Pandas快速上手(一):基本操作
作者:互联网
本文包含一些 Pandas 的基本操作,旨在快速上手 Pandas 的基本操作。
读者最好有 NumPy 的基础,如果你还不熟悉 NumPy,建议您阅读NumPy基本操作快速熟悉。
Pandas 数据结构
Pandas 有两个核心的数据结构:Series 和 DataFrame。
Series
Series 是一维的类数组对象,包含一个值序列以及对应的索引。
1 obj = pd.Series([6, 66, 666, 6666]) 2 obj
0 6 1 66 2 666 3 6666 dtype: int64
此时索引默认为 0 到 N。我们可以分别访问 Series 的值和索引:
1 obj.values 2 obj.index # 类似于 range(4)
array([ 6, 66, 666, 6666]) RangeIndex(start=0, stop=4, step=1)
索引可以用标签来指定:
1 obj2 = pd.Series([6, 66, 666, 6666], index=['d', 'b', 'a', 'c']) 2 obj2 3 obj2.index
d 6 b 66 a 666 c 6666 dtype: int64 Index(['d', 'b', 'a', 'c'], dtype='object')
可以使用标签索引来访问 Series 的值,这有点像 NumPy 和字典的结合。
1 obj2['a'] 2 obj2['d'] = 66666 3 obj2[['c', 'a', 'd']]
666 c 6666 a 666 d 66666 dtype: int64
Series 可以使用很多类似于 NumPy 的操作:
1 obj2[obj2 > 100] 2 obj2 / 2 3 np.sqrt(obj2)
d 66666 a 666 c 6666 dtype: int64 d 33333.0 b 33.0 a 333.0 c 3333.0 dtype: float64 d 258.197599 b 8.124038 a 25.806976 c 81.645576 dtype: float64
判断某索引是否存在:
1 'b' in obj2 2 'e' in obj2
True False
可以直接将字典传入 Series 来创建 Series 对象:
1 sdata = {'Ohio': 35000, 'Texas': 71000, 'Oregon': 16000, 'Utah': 5000} 2 obj3 = pd.Series(sdata) 3 obj3 4 states = ['Texas', 'California', 'Ohio', 'Oregon'] 5 obj4 = pd.Series(sdata, index=states) # 指定了索引及其顺序 6 obj4
Ohio 35000 Texas 71000 Oregon 16000 Utah 5000 dtype: int64 Texas 71000.0 California NaN Ohio 35000.0 Oregon 16000.0 dtype: float64
通过 isnull 和 notnull 可以检测是否有空值,既可以使用 Pandas 函数也可以使用 Series 方法:
1 pd.isnull(obj4) 2 pd.notnull(obj4) 3 obj4.isnull() 4 obj4.notnull()
Texas False California True Ohio False Oregon False dtype: bool Texas True California False Ohio True Oregon True dtype: bool Texas False California True Ohio False Oregon False dtype: bool Texas True California False Ohio True Oregon True dtype: bool
Series 的数据对齐,类似于数据库的连接操作:
1 obj3 2 obj4 3 obj3 + obj4
Ohio 35000 Texas 71000 Oregon 16000 Utah 5000 dtype: int64 Texas 71000.0 California NaN Ohio 35000.0 Oregon 16000.0 dtype: float64 California NaN Ohio 70000.0 Oregon 32000.0 Texas 142000.0 Utah NaN dtype: float64
Series 及其索引都有一个 name 属性:
1 obj4.name = 'population' 2 obj4.index.name = 'state' 3 obj4
state Texas 71000.0 California NaN Ohio 35000.0 Oregon 16000.0 Name: population, dtype: float64
Series 的索引可以原地(in-place)修改:
1 obj 2 obj.index = ['Bob', 'Steve', 'Jeff', 'Ryan'] 3 obj
0 6 1 66 2 666 3 6666 dtype: int64 Bob 6 Steve 66 Jeff 666 Ryan 6666 dtype: int64
DataFrame
DataFrame 表示一张矩阵数据表,其中包含有序的列集合,每列都可以表示不同的数据类型。
DataFrame 有行索引和列索引,可以把它当做是 Series 的字典,该字典共享同一套行索引。
