python学习笔记（四）：pandas基础，,pandas 基础s

文章由Byrx.net分享于2020-12-16 08:12:17评论（653）

python学习笔记（四）：pandas基础，,pandas 基础s


pandas 基础

serise

import pandas as pdfrom pandas import Series, DataFrameobj = Series([4, -7, 5, 3])obj
0    41   -72    53    3dtype: int64
obj.values
array([ 4, -7,  5,  3], dtype=int64)
obj.index
RangeIndex(start=0, stop=4, step=1)
obj[[1,3]]# 跳着选取数据
1   -73    3dtype: int64
obj[1:3]
1   -72    5dtype: int64
pd.isnull(obj)
0    False1    False2    False3    Falsedtype: bool
reindex可以用来插值
obj.reindex(range(5), method = 'ffill')
0    41   -72    53    34    3dtype: int64
标签切片是闭区间的

dataframe

data = {'state': ['asd','qwe','sdf','ert'],       'year': [2000, 2001, 2002, 2003],       'pop': [1.5,1.7,3.6,2.4]}data = DataFrame(data)data
pop state year
0 1.5 asd 2000
1 1.7 qwe 2001
2 3.6 sdf 2002
3 2.4 ert 2003 
data.year# 比r里提取列要方便点
0    20001    20012    20023    2003Name: year, dtype: int64
data['debt'] = range(4)data
pop state year debt
0 1.5 asd 2000 0
1 1.7 qwe 2001 1
2 3.6 sdf 2002 2
3 2.4 ert 2003 3
index是不能修改的
a = data.indexa[1] = 6
---------------------------------------------------------------------------TypeError                                 Traceback (most recent call last)<ipython-input-9-57677294f950> in <module>()      1 a = data.index----> 2 a[1] = 6F:\Anaconda\lib\site-packages\pandas\core\indexes\base.py in __setitem__(self, key, value)   1668    1669     def __setitem__(self, key, value):-> 1670         raise TypeError("Index does not support mutable operations")   1671    1672     def __getitem__(self, key):TypeError: Index does not support mutable operations
data.columns
Index(['pop', 'state', 'year', 'debt'], dtype='object')
.ix标签索引功能，输入行和列不加.ix只能选取其中的某列或某行，不能列与行同时选取
data[:3]
pop state year debt
0 1.5 asd 2000 0
1 1.7 qwe 2001 1
2 3.6 sdf 2002 2 
data.ix[:,:3]
pop state year
0 1.5 asd 2000
1 1.7 qwe 2001
2 3.6 sdf 2002
3 2.4 ert 2003
删除某列用drop，axis = 0表示行，1表示列删除后原数据不变
data.drop(0,axis=0)
pop state year debt
1 1.7 qwe 2001 1
2 3.6 sdf 2002 2
3 2.4 ert 2003 3 
data.drop('year', axis=1)
pop state debt
0 1.5 asd 0
1 1.7 qwe 1
2 3.6 sdf 2
3 2.4 ert 3 
data
pop state year debt
0 1.5 asd 2000 0
1 1.7 qwe 2001 1
2 3.6 sdf 2002 2
3 2.4 ert 2003 3 
import numpy as npdf = DataFrame(np.arange(9).reshape(3, 3))df
0 1 2
0 0 1 2
1 3 4 5
2 6 7 8
applymap()可以对dataframe每一个元素运用函数apply()可以对每一维数组运用函数
df.applymap(lambda x: '%.2f' % x)
0 1 2
0 0.00 1.00 2.00
1 3.00 4.00 5.00
2 6.00 7.00 8.00 
data.sort_values(by='pop')# 对某一列排序
pop state year debt
0 1.5 asd 2000 0
1 1.7 qwe 2001 1
3 2.4 ert 2003 3
2 3.6 sdf 2002 2 
data.describe()
pop year debt
count 4.000000 4.000000 4.000000
mean 2.300000 2001.500000 1.500000
std 0.948683 1.290994 1.290994
min 1.500000 2000.000000 0.000000
25% 1.650000 2000.750000 0.750000
50% 2.050000 2001.500000 1.500000
75% 2.700000 2002.250000 2.250000
max 3.600000 2003.000000 3.000000 
df.isin([1])
0 1 2
0 False True False
1 False False False
2 False False False
None、NaN会被当作NA处理df.shape不加括号相当于dim()
df.shape
(3, 3)
dropna删除缺失值
df.ix[:1, :1] = Nonedf
0 1 2
0 NaN NaN 2
1 NaN NaN 5
2 6.0 7.0 8
填充缺失值可以调用字典，不同行添加不同值
df.fillna({0:11, 1:22})
0 1 2
0 11.0 22.0 2
1 11.0 22.0 5
2 6.0 7.0 8 
df
0 1 2
0 NaN NaN 2
1 NaN NaN 5
2 6.0 7.0 8 
df.fillna({0:11, 1:22}, inplace=True)
0 1 2
0 11.0 22.0 2
1 11.0 22.0 5
2 6.0 7.0 8 
df
0 1 2
0 11.0 22.0 2
1 11.0 22.0 5
2 6.0 7.0 8
inplace修改对象不产生副本

python学习笔记（四）：pandas基础

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