数据合并，拼接和变型
分层索引
分层索引是Pandas的一个重要特性，能够让我们在一个轴上拥有多层索引。简单来说，它提供了一个在降纬模式下处理多维数据的能力。
如下是一个多层索引的例子
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| >>> data = pd.Series(np.random.randn(9),
... index=[['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd', 'd'], [1,2,3,1,3,1,2,2,3]])
>>> data
a 1 1.007189
2 -1.296221
3 0.274992
b 1 0.228913
3 1.352917
c 1 0.886429
2 -2.001637
d 2 -0.371843
3 1.669025
dtype: float64
>>> data.index
MultiIndex(levels=[['a', 'b', 'c', 'd'], [1, 2, 3]],
labels=[[0, 0, 0, 1, 1, 2, 2, 3, 3], [0, 1, 2, 0, 2, 0, 1, 1, 2]])
# 访问外层索引
>>> data['b']
1 0.228913
3 1.352917
dtype: float64
>>> data['b':'c']
b 1 0.228913
3 1.352917
c 1 0.886429
2 -2.001637
# 访问内层索引
>>> data.loc[:, 2]
a -1.296221
c -2.001637
d -0.371843
dtype: float64
|
多层索引在数据变型和透视表中非常常见，可以使用unstack方法，将Series转换成DataFrame
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| >>> data.unstack()
1 2 3
a 1.007189 -1.296221 0.274992
b 0.228913 NaN 1.352917
c 0.886429 -2.001637 NaN
d NaN -0.371843 1.669025
|
与unstack方法相反作用的就是stack
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| >>> data.unstack().stack()
a 1 1.007189
2 -1.296221
3 0.274992
b 1 0.228913
3 1.352917
c 1 0.886429
2 -2.001637
d 2 -0.371843
3 1.669025
dtype: float64
|
DataFrame的任何一个轴都可以有多个索引
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| >>> frame = pd.DataFrame(np.arange(12).reshape((4, 3)), index=[['a', 'a', 'b','b'], [1, 2, 1, 2]], columns=[['Ohio', 'Ohio', 'Colorado'], ['Green', 'Red', 'Green']])
>>> frame
Ohio Colorado
Green Red Green
a 1 0 1 2
2 3 4 5
b 1 6 7 8
2 9 10 11
# 可以给每层索引取一个名字
>>> frame.index.names = ["key1", "key2"]
>>> frame.columns.names = ["state", "color"]
>>> frame
state Ohio Colorado
color Green Red Green
key1 key2
a 1 0 1 2
2 3 4 5
b 1 6 7 8
2 9 10 11
# 通过外层索引选择列
>>> frame['Ohio']
color Green Red
key1 key2
a 1 0 1
2 3 4
b 1 6 7
2 9 10
# 通过内层索引选择列
>>> frame['Ohio']['Green']
key1 key2
a 1 0
2 3
b 1 6
2 9
|
分层排序和分类
使用swaplevel，我们可以重新对一个轴上的多层索引排序
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| >>> frame
state Ohio Colorado
color Green Red Green
key1 key2
a 1 0 1 2
2 3 4 5
b 1 6 7 8
2 9 10 11
>>> frame.swaplevel('key1', 'key2')
state Ohio Colorado
color Green Red Green
key2 key1
1 a 0 1 2
2 a 3 4 5
1 b 6 7 8
2 b 9 10 11
|
我们也可以在一个分层上对数据进行分类
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| >>> frame.sort_index(level=1)
state Ohio Colorado
color Green Red Green
key1 key2
a 1 0 1 2
b 1 6 7 8
a 2 3 4 5
b 2 9 10 11
>>> frame.swaplevel(0, 1).sort_index(level=0)
state Ohio Colorado
color Green Red Green
key2 key1
1 a 0 1 2
b 6 7 8
2 a 3 4 5
b 9 10 11
|
分层统计
许多关于Series和DataFrame的统计操作，都接受一个level参数来对层上进行统计
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| >>> frame
state Ohio Colorado
color Green Red Green
key1 key2
a 1 0 1 2
2 3 4 5
b 1 6 7 8
2 9 10 11
>>> frame.sum(level='key2')
state Ohio Colorado
color Green Red Green
key2
1 6 8 10
2 12 14 16
>>> frame.sum(level='color', axis=1)
color Green Red
key1 key2
a 1 2 1
2 8 4
b 1 14 7
2 20 10
|
利用DataFrame的列作为索引
使用set_index方法，可以将DataFrame的列作为索引
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| >>> frame = pd.DataFrame({'a': range(7), 'b': range(7, 0, -1),
... 'c': ['one', 'one', 'one', 'two', 'two',
... 'two', 'two'],
... 'd':[0,1,2,0,1,2,3]})
...
