pandas表处理下透视表实现pivot_table

摘要: pivot_table，透视表，其功能类似于excel和sql中的透视表，方便

为了将列表转换为二维透视表，之前自己写过代码，今天看到pandas直接有这个方法，感觉简单顺手多了，故重写了社会特征属性的人口矩阵和距离矩阵分离的代码，首先介绍一下pivot_table：

pandas.tools.pivot.pivot_table(data, values=None, index=None, columns=None, aggfunc='mean',fill_value=None, margins=False, dropna=True)
value为显示的值，index为行，columns为列,如下面这个

df = pd.DataFrame({'A' : ['one', 'one', 'two', 'three'] * 3,
'B' : ['A', 'B', 'C'] * 4,'C' : ['foo', 'foo', 'foo', 'bar', 'bar', 'bar'] * 2,
'D' : np.random.randn(12),'E' : np.random.randn(12)})

pd.pivot_table(df, values='D', index=['A', 'B'], columns=['C'])

这样就是df表中以D作为数值域，A,B为行，C为列的数据视图。

下面是过剩通勤的社会特征分类的修正代码，换成pivot_table就简单了很多：

# authors = Kanonpy
# coding=UTF-8
import pandas as pd
import numpy as np
import os
from scipy.optimize import linprog

distance = pd.read_excel('Distance.xlsx')

df = pd.read_excel('chuli.xls')

def commuteCalcu(pop,dist,name):
#保证pop和dist行列数值相等
intdistcolumns = {d:int(float(d)) for d in dist.columns}
intdistindex = {d:int(float(d)) for d in dist.index}
unicodepopcolumns = {d:unicode(d) for d in pop.columns}
unicodepopindex = {d:unicode(d) for d in pop.index}
for d in dist.columns:
if d not in pop.columns:
dist = dist.drop(d,axis=1)
#print 'the col %s in distance was del'%(str(d))

for i in dist.index:
if i not in pop.index:
dist = dist.drop(i,axis=0)
#print 'the col %s in distance was del '%(str(i))

for d in pop.columns:
if d not in dist.columns:
pop = pop.drop(d,axis=1)
#print 'the col %s in distance was del '%(str(i))

for i in pop.index:
if i not in dist.index:
pop = pop.drop(i,axis=0)
#print 'the col %s in distance was del '%(str(i))

if not os.path.exists(u'%s'%(name)):
os.mkdir(u'%s'%(name))
print u'creat %s_%s file'%(col,i)
dist.to_excel(u'%s/Population.xlsx'%(name))
pop.to_excel(u'%s/Distance.xlsx'%(name))
matrix = np.array(pop)*np.array(dist)
total_commute = matrix.sum()
commute = total_commute/np.array(pop).sum()
print u'%s 总通勤距离为 %s'%(name,unicode(total_commute))
print u'%s 通勤距离(ARC)为 %s'%(name,unicode(commute))
print u'%s 人口总数为 %s'%(name,unicode(np.array(pop).sum()))

for col in [u'性别', u'户籍', u'职业', u'收入']:
for i in df.groupby(col).size().index:
species = df[df[col]==i]
pt = pd.pivot_table(data=species,values=col,rows=u'工作地或学校地址',
cols=u'居住小区',aggfunc=np.size,fill_value=0)
if sum(pt.shape) > 10:
commuteCalcu(pt,distance,col+u'中的'+unicode(i))
else:
print '%s_%s is too small'%(col,i)

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@Sugar_Lover