Pandas学习总结
pandas提供了快速便捷处理结构化数据的大量数据结构和函数。
一、pandas数据结构介绍
Series和DataFrame
个人理解Series类似一个字典,而DataFrame类似多个字典的组合
Series
pd.Series(data=None, index=None, dtype=None)
- data:传入的数据,可以是ndarray、list等
- index:索引,必须是唯一的,且与数据的长度相等。如果没有传入索引参数,则默认会自动创建一个从0-N的整数索引
- dtype:数据的类型
(1)Series是一种类似于一维数组的对象,它由一组数据(各种NumPy数据类型)以及一组与之相关的数据标签(即索引)组成。
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
arr=[6,7,-5,9]
obj=pd.Series(arr)
obj
0 6
1 7
2 -5
3 9
dtype: int64
obj.index
RangeIndex(start=0, stop=4, step=1)
obj.values
array([ 6, 7, -5, 9], dtype=int64)
(2)你可以自定义索引index==>(类似字典map)
obj2=pd.Series([1,-2,3,4],index=['a','b','c','d'])
obj2
a 1
b -2
c 3
d 4
dtype: int64
(3)索引&根据索引修改值
obj2['a']
1
obj2['b']=-6
obj2[['b','c','d']]
b -6
c 3
d 4
dtype: int64
obj2[obj2>0]
a 1
c 3
d 4
dtype: int64
(4)运算
obj2*2
a 2
b -12
c 6
d 8
dtype: int64
np.exp(obj2)
a 2.718282
b 0.002479
c 20.085537
d 54.598150
dtype: float64
'b' in obj2
True
(5)你可以传入字典map构建Series
sdata={'Ohio':35000,'Texas':71000,'Oregon':16000,'Utah':5000}
obj3=pd.Series(sdata)
obj3
Ohio 35000
Texas 71000
Oregon 16000
Utah 5000
dtype: int64
states=['Wuhan','Texas','Oregon','Utah']
obj4=pd.Series(sdata,index=states)
obj4
Wuhan NaN
Texas 71000.0
Oregon 16000.0
Utah 5000.0
dtype: float64
pd.isnull(obj4)
Wuhan True
Texas False
Oregon False
Utah False
dtype: bool
obj3+obj4
Ohio NaN
Oregon 32000.0
Texas 142000.0
Utah 10000.0
Wuhan NaN
dtype: float64
obj4.name='这个是表格名'
obj4.index.name='index_name'
obj4
index_name
Wuhan NaN
Texas 71000.0
Oregon 16000.0
Utah 5000.0
Name: 这个是表格名, dtype: float64
DataFrame
pd.DataFrame(data=None, index=None, columns=None)
- index:行标签。如果没有传入索引参数,则默认会自动创建一个从0-N的整数索引。
- columns:列标签。如果没有传入索引参数,则默认会自动创建一个从0-N的整数索引
DataFrame是一个表格型的数据结构,它含有一组有序的列,每列可以是不同的值类型(数值、字符串、布尔值等)。虽然DataFrame是以二维结构保存数据的,但你仍然可以轻松地将其表示为更高维度的数据。
DataFrame的属性:
- shape:返回行和列的元组
- index:DataFrame的行索引列表
- columns:DataFrame的列索引列表
- values:直接获取其中array的值
- T:转置
- head(5):显示前5行内容
- tail(5):显示后5行内容
(1)通过字典创建DataFrame
data={'state':['Ohio','Ohio','Ohio','Nevada','Nevada','Nevada'],
'year':[2000,2001,2002,2001,2002,2003],
'pop':[1.5,1.7,3.6,2.4,2.9,3.2]}
frame=pd.DataFrame(data)
frame
| state | year | pop |