>>> frame
a b c d
0 0 7 one 0
1 1 6 one 1
2 2 5 one 2
3 3 4 two 0
4 4 3 two 1
5 5 2 two 2
6 6 1 two 3
>>> frame2 = frame.set_index(['c', 'd'])
>>> frame2
a b
c d
one 0 0 7
1 1 6
2 2 5
two 0 3 4
1 4 3
2 5 2
3 6 1
>>> frame2.index
MultiIndex(levels=[['one', 'two'], [0, 1, 2, 3]],
labels=[[0, 0, 0, 1, 1, 1, 1], [0, 1, 2, 0, 1, 2, 3]],
names=['c', 'd'])
# 默认情况下，作为索引的列会自动从DataFrame里移除，可以通过drop参数保留它们
>>> frame.set_index(['c', 'd'], drop=False)
a b c d
c d
one 0 0 7 one 0
1 1 6 one 1
2 2 5 one 2
two 0 3 4 two 0
1 4 3 two 1
2 5 2 two 2
3 6 1 two 3
|
reset_index函数的与set_index函数相反，多层索引会被放到列里面去
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| >>> frame2.reset_index()
c d a b
0 one 0 0 7
1 one 1 1 6
2 one 2 2 5
3 two 0 3 4
4 two 1 4 3
5 two 2 5 2
6 two 3 6 1
|
合并数据集
数据合并操作主要有以下几个方法:
pandas.merge: 基于key来合并row，类似关系型数据库的JOIN操作pandas.concat: 在轴上合并数据，类似stack操作combine_first: 合并重复的数据和填充确实的数据
类似数据库的DataFrame JOIN操作
如下一个"多对一"的例子，在df1里有多行值是a和b, 而在df2里面各自只有一行。
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| >>> df1 = pd.DataFrame({'key':['b', 'b', 'a', 'c', 'a', 'a', 'b'],
... 'data1': range(7)})
...
>>> df2 = pd.DataFrame({'key': ['a', 'b', 'd'],
... 'data2': range(3)})
...
>>> df1
data1 key
0 0 b
1 1 b
2 2 a
3 3 c
4 4 a
5 5 a
6 6 b
>>> df2
data2 key
0 0 a
1 1 b
2 2 d
# 默认是inner Join
>>> pd.merge(df1, df2)
data1 key data2
0 0 b 1
1 1 b 1
2 6 b 1
3 2 a 0
4 4 a 0
5 5 a 0
|
上面的例子里，我们并没有指明以哪一列作为join on的条件。merge函数默认使用列名相同的列作为key。最好的使用建议是通过on参数明确指明key信息
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| >>> pd.merge(df1, df2, on='key')
data1 key data2
0 0 b 1
1 1 b 1
2 6 b 1
3 2 a 0
4 4 a 0
5 5 a 0
|
如果两个DataFrame的列名都不相同，我们也可以单独指定它们
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| >>> df3 = pd.DataFrame({'lkey': ['b', 'b', 'a', 'c', 'a', 'a', 'b'],
... 'data1': range(7)})
...
>>> df4 = pd.DataFrame({'rkey': ['a', 'b', 'd'],
... 'data2': range(3)})
...