通过等长列表(或 NumPy 数组)的字典是常用的创建方式:
1 # 每个值都是长度为 6 的列表 2 data = { 3 'state': ['Ohio', 'Ohio', 'Ohio', 'Nevada', 'Nevada', 'Nevada'], 4 'year': [2000, 2001, 2002, 2001, 2002, 2003], 5 'pop': [2.5, 1.7, 3.6, 2.4, 2.9, 3.2], 6 } 7 frame = pd.DataFrame(data) 8 data
state year pop 0 Ohio 2000 2.5 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
通过 head 方法可以选择前几行:
1 frame.head() 2 frame.head(2)
state year pop 0 Ohio 2000 2.5 1 Ohio 2001 1.7 2 Ohio 2002 3.6 3 Nevada 2001 2.4 4 Nevada 2002 2.9
state year pop 0 Ohio 2000 2.5 1 Ohio 2001 1.7
指定列的序列,可以按照响应顺序来展示:
1 pd.DataFrame(data, columns=['year', 'state', 'pop'])
year state pop 0 2000 Ohio 2.5 1 2001 Ohio 1.7 2 2002 Ohio 3.6 3 2001 Nevada 2.4 4 2002 Nevada 2.9 5 2003 Nevada 3.2
同样也可以指定索引:
1 frame2 = pd.DataFrame(data, columns=['year', 'state', 'pop', 'debt'], 2 index=['one', 'two', 'three', 'four', 'five', 'six']) 3 frame2 4 frame2.columns 5 frame2.index
year state pop debt one 2000 Ohio 2.5 NaN two 2001 Ohio 1.7 NaN three 2002 Ohio 3.6 NaN four 2001 Nevada 2.4 NaN five 2002 Nevada 2.9 NaN six 2003 Nevada 3.2 NaN Index(['year', 'state', 'pop', 'debt'], dtype='object') Index(['one', 'two', 'three', 'four', 'five', 'six'], dtype='object')
通过键或者属性提取 DataFrame 的一列,得到的是 Series:
1 frame2['state'] 2 frame2.year 3 type(frame2.year)
one Ohio two Ohio three Ohio four Nevada five Nevada six Nevada Name: state, dtype: object one 2000 two 2001 three 2002 four 2001 five 2002 six 2003 Name: year, dtype: int64 pandas.core.series.Series
通过 loc 属性可以指定标签,检索行数据:
1 frame2.loc['three'] 2 type(frame2.loc['three'])
year 2002 state Ohio pop 3.6 debt NaN Name: three, dtype: object pandas.core.series.Series
对 'debt' 列进行赋值:
1 frame2['debt'] = 16.5 2 frame2 3 frame2['debt'] = np.arange(6.) 4 frame2
year state pop debt one 2000 Ohio 2.5 16.5 two 2001 Ohio 1.7 16.5 three 2002 Ohio 3.6 16.5 four 2001 Nevada 2.4 16.5 five 2002 Nevada 2.9 16.5 six 2003 Nevada 3.2 16.5 year state pop debt one 2000 Ohio 2.5 0.0 two 2001 Ohio 1.7 1.0 three 2002 Ohio 3.6 2.0 four 2001 Nevada 2.4 3.0 five 2002 Nevada 2.9 4.0 six 2003 Nevada 3.2 5.0
可以用 Series 来赋值 DataFrame 列:
1 val = pd.Series([-1.2, -1.5, -1.7], index=['two', 'four', 'five']) 2 frame2['debt'] = val 3 frame2
year state pop debt one 2000 Ohio 2.5 NaN two 2001 Ohio 1.7 -1.2 three 2002 Ohio 3.6 NaN four 2001 Nevada 2.4 -1.5 five 2002 Nevada 2.9 -1.7 six 2003 Nevada 3.2 NaN
给不存在的列赋值会创建新列:
1 frame2['eastern'] = frame2.state == 'Ohio' 2 frame2
