Pandas拼接、数据分析实操

pandas的拼接操作

pandas的拼接分为两种：

• 级联：pd.concat, pd.append
• 合并：pd.merge, pd.join

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

0. 回顾numpy的级联

nd = np.random.randint(0, 150, size=(5, 4))
nd

• 输出

  array([[ 47, 48, 144, 83],
  [ 69, 112, 87, 107],
  [ 67, 47, 74, 145],
  [115, 141, 125, 85],
  [ 65, 12, 87, 118]])

np.concatenate([nd, nd], axis=1)

• 输出

  array([[ 47, 48, 144, 83, 47, 48, 144, 83],
  [ 69, 112, 87, 107, 69, 112, 87, 107],
  [ 67, 47, 74, 145, 67, 47, 74, 145],
  [115, 141, 125, 85, 115, 141, 125, 85],
  [ 65, 12, 87, 118, 65, 12, 87, 118]])

为方便讲解，我们首先定义一个生成DataFrame的函数：

def make_df(cols, index):
    data = {col: [str(col)+str(ind) for ind in index] for col in cols}
    
    df = DataFrame(data=data, columns = cols, index=index)
    
    return df

df1 = make_df(['a', 'b', 'c'], [1, 2, 3])
df1

	a	b	c
1	a1	b1	c1
2	a2	b2	c2
3	a3	b3	c3

df2 = make_df(['a', 'b', 'c'], [4, 5, 6])
df2

	a	b	c
4	a4	b4	c4
5	a5	b5	c5
6	a6	b6	c6

1. 使用pd.concat()级联

pandas使用pd.concat函数，与np.concatenate函数类似，只是多了一些参数：

pd.concat(objs, axis=0, join='outer', join_axes=None, ignore_index=False,
          keys=None, levels=None, names=None, verify_integrity=False,
          copy=True)

1) 简单级联

和np.concatenate一样，优先增加行数（默认axis=0）

# np.concatenate(axis=1)的情况是水平的级联， np中没有index,和columns，所以只要行列相等就可以级联
# 在pd中，如果行和列不一致，但是形状相同，会级联成一个更大的df，但是不对应的值会填充
pd.concat([df1, df2], axis=1)

	a	b	c	a	b	c
1	a1	b1	c1	NaN	NaN	NaN
2	a2	b2	c2	NaN	NaN	NaN
3	a3	b3	c3	NaN	NaN	NaN
4	NaN	NaN	NaN	a4	b4	c4
5	NaN	NaN	NaN	a5	b5	c5
6	NaN	NaN	NaN	a6	b6	c6

可以通过设置axis来改变级联方向

注意index在级联时可以重复

df3 = make_df(['a', 'b', 'c'], [2, 3, 4])
df3

	a	b	c
2	a2	b2	c2
3	a3	b3	c3
4	a4	b4	c4

pd.concat([df1, df3])

	a	b	c
1	a1	b1	c1
2	a2	b2	c2
3	a3	b3	c3
2	a2	b2	c2
3	a3	b3	c3
4	a4	b4	c4

pd.concat([df1, df3], axis=1)

	a	b	c	a	b	c
1	a1	b1	c1	NaN	NaN	NaN
2	a2	b2	c2	a2	b2	c2
3	a3	b3	c3	a3	b3	c3
4	NaN	NaN	NaN	a4	b4	c4

也可以选择忽略ignore_index，重新索引

# gitignore 这个文件，会把写入的文件路径，给屏蔽，不会被上传到云端
# ignore_index的作用是对索引重新排序
pd.concat([df1, df3], axis=0, ignore_index=True)
# 在工作中，大部分的分析来源于mysql,mysql中的id都是唯一的，分表

# 分表， 每个表最大的存储限制是100w（索引条数），mysql,80w（实际中应用）

	a	b	c
0	a1	b1	c1
1	a2	b2	c2
2	a3	b3	c3
3	a2	b2	c2
4	a3	b3	c3
5	a4	b4	c4

或者使用多层索引 keys

concat([x,y],keys=[‘x’,‘y’])

df4 = pd.concat([df1, df3], keys=['期中', '期末'])
df4

		a	b	c
期中	1	a1	b1	c1
	2	a2	b2	c2
	3	a3	b3	c3
期末	2	a2	b2	c2
	3	a3	b3	c3
	4	a4	b4	c4