---|
0 | Ohio | 2000 | 1.5 |
---|
1 | Ohio | 2001 | 1.7 |
---|
2 | Ohio | 2002 | 3.6 |
---|
3 | Nevada | 2001 | 2.4 |
---|
4 | Nevada | 2002 | 2.9 |
---|
5 | Nevada | 2003 | 3.2 |
---|
frame.head()
| state | year | pop |
---|
0 | Ohio | 2000 | 1.5 |
---|
1 | Ohio | 2001 | 1.7 |
---|
2 | Ohio | 2002 | 3.6 |
---|
3 | Nevada | 2001 | 2.4 |
---|
4 | Nevada | 2002 | 2.9 |
---|
pd.DataFrame(data,columns=['year','pop','state'])
| year | pop | state |
---|
0 | 2000 | 1.5 | Ohio |
---|
1 | 2001 | 1.7 | Ohio |
---|
2 | 2002 | 3.6 | Ohio |
---|
3 | 2001 | 2.4 | Nevada |
---|
4 | 2002 | 2.9 | Nevada |
---|
5 | 2003 | 3.2 | Nevada |
---|
frame2=pd.DataFrame(data,index=['A','B','C','D','E','F'])
frame2
| state | year | pop |
---|
A | Ohio | 2000 | 1.5 |
---|
B | Ohio | 2001 | 1.7 |
---|
C | Ohio | 2002 | 3.6 |
---|
D | Nevada | 2001 | 2.4 |
---|
E | Nevada | 2002 | 2.9 |
---|
F | Nevada | 2003 | 3.2 |
---|
(2)通过列名获取一列Series
frame2['state']
A Ohio
B Ohio
C Ohio
D Nevada
E Nevada
F Nevada
Name: state, dtype: object
frame2.year
A 2000
B 2001
C 2002
D 2001
E 2002
F 2003
Name: year, dtype: int64
(3)通过index获取一行Series
frame2.loc['A']
state Ohio
year 2000
pop 1.5
Name: A, dtype: object
(4)删除一列
frame2['eastern']=frame2.state=='Ohio'
frame2
| state | year | pop | eastern |
---|
A | Ohio | 2000 | 1.5 | True |
---|
B | Ohio | 2001 | 1.7 | True |
---|
C | Ohio | 2002 | 3.6 | True |
---|
D | Nevada | 2001 | 2.4 | False |
---|
E | Nevada | 2002 | 2.9 | False |
---|
F | Nevada | 2003 | 3.2 | False |
---|
del frame2['eastern']
frame2.columns
Index(['state', 'year', 'pop'], dtype='object')
(5)DataFrame也可以做转置
frame2.T
| A | B | C | D | E | F |
---|
state | Ohio | Ohio | Ohio | Nevada | Nevada | Nevada |
---|
year | 2000 | 2001 | 2002 | 2001 | 2002 | 2003 |
---|
pop | 1.5 | 1.7 | 3.6 | 2.4 | 2.9 | 3.2 |
---|
frame2.values
array([['Ohio', 2000, 1.5],
['Ohio', 2001, 1.7],
['Ohio', 2002, 3.6],
['Nevada', 2001, 2.4],
['Nevada', 2002, 2.9],
['Nevada', 2003, 3.2]], dtype=object)
二、基本功能
重新索引
obj=pd.Series([4.5,7.2,-5.3,3.6],index=['d','b','a','c'])
obj
d 4.5
b 7.2
a -5.3
c 3.6
dtype: float64
obj2=obj.reindex(['a','b','c','d','e'])
obj2
a -5.3
