python - 计算用于卡方检验的先前机会数字

Question

所以，我正在使用一个脚本来计算一个人在行中指定日期之前的日期出现在列表中的次数，并且在第 6 列中出现 1，并且还计算了一个人的次数（列7) 出现在行中指定日期之前的日期列表中（注意它们按时间顺序排序。）（使用基于零的列引用）

示例数据集

02/01/2005,Data,Class xpv,4,11yo+,4,1,George Smith
02/01/2005,Data,Class xpv,4,11yo+,4,2,Ted James
02/01/2005,Data,Class xpv,4,11yo+,4,3,Emma Lilly
02/01/2005,Data,Class xpv,4,11yo+,4,5,George Smith
02/01/2005,Data,Class xpv,4,11yo+,6,4,Tom Phillips
03/01/2005,Data,Class tn2,4,10yo+,6,2,Tom Phillips
03/01/2005,Data,Class tn2,4,10yo+,6,5,George Smith
03/01/2005,Data,Class tn2,4,10yo+,6,3,Tom Phillips
03/01/2005,Data,Class tn2,4,10yo+,6,1,Emma Lilly
03/01/2005,Data,Class tn2,4,10yo+,6,6,George Smith
04/01/2005,Data,Class tn2,4,10yo+,6,6,Ted James
04/01/2005,Data,Class tn2,4,10yo+,6,3,Tom Phillips
04/01/2005,Data,Class tn2,4,10yo+,6,2,George Smith
04/01/2005,Data,Class tn2,4,10yo+,6,4,George Smith
04/01/2005,Data,Class tn2,4,10yo+,6,1,George Smith
04/01/2005,Data,Class tn2,4,10yo+,6,5,Tom Phillips
05/01/2005,Data,Class 22zn,2,10yo+,5,3,Emma Lilly
05/01/2005,Data,Class 22zn,2,10yo+,5,1,Ted James
05/01/2005,Data,Class 22zn,2,10yo+,5,2,George Smith
05/01/2005,Data,Class 22zn,2,10yo+,5,4,Emma Lilly
05/01/2005,Data,Class 22zn,2,10yo+,5,5,Tom Phillips

我正在使用的代码

import csv
import datetime
import copy
from collections import defaultdict

with open(r"C:\Temp\test.csv") as i, open(r"C:\Temp\resuls.csv", "wb") as o:
    rdr = csv.reader(i)
    wrt = csv.writer(o)

    data, currdate = defaultdict(lambda:[0, 0, 0, 0]), None
    for line in rdr:
        date, name = datetime.datetime.strptime(line[0], '%d/%m/%Y'), line[7]

        if date != currdate or not currdate:
            for v in data.itervalues(): v[:2] = v[2:]
            currdate = date

        wrt.writerow(line + data[name][:2])

        data[name][3] += 1
        if line[6] == "1": data[name][2] += 1

这将返回：

02/01/2005,Data,Class xpv,4,11yo+,4,1,George Smith,0,0
02/01/2005,Data,Class xpv,4,11yo+,4,2,Ted James,0,0
02/01/2005,Data,Class xpv,4,11yo+,4,3,Emma Lilly,0,0
02/01/2005,Data,Class xpv,4,11yo+,4,5,George Smith,0,0
02/01/2005,Data,Class xpv,4,11yo+,6,4,Tom Phillips,0,0
03/01/2005,Data,Class tn2,4,10yo+,6,2,Tom Phillips,0,1
03/01/2005,Data,Class tn2,4,10yo+,6,5,George Smith,1,2
03/01/2005,Data,Class tn2,4,10yo+,6,3,Tom Phillips,0,1
03/01/2005,Data,Class tn2,4,10yo+,6,1,Emma Lilly,0,1
03/01/2005,Data,Class tn2,4,10yo+,6,6,George Smith,1,2
04/01/2005,Data,Class tn2,4,10yo+,6,6,Ted James,0,1
04/01/2005,Data,Class tn2,4,10yo+,6,3,Tom Phillips,0,3
04/01/2005,Data,Class tn2,4,10yo+,6,2,George Smith,1,4
04/01/2005,Data,Class tn2,4,10yo+,6,4,George Smith,1,4
04/01/2005,Data,Class tn2,4,10yo+,6,1,George Smith,1,4
04/01/2005,Data,Class tn2,4,10yo+,6,5,Tom Phillips,0,3
05/01/2005,Data,Class 22zn,2,10yo+,5,3,Emma Lilly,1,2
05/01/2005,Data,Class 22zn,2,10yo+,5,1,Ted James,0,2
05/01/2005,Data,Class 22zn,2,10yo+,5,2,George Smith,2,7
05/01/2005,Data,Class 22zn,2,10yo+,5,4,Emma Lilly,1,2
05/01/2005,Data,Class 22zn,2,10yo+,5,5,Tom Phillips,0,5

