主要内容：
- 增删行列
- 修改数值：apply lambda
- 描述统计
- 合并、补齐数据
- 极端值处理
起始数据框：
In [76]: df = pd.DataFrame(np.random.randn(6,4), index=dates, columns=list('ABCD'))
In [77]: df
Out[77]:
A B C D
2013-01-01 -0.411674 0.273549 0.629843 1.881497
2013-01-02 1.240512 0.970725 0.033099 1.553420
2013-01-03 -0.544326 0.545738 -1.325810 0.130738
2013-01-04 1.044803 -0.117151 0.874583 2.278227
2013-01-05 -2.194728 -2.536257 0.478644 0.057728
2013-01-06 -1.092031 1.249952 1.598761 -0.153423
In [98]: df2 = df.copy()
...: df2['E'] = ['one', 'one','two','three','four','three']
...: df2
...:
Out[98]:
A B C D E
2013-01-01 -0.411674 0.273549 0.629843 1.881497 one
2013-01-02 1.240512 0.970725 0.033099 1.553420 one
2013-01-03 -0.544326 0.545738 -1.325810 0.130738 two
2013-01-04 1.044803 -0.117151 0.874583 2.278227 three
2013-01-05 -2.194728 -2.536257 0.478644 0.057728 four
2013-01-06 -1.092031 1.249952 1.598761 -0.153423 three
增删行列
增加新列column
In [110]: df2["F"] = np.arange(3,9) # 如果赋的是一个值比如3，那这一列都是3。
In [111]: df2
Out[111]:
A B C D E F
2013-01-01 -0.411674 0.273549 0.629843 1.881497 ONE 3
2013-01-02 1.240512 0.970725 0.033099 1.553420 ONE 4
2013-01-03 -0.544326 0.545738 -1.325810 0.130738 TWO 5
2013-01-04 1.044803 -0.117151 0.874583 2.278227 THREE 6
2013-01-05 -2.194728 -2.536257 0.478644 0.057728 FOUR 7
2013-01-06 -1.092031 1.249952 1.598761 -0.153423 THREE 8
改变列名
In [112]: df2.columns = list("qwerty")
In [113]: df2.columns
Out[113]: Index(['q', 'w', 'e', 'r', 't', 'y'], dtype='object')
设置index
pd.DataFrame.set_index(keys,drop=False, inplace=False)
# keys : column label or list of column labels / arrays
# 默认drop=True，设置为index的列从数据库删除。
修改数值
用lambda和apply更改数据值
In [101]: df2.loc[:,"E"].unique()
Out[101]: array(['one', 'two', 'three', 'four'], dtype=object)
In [105]: df2.loc[:,"E"]=df2.E.apply(lambda x: x.upper())
In [106]: df2.loc[:,"E"].unique()
Out[106]: array(['ONE', 'TWO', 'THREE', 'FOUR'], dtype=object)
函数应用 function application
三种情况：
- 整张表：pipe()
- 行列应用：apply()
- 元素级应用： applymap()
这一部分在【function application部分】：http://pandas.pydata.org/pandas-docs/stable/basics.html#basics-discretization。
apply()的使用
In [34]: df.apply(np.mean)
Out[34]:
A -0.274648
B -0.260124
C 0.152004
D -0.659868
dtype: float64
In [35]: df.apply(lambda x: x.max() - x.min()) # 求了个全距
Out[35]:
A 3.203982
B 4.084655
C 1.984507
D 3.190727
dtype: float64
合并重叠数据
#1
a = Series([np.nan, 2.5, np.nan, 3.5, 4.5, np.nan],
index=['f', 'e', 'd', 'c', 'b', 'a'])
b = Series(np.arange(len(a), dtype=np.float64),
index=['f', 'e', 'd', 'c', 'b', 'a'])
b[-1] = np.nan
np.where(pd.isnull(a), b, a)
#2 补齐数据
b[:-2].combine_first(a[2:])
#3
df1 = DataFrame({'a': [1., np.nan, 5., np.nan],
'b': [np.nan, 2., np.nan, 6.],
'c': range(2, 18, 4)})
df2 = DataFrame({'a': [5., 4., np.nan, 3., 7.],
'b': [np.nan, 3., 4., 6., 8.]})
df1.combine_first(df2)
# 把df1中缺失的数据NaN，补充为df2中对应的数据。
缺失值处理
df1.dropna(how='any') # 删除空数据，按行。
df1.fillna(value=5) # 填充空数据，9999。
pd.isnull(df1) # 获得df1数据库的nan布尔面具。
DataFrame.fillna(value=None, # scalar, dict, Series, or DataFrame. 【不能是list】
# dict Series等没有的值就没法填充。
method=None, # {‘backfill’, ‘bfill’, ‘pad’, ‘ffill’, None}, default None
# backfill/bfill，用后面的第一个值，ffill用前面的最有一个值。
axis=None, # {0 or ‘index’, 1 or ‘columns’}
inplace=False, # 布尔型，True或False。默认False。
limit=None,
downcast=None,
**kwargs)
limit : int, default None
If method is specified, this is the maximum number of consecutive NaN values to forward/backward fill. In other words, if there is a gap with more than this number of consecutive NaNs, it will only be partially filled. If method is not specified, this is the maximum number of entries along the entire axis where NaNs will be filled.