2) 不匹配级联

不匹配指的是级联的维度的索引不一致。例如纵向级联时列索引不一致，横向级联时行索引不一致

有3种连接方式：

• 外连接：补NaN（默认模式）

df1

	a	b	c
1	a1	b1	c1
2	a2	b2	c2
3	a3	b3	c3

df5 = make_df(['c', 'd', 'e'], [3, 4, 5])
df5

	c	d	e
3	c3	d3	e3
4	c4	d4	e4
5	c5	d5	e5

#using()
df6 = pd.concat([df1, df5], axis=0)
df6

	a	b	c	d	e
1	a1	b1	c1	NaN	NaN
2	a2	b2	c2	NaN	NaN
3	a3	b3	c3	NaN	NaN
3	NaN	NaN	c3	d3	e3
4	NaN	NaN	c4	d4	e4
5	NaN	NaN	c5	d5	e5

• 内连接：只连接匹配的项

# 回忆mysql中outer inner 不同
# 外连接 left 以左边的表中的数据为核心，右表数据不匹配，则填充null
# 内连接 join 两边的表数据不完全对应的话，会只显示能对应上的数据
# join 默认值是outer
df6 = pd.concat([df1, df5], axis=1, join='inner')
df6
# 同mysql一致

	a	b	c	c	d	e
3	a3	b3	c3	c3	d3	e3

• 连接指定轴 join_axes

df1

	a	b	c
1	a1	b1	c1
2	a2	b2	c2
3	a3	b3	c3

df7 = pd.concat([df1,df5], join_axes=[df1.columns])
df7
# join_axes的值是一个列表[df1.index]
# select df1.a df1.b df1.c from df1 left join df5 using(c);
# using(c) 相当于 on df1.c = df5.c

	a	b	c
1	a1	b1	c1
2	a2	b2	c2
3	a3	b3	c3
3	NaN	NaN	c3
4	NaN	NaN	c4
5	NaN	NaN	c5

3) 使用append()函数添加

由于在后面级联的使用非常普遍，因此有一个函数append专门用于在后面添加

append 和 concat 相似

df1.append(df2)

	a	b	c
1	a1	b1	c1
2	a2	b2	c2
3	a3	b3	c3
4	a4	b4	c4
5	a5	b5	c5
6	a6	b6	c6

df1.append(df5)

	a	b	c	d	e
1	a1	b1	c1	NaN	NaN
2	a2	b2	c2	NaN	NaN
3	a3	b3	c3	NaN	NaN
3	NaN	NaN	c3	d3	e3
4	NaN	NaN	c4	d4	e4
5	NaN	NaN	c5	d5	e5

============================================

新建一个只有张三李四王老五的期末考试成绩单ddd3，使用append()与期中考试成绩表ddd级联

============================================

2. 使用pd.merge()合并

merge与concat的区别在于，merge需要依据某一共同的行或列来进行合并

使用pd.merge()合并时，会自动根据两者相同column名称的那一列，作为key来进行合并。

注意每一列元素的顺序不要求一致

1) 一对一合并

df1

	a	b	c
1	a1	b1	c1
2	a2	b2	c2
3	a3	b3	c3

df2

	a	b	c
4	a4	b4	c4
5	a5	b5	c5
6	a6	b6	c6

# 默认的是内连接，表的两边数据都不对应
pd.merge(df1, df2)

	a	b	c

# how:{'left', 'right', 'outer', 'inner'}, default 'inner'
pd.merge(df1,df2, how='right')

	a	b	c
0	a4	b4	c4
1	a5	b5	c5
2	a6	b6	c6

pd.merge(df1, df5)
# how默认的是inner,数据对称才显示

	a	b	c	d	e
0	a3	b3	c3	d3	e3

pd.merge(df1, df5,how='right')

