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数据分析实例之2012美国联邦选举委员会

美国联邦选举委员会发布了有关政治竞选赞助方面的数据。其中包括赞助者的姓名、职业、雇主、地址以及出资额等信息。

这个数据集中一些特征：

contbr_employer:捐款雇主cand_nm:候选人contbr_occupation:捐款人职业contb_receipt_amt:捐款金额

1 前期准备

1.1 使用pandas.read_csv载入数据

import pandas as pdimport matplotlib.pyplot as pltimport numpy as npfec = pd.read_csv('./P00000001-ALL.csv') #加载数据print(fec.info())

   
    RangeIndex: 1001731 entries, 0 to 1001730Data columns (total 16 columns):cmte_id              1001731 non-null objectcand_id              1001731 non-null objectcand_nm              1001731 non-null objectcontbr_nm            1001731 non-null objectcontbr_city          1001712 non-null objectcontbr_st            1001727 non-null objectcontbr_zip           1001620 non-null objectcontbr_employer      988002 non-null objectcontbr_occupation    993301 non-null objectcontb_receipt_amt    1001731 non-null float64contb_receipt_dt     1001731 non-null objectreceipt_desc         14166 non-null objectmemo_cd              92482 non-null objectmemo_text            97770 non-null objectform_tp              1001731 non-null objectfile_num             1001731 non-null int64dtypes: float64(1), int64(1), object(14)memory usage: 122.3+ MBNone
   

1.2 观察样本记录

print(fec.iloc[123456])

cmte_id                             C00431445cand_id                             P80003338cand_nm                         Obama, Barackcontbr_nm                         ELLMAN, IRAcontbr_city                             TEMPEcontbr_st                                  AZcontbr_zip                          852816719contbr_employer      ARIZONA STATE UNIVERSITYcontbr_occupation                   PROFESSORcontb_receipt_amt                          50contb_receipt_dt                    01-DEC-11receipt_desc                              NaNmemo_cd                                   NaNmemo_text                                 NaNform_tp                                 SA17Afile_num                               772372Name: 123456, dtype: object

1.3 使用unique获得所有不同的政治候选人名单

unique_cands = fec.cand_nm.unique() #使用unique获得所有不同的政治候选人名单unique_cands

array(['Bachmann, Michelle', 'Romney, Mitt', 'Obama, Barack',       "Roemer, Charles E. 'Buddy' III", 'Pawlenty, Timothy',       'Johnson, Gary Earl', 'Paul, Ron', 'Santorum, Rick',       'Cain, Herman', 'Gingrich, Newt', 'McCotter, Thaddeus G',       'Huntsman, Jon', 'Perry, Rick'], dtype=object)

unique_cands[2]

'Obama, Barack'

1.4 使用相应的字典表示政党背景

parties = {
   'Bachmann, Michelle':'Republican',           'Cain, Herman':'Republican',           'Gingrich, Newt':'Republican',           'Huntsman, Jon':'Republican',           'John, Gary Earl':'Republican',           'McCotter, Thaddeus G':'Republican',           'Obama, Barack':'Democrat',           'Paul, Ron':'Republican',           'Pawlenty, Timothy':'Republican',           'Perry, Rick':'Republican',           "Roemer, Charles E. 'Buddy' III":'Republican',           'Romney, Mitt':'Republican',           'Santorum, Rick':'Republican'}  #表示政党背景的方式之一是使用相应的字典

1.5 在Series对象中使用map方法和上述的映射关系，从候选人姓名中计算出政党的数组

fec.cand_nm[123456:123461]

123456    Obama, Barack123457    Obama, Barack123458    Obama, Barack123459    Obama, Barack123460    Obama, BarackName: cand_nm, dtype: object

fec.cand_nm[123456:123461].map(parties)

123456    Democrat123457    Democrat123458    Democrat123459    Democrat123460    DemocratName: cand_nm, dtype: object

# 将其作为一列加入fec["party"] = fec.cand_nm.map(parties)fec["party"].value_counts()

Democrat      589127Republican    401114Name: party, dtype: int64

1.6 数据准备要点

(fec.contb_receipt_amt > 0).value_counts()