b 7.2
c 3.6
d 4.5
e NaN
dtype: float64
obj3=pd.Series(['blue','purple','yellow'],index=[0,2,4])
obj3
obj3.reindex(range(6),method='ffill')
0 blue
1 blue
2 purple
3 purple
4 yellow
5 yellow
dtype: object
frame=pd.DataFrame(np.arange(9).reshape((3,3)),index=['a','b','c'],
columns=['A','B','C'])
frame
states=['Wuhan','Putian','C']
frame.reindex(columns=states)
| Wuhan | Putian | C |
---|
a | NaN | NaN | 2 |
---|
b | NaN | NaN | 5 |
---|
c | NaN | NaN | 8 |
---|
删除某个轴上的项
obj=pd.Series(np.arange(5.),index=['a','b','c','d','e'])
obj
a 0.0
b 1.0
c 2.0
d 3.0
e 4.0
dtype: float64
new_obj=obj.drop('c')
new_obj
a 0.0
b 1.0
d 3.0
e 4.0
dtype: float64
obj.drop(['d','c'])
a 0.0
b 1.0
e 4.0
dtype: float64
df=pd.DataFrame(np.arange(16).reshape((4,4)),
index=['Ohio','Colorado','Utah','New York'],
columns=['A','B','C','D'])
df
| A | B | C | D |
---|
Ohio | 0 | 1 | 2 | 3 |
---|
Colorado | 4 | 5 | 6 | 7 |
---|
Utah | 8 | 9 | 10 | 11 |
---|
New York | 12 | 13 | 14 | 15 |
---|
df.drop(['Colorado','Utah'])
| A | B | C | D |
---|
Ohio | 0 | 1 | 2 | 3 |
---|
New York | 12 | 13 | 14 | 15 |
---|
df.drop('A',axis=1)
| B | C | D |
---|
Ohio | 1 | 2 | 3 |
---|
Colorado | 5 | 6 | 7 |
---|
Utah | 9 | 10 | 11 |
---|
New York | 13 | 14 | 15 |
---|
索引、选取和过滤
(1)Series索引
obj=pd.Series(np.arange(4.),index=['a','b','c','d'])
obj
a 0.0
b 1.0
c 2.0
d 3.0
dtype: float64
obj['b']
1.0
obj[1]
1.0
obj[2:4]
c 2.0
d 3.0
dtype: float64
(2)DataFrame索引
df=pd.DataFrame(np.arange(16).reshape((4,4)),
index=['Ohio','Colorado','Utah','New York'],
columns=['A','B','C','D'])
df
| A | B | C | D |
---|
Ohio | 0 | 1 | 2 | 3 |
---|
Colorado | 4 | 5 | 6 | 7 |
---|
Utah | 8 | 9 | 10 | 11 |
---|
New York | 12 | 13 | 14 | 15 |
---|
df['B']
Ohio 1
Colorado 5
Utah 9
New York 13
Name: B, dtype: int32
df[['A','C']]
| A | C |
---|
Ohio | 0 | 2 |
---|
Colorado | 4 | 6 |
---|
Utah | 8 | 10 |
---|
New York | 12 | 14 |
---|
df[:2]
df[df['A']>0]
| A | B | C | D |
---|
Colorado | 4 | 5 | 6 | 7 |
---|
Utah | 8 | 9 | 10 | 11 |
---|
New York | 12 | 13 | 14 | 15 |
---|
用loc和iloc进行选取(行的选取)
对于DataFrame的行的标签索引,我引入了特殊的标签运算符loc和iloc。它们可以让你用类似NumPy的标记,使用轴标签(loc)或整数索引(iloc),从DataFrame选择行和列的子集。