	a	b	c	d	e
0	a3	b3	c3	d3	e3
1	NaN	NaN	c4	d4	e4
2	NaN	NaN	c5	d5	e5

2) 多对一|一对多合并

df8 = make_df(['c', 'd', 'e'],[1, 1, 1, 4])
df8

	c	d	e
1	c1	d1	e1
1	c1	d1	e1
1	c1	d1	e1
4	c4	d4	e4

pd.merge(df1, df8)
# select * from df1 join df8 on df1.c = df8.c

	a	b	c	d	e
0	a1	b1	c1	d1	e1
1	a1	b1	c1	d1	e1
2	a1	b1	c1	d1	e1

pd.merge(df1, df8, how='left')

	a	b	c	d	e
0	a1	b1	c1	d1	e1
1	a1	b1	c1	d1	e1
2	a1	b1	c1	d1	e1
3	a2	b2	c2	NaN	NaN
4	a3	b3	c3	NaN	NaN

pd.merge(df1, df8, how='outer')
# 在工作中要使用outer,自动分配

	a	b	c	d	e
0	a1	b1	c1	d1	e1
1	a1	b1	c1	d1	e1
2	a1	b1	c1	d1	e1
3	a2	b2	c2	NaN	NaN
4	a3	b3	c3	NaN	NaN
5	NaN	NaN	c4	d4	e4

3) 多对多合并

df8

	c	d	e
1	c1	qwe	e1
1	c1	d1	e1
1	c1	d1	e1
4	c4	d4	asd

df8.iloc[0]['d'] = 'qwe'

df8['e'][4] = 'asd'

df9 = make_df(list('abc'),[1,1,4,4])
df9

	a	b	c
1	a1	b1	c1
1	a1	b1	c1
4	a4	b4	c4
4	a4	b4	c4

pd.merge(df9, df8, how='outer')

	a	b	c	d	e
0	a1	b1	c1	qwe	e1
1	a1	b1	c1	d1	e1
2	a1	b1	c1	d1	e1
3	a1	b1	c1	qwe	e1
4	a1	b1	c1	d1	e1
5	a1	b1	c1	d1	e1
6	a4	b4	c4	d4	asd
7	a4	b4	c4	d4	asd

4) key的规范化

• 使用on=显式指定哪一列为key,当有多个key相同时使用

df1

	a	b	c
1	a1	b1	c1
2	a2	b2	c2
3	a3	b3	c3

df10 = make_df(list('cde'),[1, 2, 3])
# 现在的情况是df已经生成了，那么我想把列修改一下
df10.columns = list('wde')
df10

	w	d	e
1	c1	d1	e1
2	c2	d2	e2
3	c3	d3	e3

df11 = make_df(list('bcd'),[1, 2, 3])
df11

	b	c	d
1	b1	c1	d1
2	b2	c2	d2
3	b3	c3	d3

pd.merge(df1, df11, on='b')
# mysql中一般碰到两个字段相同，但是数据类型不同
# 那么a.u = b.u  a.o - b.o
# select a.o as ao , b.0 as bo 

	a	b	c_x	c_y	d
0	a1	b1	c1	c1	d1
1	a2	b2	c2	c2	d2
2	a3	b3	c3	c3	d3

pd.merge(df1, df11, on='c')

• 使用left_on和right_on指定左右两边的列作为key，当左右两边的key都不想等时使用

# on的作用是将两个表中相同数据类型，含义一致的字段进行连接的
pd.merge(df1, df10, left_on='c', right_on='w')
#mysql

	a	b	c	w	d	e
0	a1	b1	c1	c1	d1	e1
1	a2	b2	c2	c2	d2	e2
2	a3	b3	c3	c3	d3	e3

============================================

1. 假设有两份成绩单，除了ddd是张三李四王老五之外，还有ddd4是张三和赵小六的成绩单，如何合并？

2. 如果ddd4中张三的名字被打错了，成为了张十三，怎么办？

============================================

5) 内合并与外合并

• 内合并：只保留两者都有的key（默认模式）

• 外合并 how=‘outer’：补NaN

• 左合并、右合并：how=‘left’，how=‘right’，

============================================

1. 如果只有张三赵小六语数英三个科目的成绩，如何合并？

2. 考虑应用情景，使用多种方式合并ddd与ddd4

============================================

6) 列冲突的解决

当列冲突时，即有多个列名称相同时，需要使用on=来指定哪一个列作为key，配合suffixes指定冲突列名

可以使用suffixes=自己指定后缀

pd.merge(df1, df11, on='c', suffixes=('_up','_down'))