True    991475Name: contb_receipt_amt, dtype: int64

# 分析范围限制在正向贡献中fec = fec[fec.contb_receipt_amt > 0]

# 为主要候选人准备一个仅对他们的竞选有贡献的子集fec_mrbo = fec[fec.cand_nm.isin(['Obama, Barack','Romney, Mitt'])]

2 按职业和雇主的捐献统计

根据职业分析捐献是一个常见的统计分析。例如，律师（法律代理人）倾向于捐更多的前给民主党，而企业主则更倾向于资助共和党。

2.1 首先，获得按职业的捐献总数

fec.contbr_occupation.value_counts()[:10]#获得按职业的捐献总数

RETIRED         233990NOT PROVIDED     57151ATTORNEY         34286HOMEMAKER        29931PHYSICIAN        23432ENGINEER         14334TEACHER          13990CONSULTANT       13273PROFESSOR        12555NOT EMPLOYED      9828Name: contbr_occupation, dtype: int64

2.2 通过将一种工作匹配到另一种来清理工作种类

occ_mapping = {
       'INFORMATION REQUESTED PER BEST EFFORTS':'NOT PROVIDED',    'INFORMATION REQUESTED':'NOT PROVIDED',    'INFORMATION REQUESTED (BEST EFFORTS)':'NOT PROVIDED',    'C.E.O':'CEO'}f = lambda x:occ_mapping.get(x,x) #如果没有映射，则返回xfec.contbr_occupation = fec.contbr_occupation.map(f)emp_mapping = {
       'INFORMATION REQUESTED PER BEST EFFORTS':'NOT PROVIDED',    'INFORMATION REQUESTED':'NOT PROVIDED',    'SELF':'SELF-EMPLOYED',    'SELF EMPLOYED':'SELF-EMPLOYED',}f = lambda x:emp_mapping.get(x,x) #如果没有映射，则返回xfec.contbr_employer = fec.contbr_employer.map(f)

2.3 使用pivot_table按党派和职业聚合数据，然后过滤出至少捐赠200万美元的子集

by_occupation = fec.pivot_table('contb_receipt_amt',                                index = 'contbr_occupation',                                columns = 'party',aggfunc = 'sum') #按照党派和职业聚合数据over_2mm = by_occupation[by_occupation.sum(1) > 2000000] #过滤出至少捐赠200万美元的子集over_2mm

2.4 以条形图的方式进行数据可视化更为简单

import matplotlib.pyplot as plt%matplotlib inlinefig = plt.figure()over_2mm.plot(kind="barh")

2.5 按候选人名称进行分组

def get_top_amounts(group,key,n = 5):    totals = group.groupby(key)['contb_receipt_amt'].sum()    return totals.nlargest(n)

2.6 按职业和雇主进行聚合

grouped = fec_mrbo.groupby('cand_nm')grouped.apply(get_top_amounts,'contbr_occupation',n = 7) #按照职业进行聚合

cand_nm        contbr_occupationObama, Barack  RETIRED              25305116.38               ATTORNEY             11141982.97               NOT PROVIDED          4866973.96               HOMEMAKER             4248875.80               PHYSICIAN             3735124.94               LAWYER                3160478.87               CONSULTANT            2459912.71Romney, Mitt   RETIRED              11508473.59               NOT PROVIDED         11396894.84               HOMEMAKER             8147446.22               ATTORNEY              5364718.82               PRESIDENT             2491244.89               EXECUTIVE             2300947.03               C.E.O.                1968386.11Name: contb_receipt_amt, dtype: float64

grouped.apply(get_top_amounts,"contbr_employer",n=7)

cand_nm        contbr_employerObama, Barack  RETIRED            22694358.85               SELF-EMPLOYED      18626807.16               NOT EMPLOYED        8586308.70               NOT PROVIDED        5053480.37               HOMEMAKER           2605408.54               STUDENT              318831.45               VOLUNTEER            257104.00Romney, Mitt   NOT PROVIDED       12059527.24               RETIRED            11506225.71               HOMEMAKER           8147196.22               SELF-EMPLOYED       7414115.22               STUDENT              496490.94               CREDIT SUISSE        281150.00               MORGAN STANLEY       267266.00Name: contb_receipt_amt, dtype: float64

