Pandas案例--人口密度分析
- 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')
abb.head(2)
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state | abbreviation | |
---|---|---|
0 | Alabama | AL |
1 | Alaska | AK |
pop = pd.read_csv('./data/state-population.csv')
pop.head(2)
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state/region | ages | year | population | |
---|---|---|---|---|
0 | AL | under18 | 2012 | 1117489.0 |
1 | AL | total | 2012 | 4817528.0 |
area = pd.read_csv('./data/state-areas.csv')
area.head(2)
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state | area (sq. mi) | |
---|---|---|
0 | Alabama | 52423 |
1 | Alaska | 656425 |
# 将人口数据和各州简称数据进行合并
abb_pop = pd.merge(abb,pop,left_on='abbreviation',right_on='state/region',how='outer')
abb_pop.head(2)
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state | abbreviation | state/region | ages | year | population | |
---|---|---|---|---|---|---|
0 | Alabama | AL | AL | under18 | 2012 | 1117489.0 |
1 | Alabama | AL | AL | total | 2012 | 4817528.0 |
# 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(2)
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state | state/region | ages | year | population | |
---|---|---|---|---|---|
0 | Alabama | AL | under18 | 2012 | 1117489.0 |
1 | Alabama | AL | total | 2012 | 4817528.0 |
# 查看存在缺失数据的列
abb_pop.isnull().any(axis=0)
state True
state/region False
ages False
year False
population True
dtype: bool
# 找到有哪些state/region使得state的值为NaN,进行去重操作
# 1.state列中哪些值为空
abb_pop['state'].isnull()
0 False
1 False
2 False
...
2542 True
2543 True
Name: state, Length: 2544, dtype: bool
# 2.可以将step1中空对应的行数据取出(state中的空值对应的行数据)
abb_pop.loc[abb_pop['state'].isnull()]
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state | state/region | ages | year | population | |
---|---|---|---|---|---|
2448 | NaN | PR | under18 | 1990 | NaN |
... | ... | ... | ... | ... | ... |
2543 | NaN | USA | total | 2012 | 313873685.0 |
96 rows × 5 columns
# 3.将对应的行数据中指定的简称列取出
abb_pop.loc[abb_pop['state'].isnull()]['state/region'].unique()
array(['PR', '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
# 1.先将USA对应的state列中的空值定位到
abb_pop['state/region'] == 'USA'
0 False
1 False
2 False
3 False
...
2541 True
2542 True
2543 True
Name: state/region, Length: 2544, dtype: bool
# 2,将布尔值作为原数据的行索引,取出USA简称对应的行数据
abb_pop.loc[abb_pop['state/region'] == 'USA']
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state | state/region | ages | year | population | |
---|---|---|---|---|---|
2496 | NaN | USA | under18 | 1990 | 64218512.0 |
... | ... | ... | ... | ... | ... |
2542 | NaN | USA | under18 | 2012 | 73708179.0 |
2543 | NaN | USA | total | 2012 | 313873685.0 |
# 3.获取符合要求行数据的行索引
indexs = abb_pop.loc[abb_pop['state/region'] == 'USA'].index
# 4.将indexs这些行中的state列的值批量赋值成united states
abb_pop.loc[indexs,'state'] = 'United Status'
# 将PR对应的state列中的空批量赋值成 PUERTO RICO
abb_pop['state/region'] == 'PR'
abb_pop.loc[abb_pop['state/region'] == 'PR']
indexs = abb_pop.loc[abb_pop['state/region'] == 'PR'].index
abb_pop.loc[indexs,'state'] = 'PUERTO RICO'
# 合并各州面积数据areas
abb_pop_area = pd.merge(abb_pop,area,how='outer')
abb_pop_area.head(3)
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state | state/region | ages | year | population | area (sq. mi) | |
---|---|---|---|---|---|---|
0 | Alabama | AL | under18 | 2012.0 | 1117489.0 | 52423.0 |
1 | Alabama | AL | total | 2012.0 | 4817528.0 | 52423.0 |
2 | Alabama | AL | under18 | 2010.0 | 1130966.0 | 52423.0 |
# 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()
# 将空值对应的行数据取出
indexs = abb_pop_area.loc[abb_pop_area['area (sq. mi)'].isnull()].index
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, 2496, 2497, 2498, 2499, 2500, 2501, 2502,
2503, 2504, 2505, 2506, 2507, 2508, 2509, 2510, 2511, 2512, 2513,
2514, 2515, 2516, 2517, 2518, 2519, 2520, 2521, 2522, 2523, 2524,
2525, 2526, 2527, 2528, 2529, 2530, 2531, 2532, 2533, 2534, 2535,
2536, 2537, 2538, 2539, 2540, 2541, 2542, 2543],
dtype='int64')
# 去除含有缺失数据的行
abb_pop_area.drop(labels=indexs,axis=0,inplace=True)
# 找出2010年的全民人口数据 条件查询
abb_pop_area.query('year == 2010 & ages == "total"')
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state | state/region | ages | year | population | area (sq. mi) | |
---|---|---|---|---|---|---|
3 | Alabama | AL | total | 2010.0 | 4785570.0 | 52423.0 |
91 | Alaska | AK | total | 2010.0 | 713868.0 | 656425.0 |
101 | Arizona | AZ | total | 2010.0 | 6408790.0 | 114006.0 |
189 | Arkansas | AR | total | 2010.0 | 2922280.0 | 53182.0 |
197 | California | CA | total | 2010.0 | 37333601.0 | 163707.0 |
2405 | Wyoming | WY | total | 2010.0 | 564222.0 | 97818.0 |
# 计算各州的人口密度
abb_pop_area['midu'] = abb_pop_area['population'] / abb_pop_area['area (sq. mi)']
abb_pop_area.head(2)
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state | state/region | ages | year | population | area (sq. mi) | midu | |
---|---|---|---|---|---|---|---|
0 | Alabama | AL | under18 | 2012.0 | 1117489.0 | 52423.0 | 21.316769 |
1 | Alabama | AL | total | 2012.0 | 4817528.0 | 52423.0 | 91.897221 |
# 排序,并找出人口密度最高的五个州 df.sort_values()
abb_pop_area.sort_values(by='midu',axis=0,ascending=False).head(5)
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state | state/region | ages | year | population | area (sq. mi) | midu | |
---|---|---|---|---|---|---|---|
391 | District of Columbia | DC | total | 2013.0 | 646449.0 | 68.0 | 9506.602941 |
385 | District of Columbia | DC | total | 2012.0 | 633427.0 | 68.0 | 9315.102941 |
387 | District of Columbia | DC | total | 2011.0 | 619624.0 | 68.0 | 9112.117647 |
431 | District of Columbia | DC | total | 1990.0 | 605321.0 | 68.0 | 8901.779412 |
389 | District of Columbia | DC | total | 2010.0 | 605125.0 | 68.0 | 8898.897059 |
abb_pop_area.groupby(by='state')['area (sq. mi)'].max().sort_values(ascending=False).head(5)
state
Alaska 656425.0
Texas 268601.0
California 163707.0
Montana 147046.0
New Mexico 121593.0
Name: area (sq. mi), dtype: float64
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