downcast : dict, default is None
a dict of item->dtype of what to downcast if possible, or the string ‘infer’ which will try to downcast to an appropriate equal type (e.g. float64 to int64 if possible)
Returns:
filled : DataFrame
- 1.96个标准差以外的
- 分组后，组内1.96个标准差以外的
- 多重分组后。。。
#---创建了两个数据表，然后合在一起---
In [116]: States = ['NY', 'NY', 'NY', 'NY', 'FL', 'FL', 'GA', 'GA', 'FL', 'FL']
...: data = [1.0, 2, 3, 4, 5, 6, 7, 8, 9, 10]
...: idx = pd.date_range('1/1/2012', periods=10, freq='MS')
...: df1 = pd.DataFrame(data, index=idx, columns=['Revenue'])
...: df1['State'] = States
...:
In [118]: data2 = [10.0, 10.0, 9, 9, 8, 8, 7, 7, 6, 6]
...: idx2 = pd.date_range('1/1/2013', periods=10, freq='MS')
...: df2 = pd.DataFrame(data2, index=idx2, columns=['Revenue'])
...: df2['State'] = States
...:
In [119]: df = pd.concat([df1,df2])
In [120]: df
Out[120]:
Revenue State
2012-01-01 1.0 NY
2012-02-01 2.0 NY
2012-03-01 3.0 NY
2012-04-01 4.0 NY
2012-05-01 5.0 FL
2012-06-01 6.0 FL
2012-07-01 7.0 GA
2012-08-01 8.0 GA
2012-09-01 9.0 FL
2012-10-01 10.0 FL
2013-01-01 10.0 NY
2013-02-01 10.0 NY
2013-03-01 9.0 NY
2013-04-01 9.0 NY
2013-05-01 8.0 FL
2013-06-01 8.0 FL
2013-07-01 7.0 GA
2013-08-01 7.0 GA
2013-09-01 6.0 FL
2013-10-01 6.0 FL
方法1：超总体均值1.96个std
In [121]: newdf = df.copy()
...:
...: newdf['x-Mean'] = abs(newdf['Revenue'] - newdf['Revenue'].mean())
...: newdf['1.96*std'] = 1.96*newdf['Revenue'].std()
...: newdf['Outlier'] = abs(newdf['Revenue'] - newdf['Revenue'].mean()) > 1.96*newdf['Revenue'].std()
...: newdf
...:
Out[121]:
Revenue State x-Mean 1.96*std Outlier
2012-01-01 1.0 NY 5.75 5.200273 True
2012-02-01 2.0 NY 4.75 5.200273 False
2012-03-01 3.0 NY 3.75 5.200273 False
2012-04-01 4.0 NY 2.75 5.200273 False
2012-05-01 5.0 FL 1.75 5.200273 False
2012-06-01 6.0 FL 0.75 5.200273 False
2012-07-01 7.0 GA 0.25 5.200273 False
2012-08-01 8.0 GA 1.25 5.200273 False
2012-09-01 9.0 FL 2.25 5.200273 False
2012-10-01 10.0 FL 3.25 5.200273 False
2013-01-01 10.0 NY 3.25 5.200273 False
2013-02-01 10.0 NY 3.25 5.200273 False
2013-03-01 9.0 NY 2.25 5.200273 False
2013-04-01 9.0 NY 2.25 5.200273 False
2013-05-01 8.0 FL 1.25 5.200273 False
2013-06-01 8.0 FL 1.25 5.200273 False
2013-07-01 7.0 GA 0.25 5.200273 False
2013-08-01 7.0 GA 0.25 5.200273 False
2013-09-01 6.0 FL 0.75 5.200273 False
2013-10-01 6.0 FL 0.75 5.200273 False
方法2：分组后超小组1.96个std
In [122]: newdf = df.copy()
...:
...: State = newdf.groupby('State') # 分类汇总的方法。groupby.transform
...:
...: newdf['Outlier'] = State.transform( lambda x: abs(x-x.mean()) > 1.96*x.std() )
...: newdf['x-Mean'] = State.transform( lambda x: abs(x-x.mean()) )
...: newdf['1.96*std'] = State.transform( lambda x: 1.96*x.std() )
...: newdf
...:
Out[122]:
Revenue State Outlier x-Mean 1.96*std
2012-01-01 1.0 NY False 5.00 7.554813
2012-02-01 2.0 NY False 4.00 7.554813
2012-03-01 3.0 NY False 3.00 7.554813
2012-04-01 4.0 NY False 2.00 7.554813
2012-05-01 5.0 FL False 2.25 3.434996
2012-06-01 6.0 FL False 1.25 3.434996
2012-07-01 7.0 GA False 0.25 0.980000
2012-08-01 8.0 GA False 0.75 0.980000
2012-09-01 9.0 FL False 1.75 3.434996
2012-10-01 10.0 FL False 2.75 3.434996
2013-01-01 10.0 NY False 4.00 7.554813
2013-02-01 10.0 NY False 4.00 7.554813