df.loc['Ohio',['A','B']]
A 0
B 1
Name: Ohio, dtype: int32
df.iloc[0:3,0:2]
df.iloc[2]
A 8
B 9
C 10
D 11
Name: Utah, dtype: int32
算术运算和数据对齐
(1)Series
s1=pd.Series([7.3,-2.5,3.4,1.5],index=['a','b','c','d'])
s2=pd.Series([-2.1,3.6,-1.5,4,3.1],index=['a','c','e','f','g'])
s1
a 7.3
b -2.5
c 3.4
d 1.5
dtype: float64
s2
a -2.1
c 3.6
e -1.5
f 4.0
g 3.1
dtype: float64
s1+s2
a 5.2
b NaN
c 7.0
d NaN
e NaN
f NaN
g NaN
dtype: float64
(2)DataFrame
df1=pd.DataFrame(np.arange(9.).reshape((3,3)),
columns=list('bcd'),
index=['Ohio','Texas','Colorado'])
df2=pd.DataFrame(np.arange(12.).reshape((4,3)),
columns=list('bde'),
index=['Utah','Ohio','Texas','Oregon'])
df1
| b | c | d |
---|
Ohio | 0.0 | 1.0 | 2.0 |
---|
Texas | 3.0 | 4.0 | 5.0 |
---|
Colorado | 6.0 | 7.0 | 8.0 |
---|
df2
| b | d | e |
---|
Utah | 0.0 | 1.0 | 2.0 |
---|
Ohio | 3.0 | 4.0 | 5.0 |
---|
Texas | 6.0 | 7.0 | 8.0 |
---|
Oregon | 9.0 | 10.0 | 11.0 |
---|
df1+df2
| b | c | d | e |
---|
Colorado | NaN | NaN | NaN | NaN |
---|
Ohio | 3.0 | NaN | 6.0 | NaN |
---|
Oregon | NaN | NaN | NaN | NaN |
---|
Texas | 9.0 | NaN | 12.0 | NaN |
---|
Utah | NaN | NaN | NaN | NaN |
---|
算术运算中的填值
在对不同索引的对象进行算术运算时,你可能希望当一个对象中某个轴标签在另一个对象中找不到时填充一个特殊值(比如0)。
df1 = pd.DataFrame(np.arange(12.).reshape((3, 4)),columns=list('abcd'))
df2 = pd.DataFrame(np.arange(20.).reshape((4, 5)),columns=list('abcde'))
df1+df2
| a | b | c | d | e |
---|
0 | 0.0 | 2.0 | 4.0 | 6.0 | NaN |
---|
1 | 9.0 | 11.0 | 13.0 | 15.0 | NaN |
---|
2 | 18.0 | 20.0 | 22.0 | 24.0 | NaN |
---|
3 | NaN | NaN | NaN | NaN | NaN |
---|
df1.add(df2,fill_value=0)
| a | b | c | d | e |
---|
0 | 0.0 | 2.0 | 4.0 | 6.0 | 4.0 |
---|
1 | 9.0 | 11.0 | 13.0 | 15.0 | 9.0 |
---|
2 | 18.0 | 20.0 | 22.0 | 24.0 | 14.0 |
---|
3 | 15.0 | 16.0 | 17.0 | 18.0 | 19.0 |
---|
方法:
add,radd:加法(+);
sub,rsub:减法(-);
div,rdiv:除法(/);
floordiv,rfloordiv:用于底除(//);
mul,rmul:用于乘法(*);
pow,rpow:用于指数(**);
DataFrame和Series之间的运算
跟不同维度的NumPy数组一样,DataFrame和Series之间算术运算也是有明确规定的。==>广播
arr=np.arange(12.).reshape((3,4))
arr
array([[ 0., 1., 2., 3.],
[ 4., 5., 6., 7.],
[ 8., 9., 10., 11.]])
arr-arr[0]
array([[0., 0., 0., 0.],
[4., 4., 4., 4.],
[8., 8., 8., 8.]])