	a	b_up	c	b_down	d
0	a1	b1	c1	b1	d1
1	a2	b2	c2	b2	d2
2	a3	b3	c3	b3	d3

============================================

假设有两个同学都叫李四，ddd5、ddd6都是张三和李四的成绩表，如何合并？

============================================

3. 案例分析：美国各州人口数据分析

首先导入文件，并查看数据样本

# 面积
areas = pd.read_csv('./data/state-areas.csv')
# 缩写
abbr = pd.read_csv('./data/state-abbrevs.csv')
# 人口
pop = pd.read_csv('./data/state-population.csv')

areas.shape

• 输出

  (52, 2)

areas.head()

	state	area (sq. mi)
0	Alabama	52423
1	Alaska	656425
2	Arizona	114006
3	Arkansas	53182
4	California	163707

abbr.shape

• 输出

  (51, 2)

abbr.head()

	state	abbreviation
0	Alabama	AL
1	Alaska	AK
2	Arizona	AZ
3	Arkansas	AR
4	California	CA

pop.shape

• 输出

  (2544, 4)

pop.head()

	state/region	ages	year	population
0	AL	under18	2012	1117489.0
1	AL	total	2012	4817528.0
2	AL	under18	2010	1130966.0
3	AL	total	2010	4785570.0
4	AL	under18	2011	1125763.0

合并pop与abbrevs两个DataFrame，分别依据state/region列和abbreviation列来合并。

为了保留所有信息，使用外合并。

abbrToPop = pd.merge(abbr, pop, left_on='abbreviation', right_on='state/region', how='outer')
abbrToPop.head()                  

	state	abbreviation	state/region	ages	year	population
0	Alabama	AL	AL	under18	2012	1117489.0
1	Alabama	AL	AL	total	2012	4817528.0
2	Alabama	AL	AL	under18	2010	1130966.0
3	Alabama	AL	AL	total	2010	4785570.0
4	Alabama	AL	AL	under18	2011	1125763.0

去除abbreviation的那一列（axis=1）

# drop()
# 一般的一执行完就打印的，这种形式的方法不对原数据产生影响
abbrToPop.drop(labels='abbreviation', axis=1, inplace=True)

abbrToPop.head()
abbrToPop.shape

• 输出

  (2544, 5)

查看存在缺失数据的列。

使用.isnull().any()，只有某一列存在一个缺失数据，就会显示True。

# NaN
abbrToPop.isnull().any()

• 输出

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

查看缺失数据

根据数据是否缺失情况显示数据，如果缺失为True，那么显示

# 怎么计算丢失数据的数量
abbrToPop['state'].isnull().sum()

• 输出

  96

找到有哪些state/region使得state的值为NaN，使用unique()查看非重复值

# 把空数据填充上
cond = abbrToPop['state'].isnull()

# state是州名，如何填充
# unique()
abbrToPop['state/region'][cond].unique()

• 输出

  array([‘PR’, ‘USA’], dtype=object)

# 我们通过翻阅资料查到的PR的全称
# Puerto Rico
# 开始赋值
cond_pr = abbrToPop['state/region'] == 'PR'

abbrToPop['state'][cond_pr] = 'Puerto Rico'

C:\Users\Administrator\Anaconda3\lib\site-packages\ipykernel_launcher.py:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  """Entry point for launching an IPython kernel.

cond_usa = abbrToPop['state/region'] == 'USA'

abbrToPop['state'][cond_usa] = 'United States'

C:\Users\Administrator\Anaconda3\lib\site-packages\ipykernel_launcher.py:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  """Entry point for launching an IPython kernel.