year state pop debt eastern one 2000 Ohio 2.5 NaN True two 2001 Ohio 1.7 -1.2 True three 2002 Ohio 3.6 NaN True four 2001 Nevada 2.4 -1.5 False five 2002 Nevada 2.9 -1.7 False six 2003 Nevada 3.2 NaN False
删除某列:
del frame2['eastern'] frame2.columns
Index(['year', 'state', 'pop', 'debt'], dtype='object')
DataFrame 取出来的列是底层数据的视图,不是拷贝。所做的修改会反映到底层数据中,如果要拷贝必须使用显式的 copy 方法。
传入嵌套字典的情况:
1 pop = { 2 'Nevada': { 3 2001: 2.4, 4 2002: 2.9, 5 }, 6 'Ohio': { 7 2000: 1.5, 8 2001: 1.7, 9 2002: 3.6, 10 } 11 } 12 frame3 = pd.DataFrame(pop) 13 frame3
Nevada Ohio 2000 NaN 1.5 2001 2.4 1.7 2002 2.9 3.6
DataFrame 的转置:
1 frame3.T
2000 2001 2002 Nevada NaN 2.4 2.9 Ohio 1.5 1.7 3.6
values 属性是包含 DataFrame 值的 ndarray:
1 frame3.values
array([[nan, 1.5], [2.4, 1.7], [2.9, 3.6]])
索引对象
Pandas 的索引对象用于存储轴标签和元数据(轴名称等)。
1 obj = pd.Series(range(3), index=['a', 'b', 'c']) 2 index = obj.index 3 index 4 index.name = 'alpha' 5 index[1:]
Index(['a', 'b', 'c'], dtype='object') Index(['b', 'c'], dtype='object', name='alpha')
索引对象是不可变对象,因此不可修改:
1 index[1] = 'd' # 报错
索引类似于固定长度的集合:
1 frame3 2 frame3.columns 3 frame3.index 4 # 类似于集合 5 'Nevada' in frame3.columns 6 2000 in frame3.index
Nevada Ohio 2000 NaN 1.5 2001 2.4 1.7 2002 2.9 3.6 Index(['Nevada', 'Ohio'], dtype='object') Int64Index([2000, 2001, 2002], dtype='int64') True True
但是索引和集合的不同之处是可以包含重复的标签:
1 dup_labels = pd.Index(['foo', 'foo', 'bar', 'bar']) 2 dup_labels 3 set_labels = set(['foo', 'foo', 'bar', 'bar']) 4 set_labels
Index(['foo', 'foo', 'bar', 'bar'], dtype='object') {'bar', 'foo'}
索引对象的一些常用方法:
- append
- difference
- intersection
- union
- isin
- delete
- drop
- insert
- is_monotonic
- is_unique
- unique
Pandas 基本功能
重新索引
reindex 用于创建一个新对象,其索引进行了重新编排。
1 obj = pd.Series([4.5, 7.2, -5.3, 3.6], index=['d', 'b', 'a', 'c']) 2 obj 3 obj2 = obj.reindex(['a', 'b', 'c', 'd', 'e']) 4 obj2
d 4.5 b 7.2 a -5.3 c 3.6 dtype: float64 a -5.3 b 7.2 c 3.6 d 4.5 e NaN dtype: float64
ffill 用于向前填充值,在重新索引时会进行插值:
1 obj3 = pd.Series(['blue', 'purple', 'yellow'], index=[0, 2, 4]) 2 obj3 3 obj3.reindex(range(6), method='ffill')
0 blue 2 purple 4 yellow dtype: object 0 blue 1 blue 2 purple 3 purple 4 yellow 5 yellow dtype: object
无论行索引还是列索引都可以重新编排:
1 frame = pd.DataFrame(np.arange(9).reshape((3, 3)), 2 index=['a', 'c', 'd'], 3 columns=['Ohio', 'Texas', 'California']) 4 frame 5 frame.reindex(['a', 'b', 'c', 'd']) 6 frame.reindex(columns=['Texas', 'Utah', 'California'])
Ohio Texas California a 0 1 2 c 3 4 5 d 6 7 8 Ohio Texas California a 0.0 1.0 2.0 b NaN NaN NaN c 3.0 4.0 5.0 d 6.0 7.0 8.0 Texas Utah California a 1 NaN 2 c 4 NaN 5 d 7 NaN 8