frame = pd.DataFrame(np.arange(12.).reshape((4, 3)),
columns=list('bde'),
index=['Utah', 'Ohio', 'Texas', 'Oregon'])
series=frame.iloc[0]
frame
| b | d | e |
---|
Utah | 0.0 | 1.0 | 2.0 |
---|
Ohio | 3.0 | 4.0 | 5.0 |
---|
Texas | 6.0 | 7.0 | 8.0 |
---|
Oregon | 9.0 | 10.0 | 11.0 |
---|
frame-series
| b | d | e |
---|
Utah | 0.0 | 0.0 | 0.0 |
---|
Ohio | 3.0 | 3.0 | 3.0 |
---|
Texas | 6.0 | 6.0 | 6.0 |
---|
Oregon | 9.0 | 9.0 | 9.0 |
---|
函数应用和映射
NumPy的ufuncs(元素级数组方法)也可用于操作pandas对象。
frame = pd.DataFrame(np.random.randn(4, 3), columns=list('bde'),
index=['Utah', 'Ohio', 'Texas', 'Oregon'])
frame
| b | d | e |
---|
Utah | -0.043022 | 1.722734 | 1.805661 |
---|
Ohio | -0.833497 | 1.536740 | 0.214056 |
---|
Texas | 0.207997 | -0.356338 | -0.814931 |
---|
Oregon | -0.450212 | 0.422857 | 1.699617 |
---|
np.abs(frame)
| b | d | e |
---|
Utah | 0.043022 | 1.722734 | 1.805661 |
---|
Ohio | 0.833497 | 1.536740 | 0.214056 |
---|
Texas | 0.207997 | 0.356338 | 0.814931 |
---|
Oregon | 0.450212 | 0.422857 | 1.699617 |
---|
f=lambda x: x.max()-x.min()
frame.apply(f)
b 1.041494
d 2.079072
e 2.620592
dtype: float64
frame.apply(f,axis='columns')
Utah 1.848683
Ohio 2.370237
Texas 1.022928
Oregon 2.149829
dtype: float64
def f(x):
return pd.Series([x.min(),x.max()],index=['min','max'])
frame.apply(f)
| b | d | e |
---|
min | -0.833497 | -0.356338 | -0.814931 |
---|
max | 0.207997 | 1.722734 | 1.805661 |
---|
排序和排名
根据条件对数据集排序(sorting)也是一种重要的内置运算。要对行或列索引进行排序(按字典顺序),可使用sort_index方法,它将返回一个已排序的新对象。
(1)Series
1)sort_index()
obj = pd.Series(range(4), index=['d', 'a', 'b', 'c'])
obj.sort_index()
a 1
b 2
c 3
d 0
dtype: int64
2)sort_values()
obj2=pd.Series([4,7,-3,2])
obj2.sort_values()
2 -3
3 2
0 4
1 7
dtype: int64
obj3= pd.Series([4, np.nan, 7, np.nan, -3, 2])
obj3.sort_values()
4 -3.0
5 2.0
0 4.0
2 7.0
1 NaN
3 NaN
dtype: float64
3)rank()
obj = pd.Series([7, -5, 7, 4, 2, 0, 4])
obj.rank()
0 6.5
1 1.0
2 6.5
3 4.5
4 3.0
5 2.0
6 4.5
dtype: float64
obj.rank(method='first')
0 6.0
1 1.0
2 7.0
3 4.0
4 3.0
5 2.0
6 5.0
dtype: float64
obj.rank(ascending=False, method='max')
0 2.0
1 7.0
2 2.0
3 4.0
4 5.0
5 6.0
6 4.0
dtype: float64
(2)DataFrame
frame = pd.DataFrame(np.arange(8).reshape((2, 4)),
index=['B', 'A'],
columns=['d', 'a', 'b', 'c'])
frame
1)sort_index()
frame.sort_index()
frame.sort_index(axis=1)
frame.sort_index(axis=1,ascending=False)
2)sort_values()
frame = pd.DataFrame({'b': [4, 7, -3, 2], 'a': [0, 1, 0, 1]})
frame
frame.sort_values(by='b')
frame.sort_values(by=['a','b'])
3)rank()
frame = pd.DataFrame({ 'a': [0, 1, 0, 1],'b': [4.3, 7, -3, 2],
'c': [-2, 5, 8, -2.5]})