最终我会想要对我生成的百分比数据执行卡方。但是，现在我想要实现的只是能够计算和求和唯一数据类（第 2 列）中任何一个人的分数机会，并将其作为新列附加到 csv 中。我不确定是否可以编辑我使用的代码以实现这一点作为一个多合一的代码。任何关于如何最好地做到这一点的建设性建议或意见将不胜感激。

我想要的输出如下：

02/01/2005,Data,Class xpv,4,11yo+,5,1,George Smith,0,0,0
02/01/2005,Data,Class xpv,4,11yo+,5,2,Ted James,0,0,0
02/01/2005,Data,Class xpv,4,11yo+,5,3,Emma Lilly,0,0,0
02/01/2005,Data,Class xpv,4,11yo+,5,5,George Smith,0,0,0
02/01/2005,Data,Class xpv,4,11yo+,5,4,Tom Phillips,0,0,0
03/01/2005,Data,Class tn2,4,10yo+,5,2,Tom Phillips,0,1,0.2, He gets 0.2 because there was a 1 in 5 chance for previous occurrences on dates prior to today. 1/5
03/01/2005,Data,Class tn2,4,10yo+,5,5,George Smith,1,2,0.4, He gets 0.4 because there was a 2 in 5 chance for previous occurrences on dates prior to today. 2/5
03/01/2005,Data,Class tn2,4,10yo+,5,3,Tom Phillips,0,1,0.2
03/01/2005,Data,Class tn2,4,10yo+,5,1,Emma Lilly,0,1,0.2
03/01/2005,Data,Class tn2,4,10yo+,5,6,George Smith,1,2,0.4
04/01/2005,Data,Class tn2,4,10yo+,6,6,Ted James,0,1,0.2
04/01/2005,Data,Class tn2,4,10yo+,6,3,Tom Phillips,0,3,0.6
04/01/2005,Data,Class tn2,4,10yo+,6,2,George Smith,1,4,0.8
04/01/2005,Data,Class tn2,4,10yo+,6,4,George Smith,1,4,0.8
04/01/2005,Data,Class tn2,4,10yo+,6,1,George Smith,1,4,0.8
04/01/2005,Data,Class tn2,4,10yo+,6,5,Tom Phillips,0,3,0.4
05/01/2005,Data,Class 22zn,2,10yo+,5,3,Emma Lilly,1,2,0.4
05/01/2005,Data,Class 22zn,2,10yo+,5,1,Ted James,0,2,0.366666667
05/01/2005,Data,Class 22zn,2,10yo+,5,2,George Smith,2,7,1.3
05/01/2005,Data,Class 22zn,2,10yo+,5,4,Emma Lilly,1,2,0.4
05/01/2005,Data,Class 22zn,2,10yo+,5,5,Tom Phillips,0,5,0.733333333

Answer 1

这不应该是您问题的完整答案（因为您尝试做的事情有点模棱两可），而只是向您展示熊猫如何自然地适应这种计算；您还可以通过名称而不是索引来调用列。

假设您有一个test.csv这样的文件：

date,x0,cls,x1,x2,x3,tag,name
02/01/2005,Data,Class xpv,4,11yo+,4,1,George Smith
02/01/2005,Data,Class xpv,4,11yo+,4,2,Ted James
02/01/2005,Data,Class xpv,4,11yo+,4,3,Emma Lilly
02/01/2005,Data,Class xpv,4,11yo+,4,5,George Smith
...