# population 查阅资料，我们先删除掉
abbrToPop.isnull().sum()

• 输出

  state 0
  state/region 0
  ages 0
  year 0
  population 0
  dtype: int64

# dropna()
# inplace表示对原表进行操作
abbrToPop.dropna(inplace=True)

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！

记住这样清除缺失数据NaN的方法！

合并各州面积数据areas，使用左合并。

思考一下为什么使用外合并？

areas.head()

	state	area (sq. mi)
0	Alabama	52423
1	Alaska	656425
2	Arizona	114006
3	Arkansas	53182
4	California	163707

abbrToPopToAreas = pd.merge(abbrToPop, areas, how='outer')
abbrToPopToAreas.head()

	state	state/region	ages	year	population	area (sq. mi)
0	Alabama	AL	under18	2012	1117489.0	52423.0
1	Alabama	AL	total	2012	4817528.0	52423.0
2	Alabama	AL	under18	2010	1130966.0	52423.0
3	Alabama	AL	total	2010	4785570.0	52423.0
4	Alabama	AL	under18	2011	1125763.0	52423.0

继续寻找存在缺失数据的列

我们会发现area(sq.mi)这一列有缺失数据，为了找出是哪一行，我们需要找出是哪个state没有数据

abbrToPopToAreas.isnull().any()

• 输出

  state False
  state/region False
  ages False
  year False
  population False
  area (sq. mi) False
  dtype: bool

cond_area = abbrToPopToAreas['area (sq. mi)'].isnull()

abbrToPopToAreas['state/region'][cond_area]

• 输出

  2476 USA
  2477 USA
  2478 USA
  2479 USA
  2480 USA
  2481 USA
  2482 USA
  2483 USA
  2484 USA
  2485 USA
  2486 USA
  2487 USA
  2488 USA
  2489 USA
  2490 USA
  2491 USA
  2492 USA
  2493 USA
  2494 USA
  2495 USA
  2496 USA
  2497 USA
  2498 USA
  2499 USA
  2500 USA
  2501 USA
  2502 USA
  2503 USA
  2504 USA
  2505 USA
  2506 USA
  2507 USA
  2508 USA
  2509 USA
  2510 USA
  2511 USA
  2512 USA
  2513 USA
  2514 USA
  2515 USA
  2516 USA
  2517 USA
  2518 USA
  2519 USA
  2520 USA
  2521 USA
  2522 USA
  2523 USA
  Name: state/region, dtype: object

total_area = areas['area (sq. mi)'].sum()
total_area

• 输出

  3790399

cond_ab = abbrToPopToAreas['state/region'] == 'USA'
abbrToPopToAreas['area (sq. mi)'][cond_ab] = total_area

C:\Users\Administrator\Anaconda3\lib\site-packages\ipykernel_launcher.py:2: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

查看数据是否缺失

abbrToPopToAreas.isnull().sum()

• 输出

  state 0
  state/region 0
  ages 0
  year 0
  population 0
  area (sq. mi) 0
  dtype: int64

找出2010年的全民人口数据,df.query(查询语句)

