人口分析案例

古城微笑少年丶 2022-01-07 09:11 264阅读 0赞

### 需求： ###

*  导入文件，查看原始数据
 *  将人口数据和各州简称数据进行合并
 *  将合并的数据中重复的abbreviation列进行删除
 *  查看存在缺失数据的列
 *  找到有哪些state/region使得state的值为NaN，进行去重操作
 *  为找到的这些state/region的state项补上正确的值，从而去除掉state这一列的所有NaN
 *  合并各州面积数据areas
 *  我们会发现area(sq.mi)这一列有缺失数据，找出是哪些行
 *  去除含有缺失数据的行
 *  找出2010年的全民人口数据
 *  计算各州的人口密度
 *  排序，并找出人口密度最高的五个州 df.sort\_values()

import numpy as np
    from pandas import DataFrame,Series
    import pandas as pd

abb = pd.read_csv('./data/state-abbrevs.csv')
    pop = pd.read_csv('./data/state-population.csv')
    area = pd.read_csv('./data/state-areas.csv')

abb.head(2)

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>state</th> 
   <th>abbreviation</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>0</td> 
   <td>Alabama</td> 
   <td>AL</td> 
  </tr> 
  <tr> 
   <td>1</td> 
   <td>Alaska</td> 
   <td>AK</td> 
  </tr> 
 </tbody> 
</table>

pop.head(2)

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>state/region</th> 
   <th>ages</th> 
   <th>year</th> 
   <th>population</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>0</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2012</td> 
   <td>1117489.0</td> 
  </tr> 
  <tr> 
   <td>1</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2012</td> 
   <td>4817528.0</td> 
  </tr> 
 </tbody> 
</table>

area.head(2)

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>state</th> 
   <th>area (sq. mi)</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>0</td> 
   <td>Alabama</td> 
   <td>52423</td> 
  </tr> 
  <tr> 
   <td>1</td> 
   <td>Alaska</td> 
   <td>656425</td> 
  </tr> 
 </tbody> 
</table>

### 1. 将人口数据和各州简称数据进行合并 ###

# 将人口数据和各州简称数据进行合并
    abb_pop = pd.merge(abb,pop,how='outer',left_on='abbreviation',right_on='state/region')
    abb_pop.head(4)

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>state</th> 
   <th>abbreviation</th> 
   <th>state/region</th> 
   <th>ages</th> 
   <th>year</th> 
   <th>population</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>0</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2012</td> 
   <td>1117489.0</td> 
  </tr> 
  <tr> 
   <td>1</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2012</td> 
   <td>4817528.0</td> 
  </tr> 
  <tr> 
   <td>2</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2010</td> 
   <td>1130966.0</td> 
  </tr> 
  <tr> 
   <td>3</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2010</td> 
   <td>4785570.0</td> 
  </tr> 
 </tbody> 
</table>

### 2.将合并的数据中重复的abbreviation列进行删除 ###

abb_pop.drop(labels='abbreviation',axis=1,inplace=True)

abb_pop.head(4)

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>state</th> 
   <th>state/region</th> 
   <th>ages</th> 
   <th>year</th> 
   <th>population</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>0</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2012</td> 
   <td>1117489.0</td> 
  </tr> 
  <tr> 
   <td>1</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2012</td> 
   <td>4817528.0</td> 
  </tr> 
  <tr> 
   <td>2</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2010</td> 
   <td>1130966.0</td> 
  </tr> 
  <tr> 
   <td>3</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2010</td> 
   <td>4785570.0</td> 
  </tr> 
 </tbody> 
</table>

### 3.查看存在缺失数据的列 ###

#查看存在缺失数据的列
    # abb_pop.loc[]
    abb_pop.isnull().any(axis=0)

state            True
    state/region    False
    ages            False
    year            False
    population       True
    dtype: bool

### 4.找到有哪些state/region使得state的值为NaN，进行去重操作 ###

#找到有哪些state/region使得state的值为NaN，进行去重操作
    
    # 判断是否为空
    abb_pop['state'].isnull()

# 定位到 state 为空的数据
    abb_pop.loc[abb_pop['state'].isnull()]

#将结果中的state/region列取出，返回的是一个Series
    abb_pop.loc[abb_pop['state'].isnull()]['state/region']
    
    #对Series进行去重
    abb_pop.loc[abb_pop['state'].isnull()]['state/region'].unique()

*  array(\['USA'\], dtype=object)

#为找到的这些state/region的state项补上正确的值，从而去除掉state这一列的所有NaN
    abb_pop.loc[abb_pop['state/region'] == 'PR']
    
    PR_indexs = abb_pop.loc[abb_pop['state/region'] == 'PR'].index
    PR_indexs

Int64Index([2448, 2449, 2450, 2451, 2452, 2453, 2454, 2455, 2456, 2457, 2458,
                2459, 2460, 2461, 2462, 2463, 2464, 2465, 2466, 2467, 2468, 2469,
                2470, 2471, 2472, 2473, 2474, 2475, 2476, 2477, 2478, 2479, 2480,
                2481, 2482, 2483, 2484, 2485, 2486, 2487, 2488, 2489, 2490, 2491,
                2492, 2493, 2494, 2495],
               dtype='int64')