>>> pd.merge(df3, df4, left_on='lkey', right_on='rkey')
data1 lkey data2 rkey
0 0 b 1 b
1 1 b 1 b
2 6 b 1 b
3 2 a 0 a
4 4 a 0 a
5 5 a 0 a
|
merge的默认方式是inner join，我们还可以指定join方式为left, right, outer
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| >>> df1
data1 key
0 0 b
1 1 b
2 2 a
3 3 c
4 4 a
5 5 a
6 6 b
>>> df2
data2 key
0 0 a
1 1 b
2 2 d
>>> pd.merge(df1, df2, how='inner')
data1 key data2
0 0 b 1
1 1 b 1
2 6 b 1
3 2 a 0
4 4 a 0
5 5 a 0
>>> pd.merge(df1, df2, how='outer')
data1 key data2
0 0.0 b 1.0
1 1.0 b 1.0
2 6.0 b 1.0
3 2.0 a 0.0
4 4.0 a 0.0
5 5.0 a 0.0
6 3.0 c NaN
7 NaN d 2.0
>>> pd.merge(df1, df2, how='left')
data1 key data2
0 0 b 1.0
1 1 b 1.0
2 2 a 0.0
3 3 c NaN
4 4 a 0.0
5 5 a 0.0
6 6 b 1.0
>>> pd.merge(df1, df2, how='right')
data1 key data2
0 0.0 b 1
1 1.0 b 1
2 6.0 b 1
3 2.0 a 0
4 4.0 a 0
5 5.0 a 0
6 NaN d 2
|
不同的Join方式表现行为如下图:
“多对多”的方式, 默认情况下，多对多的结果是“笛卡尔乘积”。
如下: df1的key列有3个b, df2的key列有2个b。因此结果里会有3x2=6个b。 不同的join方式只是会影响结果里key的出现个数，不会影响“笛卡尔乘积”的行为
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| >>> df1 = pd.DataFrame({'key': ['b', 'b', 'a', 'c', 'a', 'b'],
... 'data1': range(6)})
...
>>> df2 = pd.DataFrame({'key': ['a', 'b', 'a', 'b', 'd'],
... 'data2': range(5)})
...
>>> df1
data1 key
0 0 b
1 1 b
2 2 a
3 3 c
4 4 a
5 5 b
>>> df2
data2 key
0 0 a
1 1 b
2 2 a
3 3 b
4 4 d
>>> pd.merge(df1, df2, on='key', how='left')
data1 key data2
0 0 b 1.0
1 0 b 3.0
2 1 b 1.0
3 1 b 3.0
4 2 a 0.0
5 2 a 2.0
6 3 c NaN
7 4 a 0.0
8 4 a 2.0
9 5 b 1.0
10 5 b 3.0
# 不同的join方式，结果里的key不一样，单都是笛卡尔乘积结果
>>> pd.merge(df1, df2, on='key', how='inner')
data1 key data2
0 0 b 1
1 0 b 3
2 1 b 1
3 1 b 3
4 5 b 1
5 5 b 3
6 2 a 0
7 2 a 2
8 4 a 0
9 4 a 2
|
也可以根据多个key来进行合并操作
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| >>> left = pd.DataFrame({'key1': ['foo', 'foo', 'bar'],
... 'key2': ['one', 'two', 'one'],
... 'lval': [1, 2, 3]})
...
>>> right = pd.DataFrame({'key1': ['foo', 'foo', 'bar', 'bar'],
... 'key2': ['one', 'one', 'one', 'two'],
... 'rval': [4, 5, 6, 7]})
...
>>> left
key1 key2 lval
0 foo one 1
1 foo two 2
2 bar one 3
>>> right
key1 key2 rval
0 foo one 4
1 foo one 5
2 bar one 6
3 bar two 7
>>> pd.merge(left, right, on=['key1', 'key2'], how='outer')
key1 key2 lval rval
0 foo one 1.0 4.0
1 foo one 1.0 5.0
2 foo two 2.0 NaN
3 bar one 3.0 6.0
4 bar two NaN 7.0
|
哪些key会出现在结果里是根据join方式来决定的，可以把多个key的组合当做是一个元组来作为单个key的join方式考虑(尽管实现方式并不是这样)
注意:
当列和列合并时，DataFrame的索引对象会被丢弃
最后一个关于合并的问题是处理结果中的重复列名。可以通过suffixes参数指定
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# 默认对重复的列名是自动添加了"_x", "_y"
>>> pd.merge(left, right, on='key1')
key1 key2_x lval key2_y rval
0 foo one 1 one 4
1 foo one 1 one 5
2 foo two 2 one 4
3 foo two 2 one 5
4 bar one 3 one 6
5 bar one 3 two 7
>>> pd.merge(left, right, on='key1', suffixes=('_left', '_right'))
key1 key2_left lval key2_right rval
0 foo one 1 one 4
1 foo one 1 one 5
2 foo two 2 one 4
3 foo two 2 one 5
4 bar one 3 one 6
5 bar one 3 two 7
|
基于索引做合并(Merging on Index)
通过设置left_index=True或right_index=True（或者两个同时设置）可以实现基于索引做合并
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| >>> left1 = pd.DataFrame({'key': ['a', 'b', 'a', 'a', 'b', 'c'],
... 'value': range(6)})
...