abbrToPopToAreas

	state	state/region	ages	year	population	area (sq. mi)
0	Alabama	AL	under18	2012	1117489.0	52423.0
1	Alabama	AL	total	2012	4817528.0	52423.0
2	Alabama	AL	under18	2010	1130966.0	52423.0
3	Alabama	AL	total	2010	4785570.0	52423.0
4	Alabama	AL	under18	2011	1125763.0	52423.0
5	Alabama	AL	total	2011	4801627.0	52423.0
6	Alabama	AL	total	2009	4757938.0	52423.0
7	Alabama	AL	under18	2009	1134192.0	52423.0
8	Alabama	AL	under18	2013	1111481.0	52423.0
9	Alabama	AL	total	2013	4833722.0	52423.0
10	Alabama	AL	total	2007	4672840.0	52423.0
11	Alabama	AL	under18	2007	1132296.0	52423.0
12	Alabama	AL	total	2008	4718206.0	52423.0
13	Alabama	AL	under18	2008	1134927.0	52423.0
14	Alabama	AL	total	2005	4569805.0	52423.0
15	Alabama	AL	under18	2005	1117229.0	52423.0
16	Alabama	AL	total	2006	4628981.0	52423.0
17	Alabama	AL	under18	2006	1126798.0	52423.0
18	Alabama	AL	total	2004	4530729.0	52423.0
19	Alabama	AL	under18	2004	1113662.0	52423.0
20	Alabama	AL	total	2003	4503491.0	52423.0
21	Alabama	AL	under18	2003	1113083.0	52423.0
22	Alabama	AL	total	2001	4467634.0	52423.0
23	Alabama	AL	under18	2001	1120409.0	52423.0
24	Alabama	AL	total	2002	4480089.0	52423.0
25	Alabama	AL	under18	2002	1116590.0	52423.0
26	Alabama	AL	under18	1999	1121287.0	52423.0
27	Alabama	AL	total	1999	4430141.0	52423.0
28	Alabama	AL	total	2000	4452173.0	52423.0
29	Alabama	AL	under18	2000	1122273.0	52423.0
...	...	...	...	...	...	...
2494	United States	USA	under18	1999	71946051.0	3790399.0
2495	United States	USA	total	2000	282162411.0	3790399.0
2496	United States	USA	under18	2000	72376189.0	3790399.0
2497	United States	USA	total	1999	279040181.0	3790399.0
2498	United States	USA	total	2001	284968955.0	3790399.0
2499	United States	USA	under18	2001	72671175.0	3790399.0
2500	United States	USA	total	2002	287625193.0	3790399.0
2501	United States	USA	under18	2002	72936457.0	3790399.0
2502	United States	USA	total	2003	290107933.0	3790399.0
2503	United States	USA	under18	2003	73100758.0	3790399.0
2504	United States	USA	total	2004	292805298.0	3790399.0
2505	United States	USA	under18	2004	73297735.0	3790399.0
2506	United States	USA	total	2005	295516599.0	3790399.0
2507	United States	USA	under18	2005	73523669.0	3790399.0
2508	United States	USA	total	2006	298379912.0	3790399.0
2509	United States	USA	under18	2006	73757714.0	3790399.0
2510	United States	USA	total	2007	301231207.0	3790399.0
2511	United States	USA	under18	2007	74019405.0	3790399.0
2512	United States	USA	total	2008	304093966.0	3790399.0
2513	United States	USA	under18	2008	74104602.0	3790399.0
2514	United States	USA	under18	2013	73585872.0	3790399.0
2515	United States	USA	total	2013	316128839.0	3790399.0
2516	United States	USA	total	2009	306771529.0	3790399.0
2517	United States	USA	under18	2009	74134167.0	3790399.0
2518	United States	USA	under18	2010	74119556.0	3790399.0
2519	United States	USA	total	2010	309326295.0	3790399.0
2520	United States	USA	under18	2011	73902222.0	3790399.0
2521	United States	USA	total	2011	311582564.0	3790399.0
2522	United States	USA	under18	2012	73708179.0	3790399.0
2523	United States	USA	total	2012	313873685.0	3790399.0

2524 rows × 6 columns

abbrToPopToAreas_2010 = abbrToPopToAreas.query('year == 2010 & ages == "total"')