### 5.给对应的 空值赋值 ###

abb_pop.loc[PR_indexs,'state'] = 'ppprrr'

usa_index = abb_pop.loc[abb_pop['state/region'] == 'USA'].index
    # abb_pop.loc[usa_index,'state'] = 'Alabama'
    abb_pop.loc[usa_index,'state']

### 6.合并各州面积数据areas ###

area.head()
    abb_pop_area = pd.merge(abb_pop,area,how='outer')
    abb_pop_area.head()

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>state</th> 
   <th>state/region</th> 
   <th>ages</th> 
   <th>year</th> 
   <th>population</th> 
   <th>area (sq. mi)</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>0</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2012.0</td> 
   <td>1117489.0</td> 
   <td>52423.0</td> 
  </tr> 
  <tr> 
   <td>1</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2012.0</td> 
   <td>4817528.0</td> 
   <td>52423.0</td> 
  </tr> 
  <tr> 
   <td>2</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2010.0</td> 
   <td>1130966.0</td> 
   <td>52423.0</td> 
  </tr> 
  <tr> 
   <td>3</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2010.0</td> 
   <td>4785570.0</td> 
   <td>52423.0</td> 
  </tr> 
  <tr> 
   <td>4</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2011.0</td> 
   <td>1125763.0</td> 
   <td>52423.0</td> 
  </tr> 
 </tbody> 
</table>

### 7.我们会发现area(sq.mi)这一列有缺失数据，找出是哪些行 ###

#我们会发现area(sq.mi)这一列有缺失数据，找出是哪些行
    abb_pop_area['area (sq. mi)'].isnull()
    indexs = abb_pop_area.loc[abb_pop_area['area (sq. mi)'].isnull()].index

### 8.去除含有缺失数据的行 ###

#去除含有缺失数据的行
    abb_pop_area.drop(labels=indexs,axis=0,inplace=True)

### 9.找出2010年的全民人口数据 ###

#找出2010年的全民人口数据
    abb_pop_area.query('year == 2010 & ages == "total"')

### 10.计算各州的人口密度 ###

#计算各州的人口密度
    abb_pop_area['population']/abb_pop_area['area (sq. mi)']
    
    # 将数据添加到 总数据中
    abb_pop_area['midu'] = abb_pop_area['population']/abb_pop_area['area (sq. mi)']
    abb_pop_area.head()

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>state</th> 
   <th>state/region</th> 
   <th>ages</th> 
   <th>year</th> 
   <th>population</th> 
   <th>area (sq. mi)</th> 
   <th>midu</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>0</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2012.0</td> 
   <td>1117489.0</td> 
   <td>52423.0</td> 
   <td>21.316769</td> 
  </tr> 
  <tr> 
   <td>1</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2012.0</td> 
   <td>4817528.0</td> 
   <td>52423.0</td> 
   <td>91.897221</td> 
  </tr> 
  <tr> 
   <td>2</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2010.0</td> 
   <td>1130966.0</td> 
   <td>52423.0</td> 
   <td>21.573851</td> 
  </tr> 
  <tr> 
   <td>3</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>total</td> 
   <td>2010.0</td> 
   <td>4785570.0</td> 
   <td>52423.0</td> 
   <td>91.287603</td> 
  </tr> 
  <tr> 
   <td>4</td> 
   <td>Alabama</td> 
   <td>AL</td> 
   <td>under18</td> 
   <td>2011.0</td> 
   <td>1125763.0</td> 
   <td>52423.0</td> 
   <td>21.474601</td> 
  </tr> 
 </tbody> 
</table>

### 11.排序，并找出人口密度最高的五个州 df.sort\_values() ###

##排序，并找出人口密度最高的五个州 df.sort_values()
    # ascending=False 倒叙
    abb_pop_area.sort_values('midu',axis=0,ascending=False).head(5)

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>state</th> 
   <th>state/region</th> 
   <th>ages</th> 
   <th>year</th> 
   <th>population</th> 
   <th>area (sq. mi)</th> 
   <th>midu</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>439</td> 
   <td>District of Columbia</td> 
   <td>DC</td> 
   <td>total</td> 
   <td>2013.0</td> 
   <td>646449.0</td> 
   <td>68.0</td> 
   <td>9506.602941</td> 
  </tr> 
  <tr> 
   <td>433</td> 
   <td>District of Columbia</td> 
   <td>DC</td> 
   <td>total</td> 
   <td>2012.0</td> 
   <td>633427.0</td> 
   <td>68.0</td> 
   <td>9315.102941</td> 
  </tr> 
  <tr> 
   <td>435</td> 
   <td>District of Columbia</td> 
   <td>DC</td> 
   <td>total</td> 
   <td>2011.0</td> 
   <td>619624.0</td> 
   <td>68.0</td> 
   <td>9112.117647</td> 
  </tr> 
  <tr> 
   <td>479</td> 
   <td>District of Columbia</td> 
   <td>DC</td> 
   <td>total</td> 
   <td>1990.0</td> 
   <td>605321.0</td> 
   <td>68.0</td> 
   <td>8901.779412</td> 
  </tr> 
  <tr> 
   <td>437</td> 
   <td>District of Columbia</td> 
   <td>DC</td> 
   <td>total</td> 
   <td>2010.0</td> 
   <td>605125.0</td> 
   <td>68.0</td> 
   <td>8898.897059</td> 
  </tr> 
 </tbody> 
</table>

转载于:https://www.cnblogs.com/zhang-zi-yi/p/10749490.html