我为每一列分配了名称。您可以通过以下方式将此文件读入熊猫数据框

import pandas as pd
df = pd.DataFrame.from_csv( 'test.csv', index_col=None )

df看起来像这样：

          date    x0         cls  x1     x2  x3  tag          name
0   02/01/2005  Data   Class xpv   4  11yo+   4    1  George Smith
1   02/01/2005  Data   Class xpv   4  11yo+   4    2     Ted James
2   02/01/2005  Data   Class xpv   4  11yo+   4    3    Emma Lilly
3   02/01/2005  Data   Class xpv   4  11yo+   4    5  George Smith
...

我删除了您不使用的列（这只是为了演示，您不必删除这些列）

df.drop( labels=['x0', 'x1', 'x2', 'x3'], axis=1, inplace=True )

现在df如下所示：

          date         cls  tag          name
0   02/01/2005   Class xpv    1  George Smith
1   02/01/2005   Class xpv    2     Ted James
2   02/01/2005   Class xpv    3    Emma Lilly
3   02/01/2005   Class xpv    5  George Smith
...

假设您要查找每个人在每天之前的日期中出现的累计次数：

pv = df.pivot_table( cols='name',
                     rows='date',
                     values='tag',
                     aggfunc=len ).shift( 1 ).fillna( 0 ).cumsum( )

api 文档（请参见此处）详细说明了每种方法的作用。现在您有了如下所示的数据透视pv表

date        Emma Lilly  George Smith  Ted James  Tom Phillips
02/01/2005           0             0          0             0
03/01/2005           1             2          1             1
04/01/2005           2             4          1             3
05/01/2005           2             7          2             5

或者可以使用groupby：

df.groupby(['date', 'name'])['name'].aggregate(len).unstack( ).shift( 1 ).fillna( 0 ).cumsum( )

做同样的计算，但只针对tag == 1，你可以做

idx = df.tag == 1
pv1 = df[ idx ].pivot_table( cols='name',
                             rows='date',
                             values='tag',
                             aggfunc=len ).shift( 1 ).fillna( 0 ).cumsum( )

或使用groupby语法：

df[ df.tag == 1 ].groupby(['date', 'name'])['name'].aggregate(len).unstack( ).shift( 1 ).fillna( 0 ).cumsum( )

这将是：

date        Emma Lilly  George Smith  Ted James
02/01/2005           0             0          0
03/01/2005           0             1          0
04/01/2005           1             1          0
05/01/2005           1             2          0

为了填写这两个新列，我们编写了一个辅助函数以在缺少值时回退到 0：

def lookup( pivot_table, col, idx, fall_back=0 ):
    try:
        return pivot_table[ col ][ idx ]
    except KeyError:
        return fall_back

df[ 'cnt1' ] = [ lookup( pv1, row[ 'name' ], row[ 'date' ] ) for idx, row in df.iterrows( ) ]
df[ 'cnt' ] = [ lookup( pv, row[ 'name' ], row[ 'date' ] ) for idx, row in df.iterrows( ) ]

我们得到：

          date         cls  tag          name  cnt1  cnt
0   02/01/2005   Class xpv    1  George Smith     0    0
1   02/01/2005   Class xpv    2     Ted James     0    0
2   02/01/2005   Class xpv    3    Emma Lilly     0    0
3   02/01/2005   Class xpv    5  George Smith     0    0
4   02/01/2005   Class tn2    4  Tom Phillips     0    0
5   03/01/2005   Class tn2    2  Tom Phillips     0    1
6   03/01/2005   Class tn2    5  George Smith     1    2
7   03/01/2005   Class tn2    3  Tom Phillips     0    1
8   03/01/2005   Class tn2    1    Emma Lilly     0    1
9   03/01/2005   Class tn2    6  George Smith     1    2
10  04/01/2005   Class tn2    6     Ted James     0    1
11  04/01/2005   Class tn2    3  Tom Phillips     0    3
12  04/01/2005   Class tn2    2  George Smith     1    4
13  04/01/2005   Class tn2    4  George Smith     1    4
14  04/01/2005   Class tn2    1  George Smith     1    4
15  04/01/2005   Class tn2    5  Tom Phillips     0    3
16  05/01/2005  Class 22zn    3    Emma Lilly     1    2
17  05/01/2005  Class 22zn    1     Ted James     0    2
18  05/01/2005  Class 22zn    2  George Smith     2    7
19  05/01/2005  Class 22zn    4    Emma Lilly     1    2
20  05/01/2005  Class 22zn    5  Tom Phillips     0    5