3 捐赠金额分桶

3.1 使用cut函数将贡献者的数量按贡献大小离散化分桶

bins = np.array([0,1,10,100,1000,10000,                 100000,1000000,10000000])labels = pd.cut(fec_mrbo.contb_receipt_amt, bins) #使用cut函数将贡献者的数量按贡献大小离散化分桶labels.head()

411      (10, 100]412    (100, 1000]413    (100, 1000]414      (10, 100]415      (10, 100]Name: contb_receipt_amt, dtype: categoryCategories (8, interval[int64]): [(0, 1] < (1, 10] < (10, 100] < (100, 1000] < (1000, 10000] < (10000, 100000] < (100000, 1000000] < (1000000, 10000000]]

labels.tail()

701381      (10, 100]701382    (100, 1000]701383        (1, 10]701384      (10, 100]701385    (100, 1000]Name: contb_receipt_amt, dtype: categoryCategories (8, interval[int64]): [(0, 1] < (1, 10] < (10, 100] < (100, 1000] < (1000, 10000] < (10000, 100000] < (100000, 1000000] < (1000000, 10000000]]

3.2 将Obama和Romney的数据按名称和分类标签进行分组

grouped = fec_mrbo.groupby(['cand_nm',labels]) #将Obama和Romney的数据按名称和分类标签进行分组，以获得捐赠规模的直方图print(grouped.size().unstack(0))

cand_nm              Obama, Barack  Romney, Mittcontb_receipt_amt                               (0, 1]                       493.0          77.0(1, 10]                    40070.0        3681.0(10, 100]                 372280.0       31853.0(100, 1000]               153991.0       43357.0(1000, 10000]              22284.0       26186.0(10000, 100000]                2.0           1.0(100000, 1000000]              3.0           NaN(1000000, 10000000]            4.0           NaN

3.3 对捐款数额进行求和并在桶内进行归一化，以便对按候选人划分的捐款总额百分比进行可视化

bucket_sums = grouped.contb_receipt_amt.sum().unstack(0)normed_sums = bucket_sums.div(bucket_sums.sum(axis = 1),axis = 0) #对捐款数额进行归一化print(normed_sums)

cand_nm              Obama, Barack  Romney, Mittcontb_receipt_amt                               (0, 1]                    0.805182      0.194818(1, 10]                   0.918767      0.081233(10, 100]                 0.910769      0.089231(100, 1000]               0.710176      0.289824(1000, 10000]             0.447326      0.552674(10000, 100000]           0.823120      0.176880(100000, 1000000]         1.000000           NaN(1000000, 10000000]       1.000000           NaN

normed_sums[:-2].plot(kind = 'barh')

注：排除了两个最大的箱体，因为这些箱体不是由个人捐赠的。

这种分析可以通过多种方式进行改进和提高。例如，你可以通过捐助者姓名和邮政编码聚合捐款，以便为那些进行很多次小额捐赠的人进行调整，他们并不会进行大型捐赠。

4 按州进行捐赠统计

4.1 将数据按照候选人和州进行聚合

grouped = fec_mrbo.groupby(['cand_nm','contbr_st']) #按照候选人和州进行聚合totals = grouped.contb_receipt_amt.sum().unstack(0).fillna(0)totals = totals[totals.sum(1) > 100000]print(totals[:10])

cand_nm    Obama, Barack  Romney, Mittcontbr_st                             AK             281840.15      86204.24AL             543123.48     527303.51AR             359247.28     105556.00AZ            1506476.98    1888436.23CA           23824984.24   11237636.60CO            2132429.49    1506714.12CT            2068291.26    3499475.45DC            4373538.80    1025137.50DE             336669.14      82712.00FL            7318178.58    8338458.81

4.2 将每一行除以捐款总额，得到每个候选人按州的捐赠总额的相对百分比

percent = totals.div(totals.sum(1),axis = 0) #将每一行除以捐款总额，就可以得到每个候选人按州的捐赠总额的相对百分比print(percent[:10])

cand_nm    Obama, Barack  Romney, Mittcontbr_st                             AK              0.765778      0.234222AL              0.507390      0.492610AR              0.772902      0.227098AZ              0.443745      0.556255CA              0.679498      0.320502CO              0.585970      0.414030CT              0.371476      0.628524DC              0.810113      0.189887DE              0.802776      0.197224FL              0.467417      0.532583

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