令人难以置信的节省时间！

<强>输出：
a_list的大小：49998随机唯一值
百分位_1（你给定的df - scipy）
计算百分位数104次 - 在0：00：07.777022中的104条记录

percentile_9（使用给定df的PercentileOfScore（rank_searchsorted_list）类
计算百分位数104次 - 在0：00：00.000609中的104条记录 _ dt src dest a b pct scipy _ 0: 2016-01-01 YYZ SFO 54812 279.28 74.81299251970079 74.8129925197 1: 2016-01-01 DFW PDX 111.35 -65.5 24.66698667946718 24.6669866795 2: 2016-02-01 YYZ SFO 64.84 342.35 76.4810592423697 76.4810592424 3: 2016-02-01 DFW PDX 63.81 61.64 63.84655386215449 63.8465538622 ... 24: 2017-01-01 YYZ SFO 97.04 338.28 76.3570542821712 76.3570542822 25: 2017-01-01 DFW PDX 133.94 -129.69 21.4668586743469 21.4668586743

查看scipy.percentileofscore的实施情况我发现整个list( a ) 在percentileofscore的每次通话中复制，插入，排序，搜索 - 。

我实施了自己的class PercentileOfScore

import numpy as np
class PercentileOfScore(object):

    def __init__(self, aList):
        self.a = np.array( aList )
        self.a.sort()
        self.n = float(len(self.a))
        self.pct = self.__rank_searchsorted_list
    # end def __init__

    def __rank_searchsorted_list(self, score_list):
        adx = np.searchsorted(self.a, score_list, side='right')
        pct = []
        for idx in adx:
            # Python 2.x needs explicit type casting float(int)
            pct.append( (float(idx) / self.n) * 100.0 )

        return pct
    # end def _rank_searchsorted_list
# end class PercentileOfScore

我不认为def percentile_7符合您的需求。 dt将不予考虑。

PctOS = None
def percentile_7(df_flat):
    global PctOS
    result = {}
    for k in df_flat.pair_dict.keys():
        # df_flat.pair_dict = { 'src.dst': [b,b,...bn] }
        result[k] = PctOS.pct( df_flat.pair_dict[k] )

    return result
# end def percentile_7

在您的手动示例中，您使用整个df.a。在此示例中dt_flat.a_list，但我不确定这是否是您想要的？

from PercentileData import DF_flat
def main():
    # DF_flat.data = {'dt.src.dest':[a,b]}
    df_flat = DF_flat()

    # Instantiate Global PctOS
    global PctOS
    # df_flat.a_list = [a,a,...an]
    PctOS = PercentileOfScore(df_flat.a_list)

    result = percentile_7(df_flat)
    # result = dict{'src.dst':[pct,pct...pctn]}

使用Python测试：3.4.2和2.7.9 - numpy：1.8.2

Answer 2

假设您有一对配对列表，请说pairs = [[a,b], [c,d], ...]并定义df，

    r = stats.percentileofscore(z,b)
    return r

for pair in pairs:
    # get the corresponding rows for each pair
    bvalues = df.loc[(df['src']==pair[0])&(df['dest']==pair[1])][['a', 'b']]
    # apply the percentileofscore map
    b_vector_df['p0_a_percentile_b'] = bvalues.b.apply(lambda x: stats.percentileofscore(bvalues.a, x))

我不完全确定目标是什么。我的理解是，您为每个 src ， dest 对读取 b 值，并查找相应的 a 值，然后计算 a 值的百分位数。如果这有帮助，请告诉我:)）

编辑：假设您只使用五列date, src, dest, a, and b，您可以考虑使用仅包含这5列的数据框副本。它减少了每个提取步骤所需的工作量。我觉得仅使用您需要的大量数据更有效。 Selecting rows from a Dataframe based on values in multiple columns in pandas是一个可能与您相关的讨论。

Answer 3

您可以一次分组多个列。

# takes the b value at a specified point
# and returns its percentile of the full a array
def b_pct(df, p0):
    bval = df.b[df.dt==p0]
    assert bval.size == 1, 'can have only one entry per timestamp'
    bval = bval.values[0]
    # compute the percentile
    return (df.a < bval).sum() / len(df.a)

# splits the full dataframe up into groups by (src, dest) trajectory and
# returns a dataframe of the form src, dest, percentile
def trajectory_b_percentile(df, p0):
    percentile_df = pd.DataFrame([pd.Series([s, d, b_pct(g, p0)],
                                            index=['src', 'dest', 'percentile'])
                                  for ((s, d), g) in df.groupby(('src', 'dest'))])
    return percentile_df

为了比较，上面的代码吐出

           dt  src dest       a       b  p0_a_percentile_b
24 2017-01-01  YYZ  SFO   97.04  338.28          23.076923
25 2017-01-01  DFW  PDX  133.94 -129.69          46.153846

而``trajectory_b_percentile＆＃39;返回

   src dest  percentile
0  DFW  PDX   46.1538
1  YYZ  SFO   23.0769

我没有看到25个条目的任何加速，但它应该是更明显的。

Answer 4

似乎通过将所有内容转换为numpy数组并将百分位数构建为numpy数组来获得另一个相当大的加速：

# Get airport strings as indices
_, ir = np.unique(df['src'].values,  return_inverse=True)
_, ic = np.unique(df['dest'].values, return_inverse=True)

# Get a and b columns
a = df['a'].values
b = df['b'].values

# Compute percentile scores in a numpy array
prc = np.zeros(a.shape)
for i in range(0, a.shape[0]):
    prc[i] = stats.percentileofscore(a[np.logical_and(ir==ir[i], ic==ic[i])], b[i])