2013-03-01 9.0 NY False 3.00 7.554813
2013-04-01 9.0 NY False 3.00 7.554813
2013-05-01 8.0 FL False 0.75 3.434996
2013-06-01 8.0 FL False 0.75 3.434996
2013-07-01 7.0 GA False 0.25 0.980000
2013-08-01 7.0 GA False 0.25 0.980000
2013-09-01 6.0 FL False 1.25 3.434996
2013-10-01 6.0 FL False 1.25 3.434996
#---函数式的写法---
newdf = df.copy()
State = newdf.groupby('State')
def s(group):
group['x-Mean'] = abs(group['Revenue'] - group['Revenue'].mean())
group['1.96*std'] = 1.96*group['Revenue'].std()
group['Outlier'] = abs(group['Revenue'] - group['Revenue'].mean()) > 1.96*group['Revenue'].std()
return group
Newdf2 = State.apply(s)
Newdf2
方法3：多重分组之后，1.96
In [123]: newdf = df.copy()
...:
...: StateMonth = newdf.groupby(['State', lambda x: x.month])
...:
...: newdf['Outlier'] = StateMonth.transform( lambda x: abs(x-x.mean()) > 1.96*x.std() )
...: newdf['x-Mean'] = StateMonth.transform( lambda x: abs(x-x.mean()) )
...: newdf['1.96*std'] = StateMonth.transform( lambda x: 1.96*x.std() )
...: newdf
...:
Out[123]:
Revenue State Outlier x-Mean 1.96*std
2012-01-01 1.0 NY False 4.5 12.473364
2012-02-01 2.0 NY False 4.0 11.087434
2012-03-01 3.0 NY False 3.0 8.315576
2012-04-01 4.0 NY False 2.5 6.929646
2012-05-01 5.0 FL False 1.5 4.157788
2012-06-01 6.0 FL False 1.0 2.771859
2012-07-01 7.0 GA False 0.0 0.000000
2012-08-01 8.0 GA False 0.5 1.385929
2012-09-01 9.0 FL False 1.5 4.157788
2012-10-01 10.0 FL False 2.0 5.543717
2013-01-01 10.0 NY False 4.5 12.473364
2013-02-01 10.0 NY False 4.0 11.087434
2013-03-01 9.0 NY False 3.0 8.315576
2013-04-01 9.0 NY False 2.5 6.929646
2013-05-01 8.0 FL False 1.5 4.157788
2013-06-01 8.0 FL False 1.0 2.771859
2013-07-01 7.0 GA False 0.0 0.000000
2013-08-01 7.0 GA False 0.5 1.385929
2013-09-01 6.0 FL False 1.5 4.157788
2013-10-01 6.0 FL False 2.0 5.543717
In [124]: StateMonth
Out[124]: <pandas.core.groupby.DataFrameGroupBy object at 0x0000017F52F4DAC8>
方法4：四分位数。对于非高斯分布/正态分布的数据。
# make a copy of original df
In [126]: newdf = df.copy()
...:
...: State = newdf.groupby('State')
...:
...: newdf['Lower'] = State['Revenue'].transform( lambda x: x.quantile(q=.25) - (1.5*(x.quantile(q=.75)-x.quantile(q=.25))) )
...: newdf['Upper'] = State['Revenue'].transform( lambda x: x.quantile(q=.75) + (1.5*(x.quantile(q=.75)-x.quantile(q=.25))) )
...: newdf['Outlier'] = (newdf['Revenue'] < newdf['Lower']) | (newdf['Revenue'] > newdf['Upper'])
...: newdf
...:
Out[126]:
Revenue State Lower Upper Outlier
2012-01-01 1.0 NY -7.000 19.000 False
2012-02-01 2.0 NY -7.000 19.000 False
2012-03-01 3.0 NY -7.000 19.000 False
2012-04-01 4.0 NY -7.000 19.000 False
2012-05-01 5.0 FL 2.625 11.625 False
2012-06-01 6.0 FL 2.625 11.625 False
2012-07-01 7.0 GA 6.625 7.625 False
2012-08-01 8.0 GA 6.625 7.625 True
2012-09-01 9.0 FL 2.625 11.625 False
2012-10-01 10.0 FL 2.625 11.625 False
2013-01-01 10.0 NY -7.000 19.000 False
2013-02-01 10.0 NY -7.000 19.000 False
2013-03-01 9.0 NY -7.000 19.000 False
2013-04-01 9.0 NY -7.000 19.000 False
2013-05-01 8.0 FL 2.625 11.625 False
2013-06-01 8.0 FL 2.625 11.625 False
2013-07-01 7.0 GA 6.625 7.625 False
2013-08-01 7.0 GA 6.625 7.625 False
2013-09-01 6.0 FL 2.625 11.625 False
2013-10-01 6.0 FL 2.625 11.625 False
网友评论