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python学习笔记(四):pandas基础

程序员文章站 2022-05-11 13:55:44
pandas 基础 serise 0 4 1 7 2 5 3 3 dtype: int64 array([ 4, 7, 5, 3], dtype=int64) RangeIndex(start=0, stop=4, step=1) 1 7 3 3 dtype: int64 1 7 2 5 dtype ......

pandas 基础

serise

import pandas as pd
from pandas import Series, DataFrame
obj = Series([4, -7, 5, 3])
obj
0    4
1   -7
2    5
3    3
dtype: 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   -7
3    3
dtype: int64
obj[1:3]
1   -7
2    5
dtype: int64
pd.isnull(obj)
0    False
1    False
2    False
3    False
dtype: bool
  • reindex可以用来插值
obj.reindex(range(5), method = 'ffill')
0    4
1   -7
2    5
3    3
4    3
dtype: 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    2000
1    2001
2    2002
3    2003
Name: 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.index
a[1] = 6
---------------------------------------------------------------------------

TypeError                                 Traceback (most recent call last)

<ipython-input-9-57677294f950> in <module>()
      1 a = data.index
----> 2 a[1] = 6


F:\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 np
df = 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] = None
df
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