>>> right1 = pd.DataFrame({'group_val': [3.5, 7]}, index=['a', 'b'])
>>> left1
key value
0 a 0
1 b 1
2 a 2
3 a 3
4 b 4
5 c 5
>>> right1
group_val
a 3.5
b 7.0
>>> pd.merge(left1, right1, left_on='key', right_index=True)
key value group_val
0 a 0 3.5
2 a 2 3.5
3 a 3 3.5
1 b 1 7.0
4 b 4 7.0
>>> pd.merge(left1, right1, left_on='key', right_index=True, how='outer')
key value group_val
0 a 0 3.5
2 a 2 3.5
3 a 3 3.5
1 b 1 7.0
4 b 4 7.0
5 c 5 NaN
|
多层索引的合并会更复杂一些
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| >>> lefth = pd.DataFrame({'key1': ['Ohio', 'Ohio', 'Ohio', 'Nevada', 'Nevada'],
... 'key2': [2000, 2001, 2002, 2001, 2002],
... 'data': np.arange(5.)})
...
>>> righth = pd.DataFrame(np.arange(12).reshape((6, 2)), index=[['Nevada', 'Nevada', 'Ohio', 'Ohio', 'Ohio', 'Ohio'], [2001, 2000, 2000, 2000, 2001, 2002]],
... columns=['event1', 'event2'])
...
>>> lefth
data key1 key2
0 0.0 Ohio 2000
1 1.0 Ohio 2001
2 2.0 Ohio 2002
3 3.0 Nevada 2001
4 4.0 Nevada 2002
>>> righth
event1 event2
Nevada 2001 0 1
2000 2 3
Ohio 2000 4 5
2000 6 7
2001 8 9
2002 10 11
>>> pd.merge(lefth, righth, left_on=['key1', 'key2'], right_index=True)
data key1 key2 event1 event2
0 0.0 Ohio 2000 4 5
0 0.0 Ohio 2000 6 7
1 1.0 Ohio 2001 8 9
2 2.0 Ohio 2002 10 11
3 3.0 Nevada 2001 0 1
>>> pd.merge(lefth, righth, left_on=['key1', 'key2'], right_index=True, how='outer')
data key1 key2 event1 event2
0 0.0 Ohio 2000 4.0 5.0
0 0.0 Ohio 2000 6.0 7.0
1 1.0 Ohio 2001 8.0 9.0
2 2.0 Ohio 2002 10.0 11.0
3 3.0 Nevada 2001 0.0 1.0
4 4.0 Nevada 2002 NaN NaN
4 NaN Nevada 2000 2.0 3.0
|
同时基于两边的索引合并
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| >>> left2 = pd.DataFrame([[1., 2.], [3., 4.], [5., 6.]], index=['a', 'c', 'e'], columns=['Ohio', 'Nevada'])
>>> right2 = pd.DataFrame([[7., 8.], [9., 10.], [11., 12.], [13, 14]], index=['b', 'c', 'd', 'e'], columns=['Missouri', 'Alabama'])
>>> left2
Ohio Nevada
a 1.0 2.0
c 3.0 4.0
e 5.0 6.0
>>> right2
Missouri Alabama
b 7.0 8.0
c 9.0 10.0
d 11.0 12.0
e 13.0 14.0
>>> pd.merge(left2, right2, how='outer', left_index=True, right_index=True)
Ohio Nevada Missouri Alabama
a 1.0 2.0 NaN NaN
b NaN NaN 7.0 8.0
c 3.0 4.0 9.0 10.0
d NaN NaN 11.0 12.0
e 5.0 6.0 13.0 14.0
|
另外对于基于索引合并，DataFrame还提供了一种简单的方法
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| >>> left2.join(right2, how='outer')
Ohio Nevada Missouri Alabama
a 1.0 2.0 NaN NaN
b NaN NaN 7.0 8.0
c 3.0 4.0 9.0 10.0
d NaN NaN 11.0 12.0
e 5.0 6.0 13.0 14.0
|
一次join多个DataFrame
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| >>> another = pd.DataFrame([[7., 8.], [9., 10.], [11., 12.], [16., 17.]], index=['a', 'c', 'e', 'f'], columns=['New York', 'Oregon'])
>>> another
New York Oregon
a 7.0 8.0
c 9.0 10.0
e 11.0 12.0
f 16.0 17.0