abbrToPopToAreas_2010

	state	state/region	ages	year	population	area (sq. mi)
3	Alabama	AL	total	2010	4785570.0	52423.0
91	Alaska	AK	total	2010	713868.0	656425.0
101	Arizona	AZ	total	2010	6408790.0	114006.0
189	Arkansas	AR	total	2010	2922280.0	53182.0
197	California	CA	total	2010	37333601.0	163707.0
283	Colorado	CO	total	2010	5048196.0	104100.0
293	Connecticut	CT	total	2010	3579210.0	5544.0
379	Delaware	DE	total	2010	899711.0	1954.0
389	District of Columbia	DC	total	2010	605125.0	68.0
475	Florida	FL	total	2010	18846054.0	65758.0
485	Georgia	GA	total	2010	9713248.0	59441.0
570	Hawaii	HI	total	2010	1363731.0	10932.0
581	Idaho	ID	total	2010	1570718.0	83574.0
666	Illinois	IL	total	2010	12839695.0	57918.0
677	Indiana	IN	total	2010	6489965.0	36420.0
762	Iowa	IA	total	2010	3050314.0	56276.0
773	Kansas	KS	total	2010	2858910.0	82282.0
858	Kentucky	KY	total	2010	4347698.0	40411.0
869	Louisiana	LA	total	2010	4545392.0	51843.0
954	Maine	ME	total	2010	1327366.0	35387.0
965	Montana	MT	total	2010	990527.0	147046.0
1050	Nebraska	NE	total	2010	1829838.0	77358.0
1061	Nevada	NV	total	2010	2703230.0	110567.0
1146	New Hampshire	NH	total	2010	1316614.0	9351.0
1157	New Jersey	NJ	total	2010	8802707.0	8722.0
1242	New Mexico	NM	total	2010	2064982.0	121593.0
1253	New York	NY	total	2010	19398228.0	54475.0
1338	North Carolina	NC	total	2010	9559533.0	53821.0
1349	North Dakota	ND	total	2010	674344.0	70704.0
1434	Ohio	OH	total	2010	11545435.0	44828.0
1445	Oklahoma	OK	total	2010	3759263.0	69903.0
1530	Oregon	OR	total	2010	3837208.0	98386.0
1541	Maryland	MD	total	2010	5787193.0	12407.0
1626	Massachusetts	MA	total	2010	6563263.0	10555.0
1637	Michigan	MI	total	2010	9876149.0	96810.0
1722	Minnesota	MN	total	2010	5310337.0	86943.0
1733	Mississippi	MS	total	2010	2970047.0	48434.0
1818	Missouri	MO	total	2010	5996063.0	69709.0
1829	Pennsylvania	PA	total	2010	12710472.0	46058.0
1914	Rhode Island	RI	total	2010	1052669.0	1545.0
1925	South Carolina	SC	total	2010	4636361.0	32007.0
2010	South Dakota	SD	total	2010	816211.0	77121.0
2021	Tennessee	TN	total	2010	6356683.0	42146.0
2106	Texas	TX	total	2010	25245178.0	268601.0
2117	Utah	UT	total	2010	2774424.0	84904.0
2202	Vermont	VT	total	2010	625793.0	9615.0
2213	Virginia	VA	total	2010	8024417.0	42769.0
2298	Washington	WA	total	2010	6742256.0	71303.0
2309	West Virginia	WV	total	2010	1854146.0	24231.0
2394	Wisconsin	WI	total	2010	5689060.0	65503.0
2405	Wyoming	WY	total	2010	564222.0	97818.0
2470	Puerto Rico	PR	total	2010	3721208.0	3515.0
2519	United States	USA	total	2010	309326295.0	3790399.0

对查询结果进行处理，以state列作为新的行索引:set_index

# 工作中会使用id作为列的索引
# set_index()
abbrToPopToAreas_2010.set_index('state',inplace=True)
abbrToPopToAreas_2010.head()

	state/region	ages	year	population	area (sq. mi)
state
Alabama	AL	total	2010	4785570.0	52423.0
Alaska	AK	total	2010	713868.0	656425.0
Arizona	AZ	total	2010	6408790.0	114006.0
Arkansas	AR	total	2010	2922280.0	53182.0
California	CA	total	2010	37333601.0	163707.0

计算人口密度population/area。注意是Series/Series，其结果还是一个Series。

density_2010 = abbrToPopToAreas_2010['population'] / abbrToPopToAreas_2010['area (sq. mi)']

density_2010.head()

• 输出

  state
  Alabama 91.287603
  Alaska 1.087509
  Arizona 56.214497
  Arkansas 54.948667
  California 228.051342
  dtype: float64

2010年的人口密度融合到表中

abbrToPopToAreas_2010['density_2010']=density_2010

C:\Users\Administrator\Anaconda3\lib\site-packages\ipykernel_launcher.py:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  """Entry point for launching an IPython kernel.