在包含24000个条目的数据框上（参见下面的结构），运行%%timeit给出

1 loop, best of 3: 2.17 s per loop

然而，原始版本

df['p0_a_percentile_b'] = \
df.apply(lambda x: b_percentile_a(df,x.src,x.dest,x.b), axis=1)

产量

1 loop, best of 3: 1min 2s per loop

慢得多。我还检查了两个片段通过运行np.all(prc == df.p0_a_percentile_b.values)产生相同的输出，产生True。

附录：

我构建了一个数据框来测试这个，在这里我分享了重现性的过程。我使用100个独特的机场名称获取了2000对机场，然后每对生成12个数据帧行，然后生成随机的a和b列。

import pandas as pd
import numpy as np
import scipy.stats as stats
import numpy.matlib as mat

# Construct dataframe

T=12
N_airports = 100
N_entries = 2000
airports = np.arange(0, N_airports).astype('string')

src  = mat.repmat(airports[np.random.randint(N_airports, size=(N_entries, ))], 1, T)
dest = mat.repmat(airports[np.random.randint(N_airports, size=(N_entries, ))], 1, T)
a    = np.random.uniform(size=N_entries*T)
b    = np.random.uniform(size=N_entries*T)

df = pd.DataFrame(np.vstack((src, dest, a, b)).T, columns=['src', 'dest', 'a', 'b'])

Answer 5

如果这代表您的数据模型，请验证并发表评论！

6 ^ 6对[AAA-ZZZ] = 46,656。通常，每个 PAIR 都有12 RECORDS

这是 PAIR 的 RECORD （0）（DFW PDX）

   dt        src dest       a           b
0: 2016-01-01 DFW PDX   111.35       -65.5

这是 SET （DFW PDX）= PAIR 的记录（DFW PDX）

   dt        src dest       a           b
0: 2016-01-01 DFW PDX   111.35       -65.5
1: 2016-02-01 DFW PDX    63.81       61.64
2: 2016-03-01 DFW PDX  -207.49      115.89
3: 2016-04-01 DFW PDX   -56.43        2.35
4: 2016-05-01 DFW PDX  -296.62       62.12
5: 2016-06-01 DFW PDX  -376.26     -142.78
6: 2016-07-01 DFW PDX  -821.73     -214.36
7: 2016-08-01 DFW PDX -1410.09     -207.62
8: 2016-09-01 DFW PDX  -326.26     -132.53
9: 2016-10-01 DFW PDX   -24.44       13.47
10:2016-11-01 DFW PDX    -5.24      -144.6
11:2016-12-01 DFW PDX   195.21       -50.2
12:2017-01-01 DFW PDX   133.94     -129.69

示例：计算 RECORD 的百分位数（0）
```
   dt        src dest       a           b
0: 2016-01-01 DFW PDX   111.35       -65.5
```
伪代码： stats.percentileofscore（SET（DFW PDX）[a0 ... a12]， - 65.5）= 46.15
示例：计算 SET 的百分位数（DFW PDX）


伪代码
      用于SET（DFW PDX）中的记录：
      stats.percentileofscore（SET（DFW PDX）[a0 ... a12]，record.b）
     输出：pct0 ... pct12

使用rank_searchsorted_list不需要'来记录'：
      rank_searchsorted_list（SET（DFW PDX）[a0 ... a12]，SET（DFW PDX）[b0 ... b12]）
     输出：[pct0 ... pct12]

这是 SET （DFW PDX）矢量化

OBJECT = {'DFW PDX':[
['2016-01-01', '2016-02-01', '2016-03-01', '2016-04-01', '2016-05-01', '2016-06-01', '2016-07-01', '2016-08-01', '2016-09-01', '2016-10-01', '2016-11-01', '2016-12-01', '2017-01-01']
[111.35, 63.81, -207.49, -56.43, -296.62, -376.26, -821.73, -1410.09, -326.26, -24.44, -5.24, 195.21, 133.94]
[-65.5, 61.64, 115.89, 2.35, 62.12, -142.78, -214.36, -207.62, -132.53, 13.47, -144.6, -50.2, -129.69]
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
]}

示例：计算 OBJECT （DFW PDX）的百分位数
使用 stats.percentileofscore ：

a = 1; b = 2
for b_value in OBJECT['DFW PDX'][b]:
    stats.percentileofscore( OBJECT['DFW PDX'][a], b_value)
Output: pct0...pct12

使用 rank_searchsorted_list 对于中的b_value不需要'：

a = 1; b = 2; pct = 3 vector = OBJECT['DFW PDX'] vector[pct] = rank_searchsorted_list( vector[a], vector[b] )

输出：

dt src dest a b pct scipy 0: 2016-01-01 DFW PDX 111.35 -65.5 46.15 46.15 1: 2016-02-01 DFW PDX 63.81 61.64 69.23 69.23 2: 2016-03-01 DFW PDX -207.49 115.89 84.61 84.61 3: 2016-04-01 DFW PDX -56.43 2.35 69.23 69.23 4: 2016-05-01 DFW PDX -296.62 62.12 69.23 69.23 5: 2016-06-01 DFW PDX -376.26 -142.78 46.15 46.15 6: 2016-07-01 DFW PDX -821.73 -214.36 38.46 38.46 7: 2016-08-01 DFW PDX -1410.09 -207.62 38.46 38.46 8: 2016-09-01 DFW PDX -326.26 -132.53 46.15 46.15 9: 2016-10-01 DFW PDX -24.44 13.47 69.23 69.23 10:2016-11-01 DFW PDX -5.24 -144.6 46.15 46.15 11:2016-12-01 DFW PDX 195.21 -50.2 53.84 53.84 12:2017-01-01 DFW PDX 133.94 -129.69 46.15 46.15

请验证并确认计算出的百分位数！

向量化A列B列的百分位值（对于组）

5 个答案:

附录：