>>> left2.join([right2, another])
Ohio Nevada Missouri Alabama New York Oregon
a 1.0 2.0 NaN NaN 7.0 8.0
c 3.0 4.0 9.0 10.0 9.0 10.0
e 5.0 6.0 13.0 14.0 11.0 12.0
|
基于轴连接
Numpy里有concatenate函数来连接数组
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| >>> arr= np.arange(12).reshape((3, 4))
>>> arr
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> np.concatenate([arr, arr])
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> np.concatenate([arr, arr], axis=1)
array([[ 0, 1, 2, 3, 0, 1, 2, 3],
[ 4, 5, 6, 7, 4, 5, 6, 7],
[ 8, 9, 10, 11, 8, 9, 10, 11]])
|
Pandas里面, 由于轴上多了索引信息，合并也更复杂一些。首先我们来看Series的合并
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| >>> s1 = pd.Series([0, 1], index=['a', 'b'])
>>> s2 = pd.Series([2, 3, 4], index=['c', 'd', 'e'])
>>> s3 = pd.Series([5, 6], index=['f', 'g'])
>>> pd.concat([s1, s2, s3])
a 0
b 1
c 2
d 3
e 4
f 5
g 6
dtype: int64
>>> pd.concat([s1, s2, s3], axis=1)
0 1 2
a 0.0 NaN NaN
b 1.0 NaN NaN
c NaN 2.0 NaN
d NaN 3.0 NaN
e NaN 4.0 NaN
f NaN NaN 5.0
g NaN NaN 6.0
|
默认的合并方式是"outer"join, 我们也可以通过join参数指定innerjoin
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| >>> s4 = pd.concat([s1, s3])
>>> s4
a 0
b 1
f 5
g 6
dtype: int64
>>> pd.concat([s1, s3], axis=1)
0 1
a 0.0 NaN
b 1.0 NaN
f NaN 5.0
g NaN 6.0
>>> pd.concat([s1, s4], axis=1, join='inner')
0 1
a 0 0
b 1 1
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通过参数join_axes, 我们可以指定join的列
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| >>> pd.concat([s1, s1, s3], axis=1, keys=['one', 'two', 'three'])
one two three
a 0.0 0.0 NaN
b 1.0 1.0 NaN
f NaN NaN 5.0
g NaN NaN 6.0
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通过参数keys, 可以创建多层索引
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| # keys参数的值和要合并的列表一一对应
>>> result = pd.concat([s1, s1, s3], keys=['one', 'two', 'three'])
>>> result
one a 0
b 1
two a 0
b 1
three f 5
g 6
dtype: int64
>>> result.unstack()
a b f g
one 0.0 1.0 NaN NaN
two 0.0 1.0 NaN NaN
three NaN NaN 5.0 6.0
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通过设置axis=1， 也可以创建列索引
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|
>>> pd.concat([s1, s1, s3], axis=1)
0 1 2
a 0.0 0.0 NaN
b 1.0 1.0 NaN
f NaN NaN 5.0
g NaN NaN 6.0
>>> pd.concat([s1, s1, s3], axis=1, keys=['one', 'two', 'three'])
one two three
a 0.0 0.0 NaN
b 1.0 1.0 NaN
f NaN NaN 5.0
g NaN NaN 6.0
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同样的逻辑，可以扩展到DataFrame
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| >>> df1 = pd.DataFrame(np.arange(6).reshape(3, 2), index=['a', 'b', 'c'], columns=['one', 'two'])
>>> df2 = pd.DataFrame(5 + np.arange(4).reshape(2, 2), index=['a', 'c'],columns=['three', 'four'])
>>> df1
one two
a 0 1
b 2 3
c 4 5
>>> df2
three four
a 5 6
c 7 8