如果我知道你是如何计算最后一列的，我可以继续下去。例如，为什么“Tom Phillips”在第 6 行得到 0.2？！

编辑：好的，让我们继续。我们需要找出每个人在每个日期出现的次数；这是另一个数据透视表：

appr = df.pivot_table( cols='name',
                       rows='date',
                       values='tag',
                       aggfunc=len ).fillna( 0 )

或者

df.groupby( ['date', 'name'] )['name'].aggregate(len).unstack( ).fillna( 0 )

输出：

date        Emma Lilly  George Smith  Ted James  Tom Phillips
02/01/2005           1             2          1             1
03/01/2005           1             2          0             2
04/01/2005           0             3          1             2
05/01/2005           2             1          1             1

以及每个日期有多少人出现：

total_appr = appr.sum( axis=1 )

输出：

date
02/01/2005    5
03/01/2005    5
04/01/2005    6
05/01/2005    5

要计算累积分数，您可以简单地将每一行除以总数，移位一（因为我们查找以前的日期）并计算累积和：

frac = appr.apply( lambda x: x / total_appr ).shift( 1 ).fillna( 0 ).cumsum( )
df[ 'frac' ] = [ frac[ row[ 'name' ] ][ row[ 'date' ] ] for idx, row in df.iterrows( ) ]

现在df如下所示：

          date         cls  tag          name  cnt1  cnt      frac
0   02/01/2005   Class xpv    1  George Smith     0    0  0.000000
1   02/01/2005   Class xpv    2     Ted James     0    0  0.000000
2   02/01/2005   Class xpv    3    Emma Lilly     0    0  0.000000
3   02/01/2005   Class xpv    5  George Smith     0    0  0.000000
4   02/01/2005   Class tn2    4  Tom Phillips     0    0  0.000000
5   03/01/2005   Class tn2    2  Tom Phillips     0    1  0.200000
6   03/01/2005   Class tn2    5  George Smith     1    2  0.400000
7   03/01/2005   Class tn2    3  Tom Phillips     0    1  0.200000
8   03/01/2005   Class tn2    1    Emma Lilly     0    1  0.200000
9   03/01/2005   Class tn2    6  George Smith     1    2  0.400000
10  04/01/2005   Class tn2    6     Ted James     0    1  0.200000
11  04/01/2005   Class tn2    3  Tom Phillips     0    3  0.600000
12  04/01/2005   Class tn2    2  George Smith     1    4  0.800000
13  04/01/2005   Class tn2    4  George Smith     1    4  0.800000
14  04/01/2005   Class tn2    1  George Smith     1    4  0.800000
15  04/01/2005   Class tn2    5  Tom Phillips     0    3  0.600000
16  05/01/2005  Class 22zn    3    Emma Lilly     1    2  0.400000
17  05/01/2005  Class 22zn    1     Ted James     0    2  0.366667
18  05/01/2005  Class 22zn    2  George Smith     2    7  1.300000
19  05/01/2005  Class 22zn    4    Emma Lilly     1    2  0.400000
20  05/01/2005  Class 22zn    5  Tom Phillips     0    5  0.933333

在最后一列的两行中，我的数字与您的数字不同。所以要么我计算错了，要么你计算错了这两个数字。

Answer 2

这应该非常简单，只是不清楚您所说的“唯一数据类中任何一个人的小概率”是什么意思。例如，您的数据以 data class 的 5 行开头xpv，其中George Smith出现两次。你想为乔治史密斯看到什么样的“小概率”？您想为其他人（出现一次）看到什么？为什么您的示例输出xpv在行旁边只显示零？

答案可能取决于日期类是否在以后出现，以及这对您的计算是否重要；但如果你能解释你是如何计算前 5 个值的，那么其余的可能就会变得清楚。（如果不是，请解释第二组，其中值确实变为非零。）

PS。也许在评论中的讨论中以某种方式解决了这个问题，但是 TL；DNR。如果你能改进问题，就很容易给出正确的答案。