根据轴删除数据
使用 drop 方法对数据行或列进行删除:
1 obj = pd.Series(np.arange(5.), index=['a', 'b', 'c', 'd', 'e']) 2 obj 3 new_obj = obj.drop('c') 4 new_obj 5 obj.drop(['d', 'c'])
a 0.0 b 1.0 c 2.0 d 3.0 e 4.0 dtype: float64 a 0.0 b 1.0 d 3.0 e 4.0 dtype: float64 a 0.0 b 1.0 e 4.0 dtype: float64
1 data = pd.DataFrame(np.arange(16).reshape((4, 4)), 2 index=['Ohio', 'Colorado', 'Utah', 'New York'], 3 columns=['one', 'two', 'three', 'four']) 4 data 5 data.drop(['Colorado', 'Ohio']) 6 data.drop('two', axis=1) 7 data.drop(['two', 'four'], axis='columns')
one two three four Ohio 0 1 2 3 Colorado 4 5 6 7 Utah 8 9 10 11 New York 12 13 14 15 one two three four Utah 8 9 10 11 New York 12 13 14 15 one three four Ohio 0 2 3 Colorado 4 6 7 Utah 8 10 11 New York 12 14 15 one three Ohio 0 2 Colorado 4 6 Utah 8 10 New York 12 14
原地删除:
1 obj 2 obj.drop('c', inplace=True) 3 obj
a 0.0 b 1.0 c 2.0 d 3.0 e 4.0 dtype: float64 a 0.0 b 1.0 d 3.0 e 4.0 dtype: float64
索引、选择、过滤
Series 的索引类似于 NumPy,只不过 Series 还可以用索引标签,不一定是整型索引。
1 obj = pd.Series(np.arange(4.), index=['a', 'b', 'c', 'd']) 2 obj 3 obj['b'] 4 obj[1] 5 obj[2:4] 6 obj[['b', 'a', 'd']] 7 obj[[1, 3]] 8 obj[obj < 2]
a 0.0 b 1.0 c 2.0 d 3.0 dtype: float64 1.0 1.0 c 2.0 d 3.0 dtype: float64 b 1.0 a 0.0 d 3.0 dtype: float64 b 1.0 d 3.0 dtype: float64 a 0.0 b 1.0 dtype: float64
Series 的切片和原生 Python 有所不同,是闭区间(Python 是左闭右开区间)。
1 obj['b':'c']
b 1.0 c 2.0 dtype: float64
使用切片进行赋值:
1 obj['b':'c'] = 5 2 obj
a 0.0 b 5.0 c 5.0 d 3.0 dtype: float64
选择 DataFrame 的若干列:
1 data = pd.DataFrame(np.arange(16).reshape((4, 4)), 2 index=['Ohio', 'Colorado', 'Utah', 'New York'], 3 columns=['one', 'two', 'three', 'four']) 4 data 5 data['two'] 6 type(data['two']) 7 data[['three', 'one']] 8 type(data[['three', 'one']])
one two three four Ohio 0 1 2 3 Colorado 4 5 6 7 Utah 8 9 10 11 New York 12 13 14 15 Ohio 1 Colorado 5 Utah 9 New York 13 Name: two, dtype: int64 pandas.core.series.Series three one Ohio 2 0 Colorado 6 4 Utah 10 8 New York 14 12 pandas.core.frame.DataFrame
使用切片和布尔数组选择 DataFrame:
1 data[:2] 2 data[data['three'] > 5]
one two three four Ohio 0 1 2 3 Colorado 4 5 6 7 one two three four Colorado 4 5 6 7 Utah 8 9 10 11 New York 12 13 14 15
DataFrame 语法上很像二维的 NumPy 数组,使用布尔 DataFrame:
1 data < 5 2 data[data < 5] = 0 3 data
one two three four Ohio True True True True Colorado True False False False Utah False False False False New York False False False False one two three four Ohio 0 0 0 0 Colorado 0 5 6 7 Utah 8 9 10 11 New York 12 13 14 15
