pandas 列操作变量转换

蔚落 2023-06-08 15:56 44阅读 0赞

import pandas as pd  
import numpy  
import math

# variable conversion #

## calculate new variable ##

### new variable is constant ###

‘’’ df\[‘var\_name’\] = name’’’

df1 = pd.read_csv('123.csv',encoding='utf-8')
    df1.new_var = 1 # do nothing .only use to var that already exist
    df1['cons'] = 1 # right
    print(df1.head(10))

### operation of vars ###

‘’’  
df\[var\_name\] = df\[‘old\_var’\] \* 100  
df\[var\_name\] = df.old\_var \* 100  
‘’’

df1['新名次'] = df1['名次'] * 10
    print(df1.head(10))
    
    df1['方根'] = round(numpy.sqrt(df1['新名次']),2)
    print(df1.head(10))
    #下面错误用法
    df1['方根2'] =pd.Series(math.sqrt(i) for i in df1['新名次'])#为什么不行？

### batch operation ###

‘’’  
df.apply(  
func:function or lambda for row or columns  
axis = 0 :row or columns ;1 for row

)

df\[‘new\_name’\] = df.oldvar.apply(func)

‘’’

df1['方根2'] = df1['新名次'].apply(math.sqrt)
    df1['名次3'] = df1['新名次'].apply(lambda x : x*x)
    print(df1.head(10))

‘’‘df.applymap(func)’’’ \#use for all cell  
‘’‘creat new df ,use df.assign(new\_var = expression)’’’

df2 = df1.assign(new_var = df1['名次3'].apply(math.log10))
    print(df2.head(10))

## insert var ##

‘’’  
df.insert(  
loc: 0 <= loc <=len(columns)  
columns:new\_name of new column  
value:series or like list structure of values to pour into  
allow\_duplicates = False: if or not allow new\_columns have duplication of name  
)

‘’’

df2.insert(1,'neworder',[i for i in range(11,67)])
    print(df2.head(10))

## edit/replace variable value ##

‘’‘It’s essentially a matter of locating cells ‘’’

print(df2.地址.isin(['长安'])) # series of True and False
    print(df2[df2.地址.isin(['长安'])])
    print(df2.地址[df2.地址.isin(['长安'])])#定位到了
    df2.地址[df2.地址.isin(['长安'])] = '小杨家' #警告 常数赋给切片容易导致错误
    print(df2)
    df2.地址[50] = 'test'   #警告 常数赋给切片容易导致错误 []被认为是切片
    print(df2[45:])
    
    df2['地址'][50] = 'restore'
    print(df2[45:])
    
    df2.loc[50,'地址'] = '威哥' #精准引用 50 为索引的数值 iloc 是绝对位置
    print(df2[45:])

### 对应数值的替换 ###

‘’’  
df.replace(  
to\_replace = None:被替换值，可以是str/regex/list/dict/Series/numeric/None  
value = None:new value  
inplace = False  
)  
‘’’

print(df2['地址'].replace('朝阳','晓阳'))
    print(df2.replace('朝阳','晓阳'))
    print(df2.地址.replace(['大兴','昌平'],['丰台','大文']))
    print(df2.地址.replace({ '大兴':'嗯嗯嗯','丰台':'大文'}))

### 筛选数据替换 ###

‘’‘1. regex ‘’’  
‘’’ df.replace(regex,newvalue) ‘’’ \#一定要进行一下效率对比

‘’‘2. 行筛选方式’’’  
‘’‘行筛选得到索引->loc定位替换 | 也可以直接筛选出单元格进行数值替换’’’  
df3 = df2.地址.iloc\[0:5\] = ‘雄安’ \#替换没问题，df3只是str，传递赋值？？？？

print(df2)
    
    df2.loc[3:5,'地址'] = '沙河'
    print(df2)
    
    df2.loc[df2[df2.评分 ==5].index,'地址'] = '杨浦' #定位索引速度快吧
    print(df2)
    
    df2.loc[df2.评分 < 5,'地址'] = '松江' #也是可以的 标准的引用替换方式
    print(df2 )

#### 变形 ####

df2.地址[df2.评分 < 5] = '我乖'
    print(df2)

#### 注意区分拷贝与直接赋值（尽量使用loc方式，偷懒可用变形，但是最标准的才是，隐患最小的，不用投机方法） ####

df2[df2.评分< 4 ].地址 = '阳阳' #赋值到了拷贝里，直接再赋值，又成了传递赋值
    
    print(df2)

=先搞列，最后操作再搞行，否则容易因为长度的各种变化把自己搞死

### query ###

df2.loc[df2.query("评分 < 5 and 名次 < 20").index,'地址'] = '恩惠'
    print(df2)
    df2.query("评分 < 5 and 名次 < 20") #是整个df 的查询视图
    print(df2)
    
    df2.drop(columns=['cons', 'new_var'],inplace = True)
    
    print(df2)
    
    df2.query("评分 < 5 and 名次 < 20")['地址'] = '视图'
    print(df2) # 原df没有变化

## 哑变量变换（用于分类变量） ##

‘’’  
pd.get\_dummier(  
data:df/col to couvert  
prefix = None: 哑变量名前缀  
prefix\_sep = ‘\_’ 前缀和序号之间的连接符  
dummy\_na = False: 是否为NaNs 专门设定一个哑变量  
columns = None:columns’ name to be couvert  
drop\_first = False :是否返回k-1个哑变量，而不是K个哑变量  
) \# return df  
‘’’

print(df2.head())
    print(pd.get_dummies(df2.地址,prefix = 'pre'))
    print(pd.get_dummies(df2,columns =[ '地址'])) #省去了合并的那一步

## 数值变量的分段 ##

‘’’  
pd.cut(  
x：col name to be cut  
bins:分段设定  
int:被等距等分的段数  
sequence of scalars:每个分段起点，必须包括最值，可不等距  
right = True:每段是否包含右边界  
labels = None ：为每个分段提供自定义标签  
include\_lowest = False:第一段是否包括最左侧边界值，与right配合使用  
)\# 数值类型为Categories的序列

pd.qcut()\#按照频次，而不是按照取值范围进行等分  
‘’’

df2['stars'] = pd.cut(df2.评分,bins = [2,4,5],right = True)
    df2.drop(columns=['方根2', '方根'],inplace = True)
    print(df2)
    
    df2['mingci'] = pd.cut(df2.名次,bins = [1,20,60],include_lowest = True,right = True)
    print(df2)