frame
| a | b | c |
---|
0 | 0 | 4.3 | -2.0 |
---|
1 | 1 | 7.0 | 5.0 |
---|
2 | 0 | -3.0 | 8.0 |
---|
3 | 1 | 2.0 | -2.5 |
---|
frame.rank(axis='columns')
| a | b | c |
---|
0 | 2.0 | 3.0 | 1.0 |
---|
1 | 1.0 | 3.0 | 2.0 |
---|
2 | 2.0 | 1.0 | 3.0 |
---|
3 | 2.0 | 3.0 | 1.0 |
---|
带有重复标签的轴索引
(1)Series
obj = pd.Series(range(5), index=['a', 'a', 'b', 'b', 'c'])
obj['a']
a 0
a 1
dtype: int64
(2)DataFrame
df = pd.DataFrame(np.random.randn(4, 3), index=['a', 'a', 'b', 'b'])
df.loc['b']
| 0 | 1 | 2 |
---|
b | -1.539688 | 0.887587 | 0.177349 |
---|
b | -1.396467 | 1.041014 | -0.638415 |
---|
三、汇总和计算描述统计
pandas对象拥有一组常用的数学和统计方法。
df = pd.DataFrame([[1.4, np.nan], [7.1, -4.5],
[np.nan, np.nan], [0.75, -1.3]],
index=['a', 'b', 'c', 'd'],
columns=['one', 'two'])
df
| one | two |
---|
a | 1.40 | NaN |
---|
b | 7.10 | -4.5 |
---|
c | NaN | NaN |
---|
d | 0.75 | -1.3 |
---|
sum()
df.sum()
one 9.25
two -5.80
dtype: float64
df.sum(axis=1)
a 1.40
b 2.60
c 0.00
d -0.55
dtype: float64
df.sum(axis=1,skipna=False)
a NaN
b 2.60
c NaN
d -0.55
dtype: float64
idxmin()和idxmax()
最大值或最小值的索引
df.idxmax()
one b
two d
dtype: object
cumsum()
累计
df.cumsum()
| one | two |
---|
a | 1.40 | NaN |
---|
b | 8.50 | -4.5 |
---|
c | NaN | NaN |
---|
d | 9.25 | -5.8 |
---|
describe()
于一次性产生多个汇总统计。
df.describe()
| one | two |
---|
count | 3.000000 | 2.000000 |
---|
mean | 3.083333 | -2.900000 |
---|
std | 3.493685 | 2.262742 |
---|
min | 0.750000 | -4.500000 |
---|
25% | 1.075000 | -3.700000 |
---|
50% | 1.400000 | -2.900000 |
---|
75% | 4.250000 | -2.100000 |
---|
max | 7.100000 | -1.300000 |
---|
方法统计及说明
(1)count:非NA值的数量;
(2)describe:针对Series和DataFrame列计算汇总统计;
(3)min、max:最小值和最大值;
(4)argmin、argmax:计算最小值和最大值索引(整数);
(5)idxmin、idxmax:计算能获得到的最小值和最大值的索引;
(6)quantile:计算样本的分位数;
(7)sum:值的总和;
(8)mean:值的平均值;
(9)median:值的算术中位数(50%分位数);
(10)mad:根据平均值计算平均绝对离差;
(11)var:样本值的方差;
(12)std:样本值的方差;
(13)skew:样本值的偏度(三阶矩);
(14)kurt:样本值的峰度(四阶矩);
(15)cumsum:样本值的累计和;
(16)cummin、cummax:样本值的累计最小值和最大值;
(17)cumprod:样本值的累计积;
(18)diff:计算一阶差分(对时间序列很有用);
(19)pct_change:计算百分数变化;
相关系数和协方差
corr()
用于计算相关系数
cov()
用于计算协方差
唯一值、值计数以及成员资格
unique() ==》唯一值
obj=pd.Series(['c', 'a', 'd', 'a', 'a', 'b', 'b', 'c', 'c'])
uniques = obj.unique()
uniques
array(['c', 'a', 'd', 'b'], dtype=object)
value_counts() ==》值的计数
value_counts用于计算一个Series中各值出现的频率
obj.value_counts()
a 3
c 3
b 2
d 1
dtype: int64
isin() ==》成员资格
mask=obj.isin(['b', 'c'])
mask
0 True
1 False
2 False
3 False
4 False
5 True
6 True
7 True
8 True
dtype: bool
obj
0 c
1 a
2 d
3 a
4 a
5 b
6 b
7 c
8 c
dtype: object
四、分组聚合案例
数据获取
从文件中读取星巴克店铺数据
starbucks = pd.read_csv("./data/directory.csv")