使用 loc 和 iloc 进行选择
loc 和 iloc 可以用于选择 DataFrame 的行和列。
1 data.loc['Colorado', ['two', 'three']]
two 5 three 6 Name: Colorado, dtype: int64
1 data.iloc[2, [3, 0, 1]] 2 data.iloc[2] 3 data.iloc[[1, 2], [3, 0, 1]]
four 11 one 8 two 9 Name: Utah, dtype: int64 one 8 two 9 three 10 four 11 Name: Utah, dtype: int64 four one two Colorado 7 0 5 Utah 11 8 9
同样可以使用切片:
1 data.loc[:'Utah', 'two'] 2 data.iloc[:, :3][data.three > 5]
Ohio 0 Colorado 5 Utah 9 Name: two, dtype: int64 one two three Colorado 0 5 6 Utah 8 9 10 New York 12 13 14
DataFrame 常用索引方法:
- df[val]:选列
- df.loc[val]:选行
- df.loc[:, val]:选列
- df.loc[val1, val2]:选列和行
- df.iloc[where]:选行
- df.iloc[:, where]:选列
- df.iloc[where_i, where_j]:选列和行
- df.at[label_i, label_j]:选某一标量
- df.iat[i, j]:选某一标量
- reindex:选列和行
- get_value, set_value:选某一标量
算术和数据对齐
前面我们介绍了 Series 的算术对齐,接下来是 DataFrame 的:
1 df1 = pd.DataFrame(np.arange(9.).reshape((3, 3)), columns=list('bcd'), 2 index=['Ohio', 'Texas', 'Colorado']) 3 df2 = pd.DataFrame(np.arange(12.).reshape((4, 3)), columns=list('bde'), 4 index=['Utah', 'Ohio', 'Texas', 'Oregon']) 5 df1 6 df2 7 df1 + df2
b c d Ohio 0.0 1.0 2.0 Texas 3.0 4.0 5.0 Colorado6.0 7.0 8.0 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 b c d e ColoradoNaN 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
没有相同行索引和列索引的情况:
1 df1 = pd.DataFrame({'A': [1, 2]}) 2 df2 = pd.DataFrame({'B': [3, 4]}) 3 df1 4 df2 5 df1 - df2
A 0 1 1 2 B 0 3 1 4 A B 0 NaN NaN 1 NaN NaN
填补值的算术方法
对于算术运算后产生空值的情况:
1 df1 = pd.DataFrame(np.arange(12.).reshape((3, 4)), 2 columns=list('abcd')) 3 df2 = pd.DataFrame(np.arange(20.).reshape((4, 5)), 4 columns=list('abcde')) 5 df2.loc[1, 'b'] = np.nan 6 df1 7 df2 8 df1 + df2
a b c d 0 0.0 1.0 2.0 3.0 1 4.0 5.0 6.0 7.0 2 8.0 9.0 10.0 11.0 a b c d e 0 0.0 1.0 2.0 3.0 4.0 1 5.0 NaN 7.0 8.0 9.0 2 10.0 11.0 12.0 13.0 14.0 3 15.0 16.0 17.0 18.0 19.0 a b c d e 0 0.0 2.0 4.0 6.0 NaN 1 9.0 NaN 13.0 15.0 NaN 2 18.0 20.0 22.0 24.0 NaN 3 NaN NaN NaN NaN NaN
可以使用 fill_value 属性来自动填空值:
1 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 5.0 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
常用的算术方法(可用于补空值):
- add, radd
- sub, rsub
- div, rdiv
- floordiv, rfloordiv
- mul, rmul
- pow, rpow
DataFrame 和 Series 之间的运算
首先,看看一维数组和二维数组的减法的情况。
1 arr = np.arange(12.).reshape((3, 4)) 2 arr 3 arr[0] 4 arr - arr[0]
array([[ 0., 1., 2., 3.], [ 4., 5., 6., 7.], [ 8., 9., 10., 11.]]) array([0., 1., 2., 3.]) array([[0., 0., 0., 0.], [4., 4., 4., 4.], [8., 8., 8., 8.]])