>>> pd.concat([df1, df2], axis=1, keys=['level1', 'level2'])
level1 level2
one two three four
a 0 1 5.0 6.0
b 2 3 NaN NaN
c 4 5 7.0 8.0
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也可以通过传递字典的方式, 字典的keys将作为keys参数。
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| >>> pd.concat({'level1': df1, 'level2': df2}, axis=1)
level1 level2
one two three four
a 0 1 5.0 6.0
b 2 3 NaN NaN
c 4 5 7.0 8.0
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还可以通过names参数指定所以的名字
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| >>> pd.concat([df1, df2], axis=1, keys=['level1', 'level2'],names=['upper', 'lower'])
upper level1 level2
lower one two three four
a 0 1 5.0 6.0
b 2 3 NaN NaN
c 4 5 7.0 8.0
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最后一种情况是如果两个需要合并的DataFrame行索引没有任何相关的数据，可以通过 ignore_index=True来合并。
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| >>> df1 = pd.DataFrame(np.random.randn(3, 4), columns=['a', 'b', 'c', 'd'])
>>> df2 = pd.DataFrame(np.random.randn(2, 3), columns=['b', 'd', 'a'])
>>> df1
a b c d
0 0.440681 0.961317 1.087380 0.906632
1 1.614141 -0.511330 -0.384905 0.691080
2 -0.526509 0.084097 -0.068141 1.002294
>>> df2
b d a
0 0.264169 -1.022511 -1.322645
1 -0.698436 -0.148560 1.582804
>>> pd.concat([df1, df2], ignore_index=True)
a b c d
0 0.440681 0.961317 1.087380 0.906632
1 1.614141 -0.511330 -0.384905 0.691080
2 -0.526509 0.084097 -0.068141 1.002294
3 -1.322645 0.264169 NaN -1.022511
4 1.582804 -0.698436 NaN -0.148560
|
合并重复的数据
有一种数据结合方法既不属于merge，也不属于concatenation。比如两个数据集，index可能完全覆盖，或覆盖一部分。这里举个例子，考虑下numpy的where函数，可以在数组上进行类似于if-else表达式般的判断：
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| >>> a = pd.Series([np.nan, 2.5, np.nan, 3.5, 4.5, np.nan], index=['f', 'e', 'd', 'c', 'b', 'a'])
>>> b = pd.Series(np.arange(len(a), dtype=np.float64), index=['f', 'e', 'd', 'c', 'b', 'a'])
>>> b[-1] = np.nan
>>> a
f NaN
e 2.5
d NaN
c 3.5
b 4.5
a NaN
dtype: float64
>>> b
f 0.0
e 1.0
d 2.0
c 3.0
b 4.0
a NaN
dtype: float64
>>> np.where(pd.isnull(a), b, a)
array([ 0. , 2.5, 2. , 3.5, 4.5, nan])
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Series的combine_first函数可以达到类似的效果
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| >>> b[:-2]
f 0.0
e 1.0
d 2.0
c 3.0
dtype: float64
>>> a[2:]
d NaN
c 3.5
b 4.5
a NaN
dtype: float64
# b中有值时就用b中的，否则就用a的，index是两个Series的 outer join
>>> b[:-2].combine_first(a[2:])
a NaN
b 4.5
c 3.0
d 2.0
e 1.0
f 0.0
dtype: float64
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DataFrame的combine_first函数也有同样的功能，在DataFrame调用combine_first时，可以认为把传递给combine_first函数的DataFrame作为调用该函数的Dataframe缺失数据的补充。
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| >>> df1 = pd.DataFrame({'a': [1., np.nan, 5., np.nan],
... 'b': [np.nan, 2., np.nan, 6.],
... 'c': range(2, 18, 4)})
...