starbucks.head()
| Brand | Store Number | Store Name | Ownership Type | Street Address | City | State/Province | Country | Postcode | Phone Number | Timezone | Longitude | Latitude |
---|
0 | Starbucks | 47370-257954 | Meritxell, 96 | Licensed | Av. Meritxell, 96 | Andorra la Vella | 7 | AD | AD500 | 376818720 | GMT+1:00 Europe/Andorra | 1.53 | 42.51 |
---|
1 | Starbucks | 22331-212325 | Ajman Drive Thru | Licensed | 1 Street 69, Al Jarf | Ajman | AJ | AE | NaN | NaN | GMT+04:00 Asia/Dubai | 55.47 | 25.42 |
---|
2 | Starbucks | 47089-256771 | Dana Mall | Licensed | Sheikh Khalifa Bin Zayed St. | Ajman | AJ | AE | NaN | NaN | GMT+04:00 Asia/Dubai | 55.47 | 25.39 |
---|
3 | Starbucks | 22126-218024 | Twofour 54 | Licensed | Al Salam Street | Abu Dhabi | AZ | AE | NaN | NaN | GMT+04:00 Asia/Dubai | 54.38 | 24.48 |
---|
4 | Starbucks | 17127-178586 | Al Ain Tower | Licensed | Khaldiya Area, Abu Dhabi Island | Abu Dhabi | AZ | AE | NaN | NaN | GMT+04:00 Asia/Dubai | 54.54 | 24.51 |
---|
进行分组聚合
count=starbucks.groupby(['Country']).count()
count.head()
| Brand | Store Number | Store Name | Ownership Type | Street Address | City | State/Province | Postcode | Phone Number | Timezone | Longitude | Latitude |
---|
Country | | | | | | | | | | | | |
---|
AD | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
---|
AE | 144 | 144 | 144 | 144 | 144 | 144 | 144 | 24 | 78 | 144 | 144 | 144 |
---|
AR | 108 | 108 | 108 | 108 | 108 | 108 | 108 | 100 | 29 | 108 | 108 | 108 |
---|
AT | 18 | 18 | 18 | 18 | 18 | 18 | 18 | 18 | 17 | 18 | 18 | 18 |
---|
AU | 22 | 22 | 22 | 22 | 22 | 22 | 22 | 22 | 0 | 22 | 22 | 22 |
---|
count['Brand'].plot(kind='bar',figsize=(20,8))
plt.show()

starbucks.groupby(['Country','State/Province']).count().head(10)
| | Brand | Store Number | Store Name | Ownership Type | Street Address | City | Postcode | Phone Number | Timezone | Longitude | Latitude |
---|
Country | State/Province | | | | | | | | | | | |
---|
AD | 7 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
---|
AE | AJ | 2 | 2 | 2 | 2 | 2 | 2 | 0 | 0 | 2 | 2 | 2 |
---|
AZ | 48 | 48 | 48 | 48 | 48 | 48 | 7 | 20 | 48 | 48 | 48 |
---|
DU | 82 | 82 | 82 | 82 | 82 | 82 | 16 | 50 | 82 | 82 | 82 |
---|
FU | 2 | 2 | 2 | 2 | 2 | 2 | 1 | 0 | 2 | 2 | 2 |
---|
RK | 3 | 3 | 3 | 3 | 3 | 3 | 0 | 3 | 3 | 3 | 3 |
---|
SH | 6 | 6 | 6 | 6 | 6 | 6 | 0 | 5 | 6 | 6 | 6 |
---|
UQ | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
---|
AR | B | 21 | 21 | 21 | 21 | 21 | 21 | 18 | 5 | 21 | 21 | 21 |
---|
C | 73 | 73 | 73 | 73 | 73 | 73 | 71 | 24 | 73 | 73 | 73 |
---|