默认情况下,DataFrame 和 Series 会匹配索引进行计算。
1 frame = pd.DataFrame(np.arange(12.).reshape((4, 3)), 2 columns=list('bde'), 3 index=['Utah', 'Ohio', 'Texas', 'Oregon']) 4 series = frame.iloc[0] 5 frame 6 series 7 frame - series
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 b 0.0 d 1.0 e 2.0 Name: Utah, dtype: float64 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
如果索引不匹配,会用外连接的方式重组索引:
1 series2 = pd.Series(range(3), index=['b', 'e', 'f']) 2 frame + series2
b d e f Utah 0.0 NaN 3.0 NaN Ohio 3.0 NaN 6.0 NaN Texas 6.0 NaN 9.0 NaN Oregon 9.0 NaN 12.0 NaN
如果希望对于所有列进行算术计算(broadcast 机制),必须使用前面介绍的算术方法。
1 series3 = frame['d'] 2 frame 3 series3 4 frame.sub(series3, axis=0)
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 Utah 1.0 Ohio 4.0 Texas 7.0 Oregon 10.0 Name: d, dtype: float64 b d e Utah -1.0 0.0 1.0 Ohio -1.0 0.0 1.0 Texas -1.0 0.0 1.0 Oregon -1.0 0.0 1.0
apply 和 map
NumPy 的通用函数(按元素的数组方法)可以用于 Pandas 对象:
1 frame = pd.DataFrame(np.random.randn(4, 3), columns=list('bde'), 2 index=['Utah', 'Ohio', 'Texas', 'Oregon']) 3 frame 4 np.abs(frame)
b d e Utah -0.590458 -0.352861 0.820549 Ohio -1.708280 0.174739 -1.081811 Texas 0.857712 0.246972 0.532208 Oregon 0.812756 1.260538 0.818304 b d e Utah 0.590458 0.352861 0.820549 Ohio 1.708280 0.174739 1.081811 Texas 0.857712 0.246972 0.532208 Oregon 0.812756 1.260538 0.818304
一个常用的操作是按列或者行调用某一个函数,apply 可以达到该功能:
1 f = lambda x: x.max() - x.min() 2 frame.apply(f)
b 2.565993 d 1.613398 e 1.902360 dtype: float64
1 frame.apply(f, axis=1)
Utah 1.411007 Ohio 1.883019 Texas 0.610741 Oregon 0.447782 dtype: float64
apply 传入的函数不一定非得返回一个标量值,可以返回 Series:
1 def f(x): 2 return pd.Series([x.min(), x.max()], index=['min', 'max']) 3 frame.apply(f)
b d e min -1.708280 -0.352861 -1.081811 max 0.857712 1.260538 0.820549
还可以传入按元素计算的函数:
1 frame 2 format = lambda x: '%.2f' % x 3 frame.applymap(format)
b d e Utah -0.590458 -0.352861 0.820549 Ohio -1.708280 0.174739 -1.081811 Texas 0.857712 0.246972 0.532208 Oregon 0.812756 1.260538 0.818304 b d e Utah -0.59 -0.35 0.82 Ohio -1.71 0.17 -1.08 Texas 0.86 0.25 0.53 Oregon 0.81 1.26 0.82
按元素应用某函数必须使用 applymap。取这个名字的原因是 Series 有一个 map 方法就是用来按元素调用函数的。
1 frame['e'].map(format)
Utah 0.82 Ohio -1.08 Texas 0.53 Oregon 0.82 Name: e, dtype: object
排序和排名
对行索引进行排列:
1 obj = pd.Series(range(4), index=['d', 'a', 'b', 'c']) 2 obj 3 obj.sort_index()
d 0 a 1 b 2 c 3 dtype: int64 a 1 b 2 c 3 d 0 dtype: int64
对列索引进行排列:
1 frame = pd.DataFrame(np.arange(8).reshape((2, 4)), 2 index=['three', 'one'], 3 columns=['d', 'a', 'b', 'c']) 4 frame 5 frame.sort_index() 6 frame.sort_index(axis=1)
d a b c three 0 1 2 3 one 4 5 6 7 d a b c one 4 5 6 7 three 0 1 2 3 a b c d three 1 2 3 0 one 5 6 7 4
按降序排列:
1 frame.sort_index(axis=1, ascending=False)
d c b a three 0 3 2 1 one 4 7 6 5