>>> df2 = pd.DataFrame({'a': [5., 4., np.nan, 3., 7.],
... 'b': [np.nan, 3., 4., 6., 8.]})
...
>>> df1
a b c
0 1.0 NaN 2
1 NaN 2.0 6
2 5.0 NaN 10
3 NaN 6.0 14
>>> df2
a b
0 5.0 NaN
1 4.0 3.0
2 NaN 4.0
3 3.0 6.0
4 7.0 8.0
# df1中缺失的数据，使用df2中的补上
>>> df1.combine_first(df2)
a b c
0 1.0 NaN 2.0
1 4.0 2.0 6.0
2 5.0 4.0 10.0
3 3.0 6.0 14.0
4 7.0 8.0 NaN
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变型和旋转(Reshaping and Pivoting)
多层索引变型
在DataFrame中，主要有两个方法用于多层索引变型
stack: 把列转换为行unstack: 把行转换为列
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| >>> data = pd.DataFrame(np.arange(6).reshape((2, 3)),
... index=pd.Index(['Ohio', 'Colorado'], name='state'),
... columns=pd.Index(['one', 'two', 'three'], name='number'))
...
>>> data
number one two three
state
Ohio 0 1 2
Colorado 3 4 5
# 把列转换为行
>>> result = data.stack()
>>> result
state number
Ohio one 0
two 1
three 2
Colorado one 3
two 4
three 5
dtype: int64
# 把行转换为列
>>> result.unstack()
number one two three
state
Ohio 0 1 2
Colorado 3 4 5
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默认情况下unstack是从最里层的列被转换为行(stack函数也类似)。也可以通过层的索引和名字来。
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| >>> result.index
MultiIndex(levels=[['Ohio', 'Colorado'], ['one', 'two', 'three']],
labels=[[0, 0, 0, 1, 1, 1], [0, 1, 2, 0, 1, 2]],
names=['state', 'number'])
# 0对应Index levels的索引
>>> result.unstack(0)
state Ohio Colorado
number
one 0 3
two 1 4
three 2 5
# state对应Index name
>>> result.unstack('state')
state Ohio Colorado
number
one 0 3
two 1 4
three 2 5
|
unstack的过程中，也可能会引入一些缺失值
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| >>> s1 = pd.Series([0, 1, 2, 3], index=['a', 'b', 'c', 'd'])
>>> s2 = pd.Series([4, 5, 6], index=['c', 'd', 'e'])
>>> data2 = pd.concat([s1, s2], keys=['one', 'two'])
>>> data2
one a 0
b 1
c 2
d 3
two c 4
d 5
e 6
dtype: int64
>>> data2.unstack()
a b c d e
one 0.0 1.0 2.0 3.0 NaN
two NaN NaN 4.0 5.0 6.0
|
stack函数默认会过滤掉NaN数据
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| >>> data2.unstack().stack()
one a 0.0
b 1.0
c 2.0
d 3.0
two c 4.0
d 5.0
e 6.0
dtype: float64
>>> data2.unstack().stack(dropna=False)
one a 0.0
b 1.0
c 2.0
d 3.0
e NaN
two a NaN
b NaN
c 4.0
d 5.0
e 6.0
dtype: float64
|
当unstack作用于DataFrame时，则是从最低层开始。
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| >>> df = pd.DataFrame({'left': result, 'right': result + 5},
... columns=pd.Index(['left', 'right'], name='side'))
...
