不區分大小寫過濾熊貓中的多列using.loc

Question

我想搜索忽略案例差異的值。 因此，例如，如果我輸入“fred”，我仍然能夠過濾所有包含 Fred 的值，即使 F 是大寫的。

這就是我目前擁有的：

def find(**kwargs):
    result = data.loc[data.rename(columns={"FirstName": "first",
                             "LastName": "last", 
                             "City": "city",
                             })[list(kwargs.keys())]
                    .eq(list(kwargs.values())).all(axis=1)]
    return result

但是，我意識到我不能在任何時候使用 .lower() 強制小寫我傳入的字符串和我過濾的值

這是我的數據示例：

FirstName    LastName   City
Fred           Bob       Austin
Billy          Bob       NYC

當我運行我的函數時，我期望這樣：

find('fred')
Output: Fred    Bob  Austin

Answer 1

import pandas as pd

data = pd.DataFrame({"FirstName": ['Fred', 'Billy'], 'LastName':['Bob','Bob'], 'City': ['A', 'D']} )


def find(**kwargs):
    result = data.loc[data.rename(columns={"FirstName": "first",
                         "LastName": "last",
                         "City": "city",
                         })[list(kwargs.keys())].apply(lambda x: x.str.lower()).eq(list(kwargs.values())).all(axis=1)]
    return result

print(find(first='fred'))

回報

  FirstName LastName City
  0      Fred      Bob    A

Answer 2

這里有兩種方法可以完成我相信你已經問過的事情，即：

• 分別基於參數first 、 last和city的任意組合對 df 列FirstName 、 LastName和City進行不區分大小寫的過濾。

方式#1

import pandas as pd
def find(**kwargs):
    df = ( data.rename(columns={"FirstName": "first",
                             "LastName": "last", 
                             "City": "city",
                             })[list(kwargs.keys())]
        .apply(lambda x: x.str.lower(), axis=1) )
    mask = df.eq(list(val.lower() for val in kwargs.values())).all(axis=1)
    return data[mask]

data = pd.DataFrame({'FirstName':['Fred','Billy'],'LastName':['Bob','Bob'],'City':['Austin','NYC']})

方式#2

import pandas as pd
from operator import and_
from functools import reduce
def find(**kwargs):
    df = data.rename(columns={"FirstName": "first",
                             "LastName": "last", 
                             "City": "city",
                             })[list(kwargs.keys())]

    valsLower = pd.Series([val.lower() for val in kwargs.values()], index=kwargs.keys())
    mask = reduce(and_, (df[col].str.lower() == valsLower[col] for col in df.columns))
    return data[mask]

data = pd.DataFrame({'FirstName':['Fred','Billy'],'LastName':['Bob','Bob'],'City':['Austin','NYC']})

測試代碼：

print( '',"data",data,sep='\n' )
print( '',"first='fred'",find(first='fred'),sep='\n' )
print( '',"first='fReD'",find(first='fred'),sep='\n' )
print( '',"last='bob'",find(last='bob'),sep='\n' )
print( '',"city='austin'",find(city='austin'),sep='\n' )
print( '',"first='fred', city='austin'",find(first='fred', city='austin'),sep='\n' )
print( '',"city='austin', first='fred'",find(first='fred', city='austin'),sep='\n' )
print( '',"last='bob', city='austin'",find(last='bob', city='austin'),sep='\n' )
print( '',"first='billy', city='austin'",find(first='billy', city='austin'),sep='\n' )

示例輸出：

data
  FirstName LastName    City
0      Fred      Bob  Austin
1     Billy      Bob     NYC

first='fred'
  FirstName LastName    City
0      Fred      Bob  Austin

first='fReD'
  FirstName LastName    City
0      Fred      Bob  Austin

last='bob'
  FirstName LastName    City
0      Fred      Bob  Austin
1     Billy      Bob     NYC

city='austin'
  FirstName LastName    City
0      Fred      Bob  Austin

first='fred', city='austin'
  FirstName LastName    City
0      Fred      Bob  Austin

city='austin', first='fred'
  FirstName LastName    City
0      Fred      Bob  Austin

last='bob', city='austin'
  FirstName LastName    City
0      Fred      Bob  Austin

first='billy', city='austin'
Empty DataFrame
Columns: [FirstName, LastName, City]
Index: []

Answer 3

使用match功能。

import re
from functools import reduce

def find(df, **kwargs):
    # Using AND condition. Modify & to | for OR condition.
    cond = reduce(lambda prev, x: prev & df[x[0]].str.match(f'{x[1]}', flags=re.IGNORECASE), 
                  kwargs.items(),
                  True)

    return df[cond]

find(df, FirstName='fre', LastName='bob')
#   FirstName LastName   City
# 0      Fred      Bob Austin