对于 Series 和 DataFrame,如果需要根据值来进行排列,使用 sort_values 方法:
1 obj = pd.Series([4, 7, -3, 2]) 2 obj.sort_values() 3 obj = pd.Series([4, np.nan, 7, np.nan, -3, 2]) 4 obj.sort_values()
2 -3 3 2 0 4 1 7 dtype: int64 4 -3.0 5 2.0 0 4.0 2 7.0 1 NaN 3 NaN dtype: float64
1 frame = pd.DataFrame({'b': [4, 7, -3, 2], 'a': [0, 1, 0, 1]}) 2 frame 3 frame.sort_values(by='b') 4 frame.sort_values(by=['a', 'b'])
b a 0 4 0 1 7 1 2 -3 0 3 2 1 b a 2 -3 0 3 2 1 0 4 0 1 7 1 b a 2 -3 0 0 4 0 3 2 1 1 7 1
排名(ranking)用于计算出排名值。
1 obj = pd.Series([7, -5, 7, 4, 2, 4, 0, 4, 4]) 2 obj.rank() 3 obj.rank(method='first') # 不使用平均值的排名方法
0 8.5 1 1.0 2 8.5 3 5.5 4 3.0 5 5.5 6 2.0 7 5.5 8 5.5 dtype: float64 0 8.0 1 1.0 2 9.0 3 4.0 4 3.0 5 5.0 6 2.0 7 6.0 8 7.0 dtype: float64
降序排列,并且按最大值指定名次:
1 obj.rank(ascending=False, method='max')
0 2.0 1 9.0 2 2.0 3 6.0 4 7.0 5 6.0 6 8.0 7 6.0 8 6.0 dtype: float64
按列进行排名:
1 frame = pd.DataFrame({'b': [4.3, 7, -3, 2], 'a': [0, 1, 0, 1],'c': [-2, 5, 8, -2.5]}) 2 frame 3 frame.rank(axis=1)
b a c 0 4.3 0 -2.0 1 7.0 1 5.0 2 -3.0 0 8.0 3 2.0 1 -2.5 b a c 0 3.0 2.0 1.0 1 3.0 1.0 2.0 2 1.0 2.0 3.0 3 3.0 2.0 1.0
带有重复标签的索引
Series 的重复索引:
1 obj = pd.Series(range(5), index=['a', 'a', 'b', 'b', 'c']) 2 obj 3 obj.index.is_unique 4 obj['a'] 5 obj['c']
a 0 a 1 b 2 b 3 c 4 dtype: int64 False a 0 a 1 dtype: int64 4
Pandas 的重复索引:
1 df = pd.DataFrame(np.random.randn(4, 3), index=['a', 'a', 'b', 'b']) 2 df 3 df.loc['b']
0 1 2 a -0.975030 2.041130 1.022168 a 0.321428 2.124496 0.037530 b 0.343309 -0.386692 -0.577290 b 0.002090 -0.890841 1.759072 0 1 2 b 0.343309 -0.386692 -0.577290 b 0.002090 -0.890841 1.759072
求和与计算描述性统计量
sum 求和:
1 df = pd.DataFrame([[1.4, np.nan], [7.1, -4.5], 2 [np.nan, np.nan], [0.75, -1.3]], 3 index=['a', 'b', 'c', 'd'], 4 columns=['one', 'two']) 5 df 6 df.sum() 7 df.sum(axis=1) 8 df.mean(axis=1, skipna=False) # 不跳过 NAN
one two a 1.40 NaN b 7.10 -4.5 c NaN NaN d 0.75 -1.3 one 9.25 two -5.80 dtype: float64 a 1.40 b 2.60 c 0.00 d -0.55 dtype: float64 a NaN b 1.300 c NaN d -0.275 dtype: float64
返回最大值或最小值对应的索引:
1 df.idxmax() 2 df.idxmin(axis=1)
one b two d dtype: object a one b two c NaN d two dtype: object
累加:
1 df.cumsum()
one two a 1.40 NaN b 8.50 -4.5 c NaN NaN d 9.25 -5.8
describe 方法生成一些统计信息:
1 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
1 # 非数值数据 2 obj = pd.Series(['a', 'a', 'b', 'c'] * 4) 3 obj 4 obj.describe()
0 a 1 a 2 b 3 c 4 a 5 a 6 b 7 c 8 a 9 a 10 b 11 c 12 a 13 a 14 b 15 c dtype: object count 16 unique 3 top a freq 8 dtype: object
一些常用的统计方法:
- count
- describe
- min, max
- argmin, argmax
- idxmin, idxmax
- quantile
- sum
- mean
- median
- mad
- prod
- var
- std
- skew
- kurt
- cumsum
- cummin, cummax
- cumprod
- diff
- pct_change
自相关和协方差
首先安装一个读取数据集的模块:
pip install pandas-datareader -i https://pypi.douban.com/simple