>>> df
side left right
state number
Ohio one 0 5
two 1 6
three 2 7
Colorado one 3 8
two 4 9
three 5 10
>>> df.unstack()
side left right
number one two three one two three
state
Ohio 0 1 2 5 6 7
Colorado 3 4 5 8 9 10
# 也可以通过索引名来指定要操作的列
>>> df.unstack('state')
side left right
state Ohio Colorado Ohio Colorado
number
one 0 3 5 8
two 1 4 6 9
three 2 5 7 10
>>> df.unstack('state').stack('side')
state Colorado Ohio
number side
one left 3 0
right 8 5
two left 4 1
right 9 6
three left 5 2
right 10 7
|
一种常见的把多个时间序列保存在数据库或csv文件的方式叫作long or stacked format。 让我们来看一些例子(例子里使用的文件可以从这里下载)
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| >>> data = pd.read_csv('examples/macrodata.csv')
>>> data.head()
year quarter realgdp realcons realinv realgovt realdpi cpi \
0 1959.0 1.0 2710.349 1707.4 286.898 470.045 1886.9 28.98
1 1959.0 2.0 2778.801 1733.7 310.859 481.301 1919.7 29.15
2 1959.0 3.0 2775.488 1751.8 289.226 491.260 1916.4 29.35
3 1959.0 4.0 2785.204 1753.7 299.356 484.052 1931.3 29.37
4 1960.0 1.0 2847.699 1770.5 331.722 462.199 1955.5 29.54
m1 tbilrate unemp pop infl realint
0 139.7 2.82 5.8 177.146 0.00 0.00
1 141.7 3.08 5.1 177.830 2.34 0.74
2 140.5 3.82 5.3 178.657 2.74 1.09
3 140.0 4.33 5.6 179.386 0.27 4.06
4 139.6 3.50 5.2 180.007 2.31 1.19
>>> periods = pd.PeriodIndex(year=data.year, quarter=data.quarter, name='date')
>>> columns = pd.Index(['realgdp', 'infl', 'unemp'], name='item')
>>> data = data.reindex(columns=columns)
>>> data.head()
item realgdp infl unemp
0 2710.349 0.00 5.8
1 2778.801 2.34 5.1
2 2775.488 2.74 5.3
3 2785.204 0.27 5.6
4 2847.699 2.31 5.2
>>> data.index = periods.to_timestamp('D', 'end')
>>> data.head()
item realgdp infl unemp
date
1959-03-31 2710.349 0.00 5.8
1959-06-30 2778.801 2.34 5.1
1959-09-30 2775.488 2.74 5.3
1959-12-31 2785.204 0.27 5.6
1960-03-31 2847.699 2.31 5.2
>>> data.stack().head()
date item
1959-03-31 realgdp 2710.349
infl 0.000
unemp 5.800
1959-06-30 realgdp 2778.801
infl 2.340
dtype: float64
>>> data.stack().head().reset_index()
date item 0
0 1959-03-31 realgdp 2710.349
1 1959-03-31 infl 0.000
2 1959-03-31 unemp 5.800
3 1959-06-30 realgdp 2778.801
4 1959-06-30 infl 2.340
>>> data.stack().head().reset_index().rename(columns={0: 'value'})
date item value
0 1959-03-31 realgdp 2710.349
1 1959-03-31 infl 0.000
2 1959-03-31 unemp 5.800
3 1959-06-30 realgdp 2778.801
4 1959-06-30 infl 2.340
>>> ldata[:10]
date item value
0 1959-03-31 realgdp 2710.349
1 1959-03-31 infl 0.000
2 1959-03-31 unemp 5.800
3 1959-06-30 realgdp 2778.801
4 1959-06-30 infl 2.340
5 1959-06-30 unemp 5.100
6 1959-09-30 realgdp 2775.488
7 1959-09-30 infl 2.740
8 1959-09-30 unemp 5.300
9 1959-12-31 realgdp 2785.204
|
上面这种ldata的格式就被称作long format for multiple time series
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| >>> pivoted = ldata.pivot('date', 'item', 'value')
>>> pivoted.head()
item infl realgdp unemp
date
1959-03-31 0.00 2710.349 5.8
1959-06-30 2.34 2778.801 5.1
1959-09-30 2.74 2775.488 5.3
1959-12-31 0.27 2785.204 5.6
1960-03-31 2.31 2847.699 5.2
|
与pivot相反的操作就是pandas.melt
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| >>> df = pd.DataFrame({'key': ['foo', 'bar', 'baz'],
... 'A': [1, 2, 3],
... 'B': [4, 5, 6],
... 'C': [7, 8, 9]})
...
>>> df
A B C key
0 1 4 7 foo
1 2 5 8 bar
2 3 6 9 baz
>>> melted = pd.melt(df, ['key'])
>>> melted
key variable value
0 foo A 1
1 bar A 2
2 baz A 3
3 foo B 4
4 bar B 5
5 baz B 6
6 foo C 7
7 bar C 8
8 baz C 9
|