排序，并找出人口密度最高的五个州sort_values()(根据值来排序)

abbrToPopToAreas_2010.sort_values(by='density_2010').tail()

	state/region	ages	year	population	area (sq. mi)	density_2010
state
Connecticut	CT	total	2010	3579210.0	5544.0	645.600649
Rhode Island	RI	total	2010	1052669.0	1545.0	681.339159
New Jersey	NJ	total	2010	8802707.0	8722.0	1009.253268
Puerto Rico	PR	total	2010	3721208.0	3515.0	1058.665149
District of Columbia	DC	total	2010	605125.0	68.0	8898.897059

找出人口密度最低的五个州

abbrToPopToAreas_2010.sort_values(by='density_2010').head()

要点总结：

• 统一用loc()索引
• 善于使用.isnull().any()找到存在NaN的列
• 善于使用.unique()确定该列中哪些key是我们需要的
• 一般使用外合并、左合并，目的只有一个：宁愿该列是NaN也不要丢弃其他列的信息

回顾：Series/DataFrame运算与ndarray运算的区别

• Series与DataFrame没有广播，如果对应index没有值，则记为NaN；或者使用add的fill_value来补缺失值
• ndarray有广播，通过重复已有值来计算

苹果股票涨跌绘制

import matplotlib.pyplot as plt
import numpy as np

读取数据

apple = pd.read_csv('./data/AAPL.csv')

apple.head()
# adj闭盘以后的调整价格

	Date	Open	High	Low	Close	Adj Close	Volume
0	1980-12-12	0.513393	0.515625	0.513393	0.513393	0.023268	117258400.0
1	1980-12-15	0.488839	0.488839	0.486607	0.486607	0.022054	43971200.0
2	1980-12-16	0.453125	0.453125	0.450893	0.450893	0.020435	26432000.0
3	1980-12-17	0.462054	0.464286	0.462054	0.462054	0.020941	21610400.0
4	1980-12-18	0.475446	0.477679	0.475446	0.475446	0.021548	18362400.0

apple.tail()

	Date	Open	High	Low	Close	Adj Close	Volume
9453	2018-06-08	191.169998	192.000000	189.770004	191.699997	191.699997	26656800.0
9454	2018-06-11	191.350006	191.970001	190.210007	191.229996	191.229996	18308500.0
9455	2018-06-12	191.389999	192.610001	191.149994	192.279999	192.279999	16911100.0
9456	2018-06-13	192.419998	192.880005	190.440002	190.699997	190.699997	21638400.0
9457	2018-06-14	191.550003	191.570007	190.220001	190.800003	190.800003	21491500.0

apple.shape

• 输出

  (9458, 7)

apple.dtypes

• 输出

  Date datetime64[ns]
  Open float64
  High float64
  Low float64
  Close float64
  Adj Close float64
  Volume float64
  dtype: object

转换一下data的数据类型
mysql中datetime pd.to_datetime()

apple['Date'] = pd.to_datetime(apple['Date'])

apple.set_index('Date', inplace=True)

---------------------------------------------------------------------------

apple.tail()

	Open	High	Low	Close	Adj Close	Volume
Date
2018-06-08	191.169998	192.000000	189.770004	191.699997	191.699997	26656800.0
2018-06-11	191.350006	191.970001	190.210007	191.229996	191.229996	18308500.0
2018-06-12	191.389999	192.610001	191.149994	192.279999	192.279999	16911100.0
2018-06-13	192.419998	192.880005	190.440002	190.699997	190.699997	21638400.0
2018-06-14	191.550003	191.570007	190.220001	190.800003	190.800003	21491500.0

绘制图形

adj_plot = apple['Adj Close'].plot()
fig = adj_plot.get_figure()
# set_size_inches（）设置图片大小，单位英寸
fig.set_size_inches(12,6)

apple.drop('Volume',axis=1,inplace=True)

app = apple.plot()
# 需要获取当前图片
fig1 = app.get_figure()
fig1 = fig1.set_size_inches(12,6)