下载一个股票行情的数据集:
1 import pandas_datareader.data as web 2 all_data = {ticker: web.get_data_yahoo(ticker) 3 for ticker in ['AAPL', 'IBM', 'MSFT', 'GOOG']} 4 price = pd.DataFrame({ticker: data['Adj Close'] 5 for ticker, data in all_data.items()}) 6 volume = pd.DataFrame({ticker: data['Volume'] 7 for ticker, data in all_data.items()})
计算价格的百分比变化:
1 returns = price.pct_change() 2 returns.tail()
AAPL IBM MSFT GOOG Date 2019-08-06 0.018930 -0.000213 0.018758 0.015300 2019-08-07 0.010355 -0.011511 0.004380 0.003453 2019-08-08 0.022056 0.018983 0.026685 0.026244 2019-08-09 -0.008240 -0.028337 -0.008496 -0.013936 2019-08-12 -0.002537 -0.014765 -0.013942 -0.011195
Series 的 corr 方法会计算两个 Series 的相关性,而 cov 方法计算协方差。
1 returns['MSFT'].corr(returns['IBM']) 2 returns.MSFT.cov(returns['IBM'])
0.48863990166304594 8.714318020797283e-05
DataFrame 的 corr 方法和 cov 会计算相关性和协方差的矩阵:
1 returns.corr() 2 returns.cov()
AAPL IBM MSFT GOOG AAPL 1.000000 0.381659 0.453727 0.459663 IBM 0.381659 1.000000 0.488640 0.402751 MSFT 0.453727 0.488640 1.000000 0.535898 GOOG 0.459663 0.402751 0.535898 1.000000 AAPL IBM MSFT GOOG AAPL 0.000266 0.000077 0.000107 0.000117 IBM 0.000077 0.000152 0.000087 0.000077 MSFT 0.000107 0.000087 0.000209 0.000120 GOOG 0.000117 0.000077 0.000120 0.000242
DataFrame 的 corrwith 方法可以计算和其他 Series 或 DataFrame 的相关性:
1 returns.corrwith(returns.IBM) 2 returns.corrwith(volume)
AAPL 0.381659 IBM 1.000000 MSFT 0.488640 GOOG 0.402751 dtype: float64 AAPL -0.061924 IBM -0.151708 MSFT -0.089946 GOOG -0.018591 dtype: float64
唯一值、值的计数、关系
unique 方法唯一值数组:
1 obj = pd.Series(['c', 'a', 'd', 'a', 'a', 'b', 'b', 'c', 'c']) 2 obj.unique()
array(['c', 'a', 'd', 'b'], dtype=object)
value_counts 方法返回值的计数:
1 obj.value_counts() 2 pd.value_counts(obj.values, sort=False) # 等价的写法
a 3 c 3 b 2 d 1 dtype: int64 c 3 b 2 d 1 a 3 dtype: int64
isin 检查 Series 中的数是否属于某个集合:
1 obj 2 mask = obj.isin(['b', 'c']) 3 mask 4 obj[mask]
0 c 1 a 2 d 3 a 4 a 5 b 6 b 7 c 8 c dtype: object 0 True 1 False 2 False 3 False 4 False 5 True 6 True 7 True 8 True dtype: bool 0 c 5 b 6 b 7 c 8 c dtype: object
get_indexer 将唯一值转换为索引(对于标签转换为数值很管用):
1 to_match = pd.Series(['c', 'a', 'b', 'b', 'c', 'a']) 2 unique_vals = pd.Series(['c', 'b', 'a']) 3 pd.Index(unique_vals).get_indexer(to_match)
array([0, 2, 1, 1, 0, 2])
计算多个列的直方图:
1 data = pd.DataFrame({'Qu1': [1, 3, 4, 3, 4], 2 'Qu2': [2, 3, 1, 2, 3], 3 'Qu3': [1, 5, 2, 4, 4]}) 4 data 5 data.apply(pd.value_counts).fillna(0)
Qu1 Qu2 Qu3 0 1 2 1 1 3 3 5 2 4 1 2 3 3 2 4 4 4 3 4 Qu1 Qu2 Qu3 1 1.0 1.0 1.0 2 0.0 2.0 1.0 3 2.0 2.0 0.0 4 2.0 0.0 2.0 5 0.0 0.0 1.0
参考
- 《Python for Data Analysis, 2nd Edition》by Wes McKinney
标签:Series,obj,dtype,NaN,pd,Ohio,基本操作,快速,Pandas 来源: https://www.cnblogs.com/noluye/p/11338271.html