同时融化Python Pandas中的多个列

问题描述 投票：0回答：3

想知道 pd.melt 是否支持熔化多个柱。我有下面的示例尝试将 value_vars 作为列表列表，但我收到错误：

ValueError: Location based indexing can only have [labels (MUST BE IN THE INDEX), slices of labels (BOTH endpoints included! Can be slices of integers if the index is integers), listlike of labels, boolean] types

使用熊猫0.23.1.

df = pd.DataFrame({'City': ['Houston', 'Austin', 'Hoover'],
                   'State': ['Texas', 'Texas', 'Alabama'],
                   'Name':['Aria', 'Penelope', 'Niko'],
                   'Mango':[4, 10, 90],
                   'Orange': [10, 8, 14], 
                   'Watermelon':[40, 99, 43],
                   'Gin':[16, 200, 34],
                   'Vodka':[20, 33, 18]},
                 columns=['City', 'State', 'Name', 'Mango', 'Orange', 'Watermelon', 'Gin', 'Vodka'])

所需输出：

      City    State       Fruit  Pounds  Drink  Ounces
0  Houston    Texas       Mango       4    Gin    16.0
1   Austin    Texas       Mango      10    Gin   200.0
2   Hoover  Alabama       Mango      90    Gin    34.0
3  Houston    Texas      Orange      10  Vodka    20.0
4   Austin    Texas      Orange       8  Vodka    33.0
5   Hoover  Alabama      Orange      14  Vodka    18.0
6  Houston    Texas  Watermelon      40    nan     NaN
7   Austin    Texas  Watermelon      99    nan     NaN
8   Hoover  Alabama  Watermelon      43    nan     NaN

我尝试过，但收到上述错误：

df.melt(id_vars=['City', 'State'], 
        value_vars=[['Mango', 'Orange', 'Watermelon'], ['Gin', 'Vodka']],var_name=['Fruit', 'Drink'], 
        value_name=['Pounds', 'Ounces'])
pandas melt valueerror
3个回答
31
投票

对每个类别使用双

melt
，然后使用
concat
，但因为重复的值会在cumcount
中为唯一的
triples添加
MultiIndex
：



df1 = df.melt(id_vars=['City', 'State'], 
              value_vars=['Mango', 'Orange', 'Watermelon'],
              var_name='Fruit', value_name='Pounds')
df2 = df.melt(id_vars=['City', 'State'], 
              value_vars=['Gin', 'Vodka'], 
              var_name='Drink', value_name='Ounces')

df1 = df1.set_index(['City', 'State', df1.groupby(['City', 'State']).cumcount()])
df2 = df2.set_index(['City', 'State', df2.groupby(['City', 'State']).cumcount()])


df3 = (pd.concat([df1, df2],axis=1)
         .sort_index(level=2)
         .reset_index(level=2, drop=True)
         .reset_index())
print (df3)
      City    State       Fruit  Pounds  Drink  Ounces
0   Austin    Texas       Mango      10    Gin   200.0
1   Hoover  Alabama       Mango      90    Gin    34.0
2  Houston    Texas       Mango       4    Gin    16.0
3   Austin    Texas      Orange       8  Vodka    33.0
4   Hoover  Alabama      Orange      14  Vodka    18.0
5  Houston    Texas      Orange      10  Vodka    20.0
6   Austin    Texas  Watermelon      99    NaN     NaN
7   Hoover  Alabama  Watermelon      43    NaN     NaN
8  Houston    Texas  Watermelon      40    NaN     NaN


    

      

      
      
      

      

        

          

            

              
1
投票

              
            

          

        

        


我今天再次遇到同样的问题，发现了这个问题，看到我已经投票了，并认为将其变成一件事可能会很好，因为这对我来说是一个反复出现的问题。


因此，我编写了一个 
multi_melt
 函数，该函数使用 jezarel 提出的方法，但适用于可迭代输入（Martin Petrov 使用的语法）。请注意，此版本“广播”标量输入：


from itertools import cycle
import pandas as pd


def is_scalar(obj):
    if isinstance(obj, str):
        return True
    elif hasattr(obj, "__iter__"):
        return False
    else:
        return True


def multi_melt(
    df: pd.DataFrame,
    id_vars=None,
    value_vars=None,
    var_name=None,
    value_name="value",
    col_level=None,
    ignore_index=True,
) -> pd.DataFrame:

    # Note: we don't broadcast value_vars ... that would seem unintuitive
    value_vars = value_vars if not is_scalar(value_vars[0]) else [value_vars]
    var_name = var_name if not is_scalar(var_name) else cycle([var_name])
    value_name = value_name if not is_scalar(value_name) else cycle([value_name])

    melted_dfs = [
        (
            df.melt(
                id_vars,
                *melt_args,
                col_level,
                ignore_index,
            ).pipe(lambda df: df.set_index([*id_vars, df.groupby(id_vars).cumcount()]))
        )
        for melt_args in zip(value_vars, var_name, value_name)
    ]

    return (
        pd.concat(melted_dfs, axis=1)
        .sort_index(level=2)
        .reset_index(level=2, drop=True)
        .reset_index()
    )



由于它不是 pandas API 的一部分，因此您必须 
pipe
 它，但否则它应该像接受可迭代的普通 
melt
 一样工作：


示例：


df = pd.DataFrame(
    {
        "City": ["Houston", "Austin", "Hoover"],
        "State": ["Texas", "Texas", "Alabama"],
        "Name": ["Aria", "Penelope", "Niko"],
        "Mango": [4, 10, 90],
        "Orange": [10, 8, 14],
        "Watermelon": [40, 99, 43],
        "Gin": [16, 200, 34],
        "Vodka": [20, 33, 18],
    },
    columns=["City", "State", "Name", "Mango", "Orange", "Watermelon", "Gin", "Vodka"],
)


df.pipe(
    multi_melt,
    id_vars=["City", "State"],
    value_vars=[["Mango", "Orange", "Watermelon"], ["Gin", "Vodka"]],
    var_name=["Fruit", "Drink"],
    value_name=["Pounds", "Ounces"],
)




结果：


      City    State       Fruit  Pounds  Drink  Ounces
0   Austin    Texas       Mango      10    Gin   200.0
1   Hoover  Alabama       Mango      90    Gin    34.0
2  Houston    Texas       Mango       4    Gin    16.0
3   Austin    Texas      Orange       8  Vodka    33.0
4   Hoover  Alabama      Orange      14  Vodka    18.0
5  Houston    Texas      Orange      10  Vodka    20.0
6   Austin    Texas  Watermelon      99    NaN     NaN
7   Hoover  Alabama  Watermelon      43    NaN     NaN
8  Houston    Texas  Watermelon      40    NaN     NaN



单一熔体：


df.pipe(
    multi_melt,
    id_vars=["City", "State"],
    value_vars=["Mango", "Orange", "Watermelon"],
    var_name="Fruit",
    value_name="Pounds",
)



      City    State       Fruit  Pounds
0   Austin    Texas       Mango      10
1   Hoover  Alabama       Mango      90
2  Houston    Texas       Mango       4
3   Austin    Texas      Orange       8
4   Hoover  Alabama      Orange      14
5  Houston    Texas      Orange      10
6   Austin    Texas  Watermelon      99
7   Hoover  Alabama  Watermelon      43
8  Houston    Texas  Watermelon      40


    

      

      
      
      

      

        

          

            

              
0
投票

              
            

          

        

        


一个高效且高性能的选项是使用 pyjanitor 中的 pivot_longer，使用正则表达式列表，并依赖于列中的现有顺序（
Mango, Gin
、
Orange, Vodka
、
Watermelon
）：


# pip install pyjanitor
import pandas as pd
import janitor

df.pivot_longer(
    index=["City", "State"],
    column_names=slice("Mango", "Vodka"),
    names_to=("Fruit", "Drink"),
    values_to=("Pounds", "Ounces"),
   names_pattern=[r"M|O|W", r"G|V"],
   )

      City    State       Fruit  Pounds  Drink  Ounces
0  Houston    Texas       Mango       4    Gin    16.0
1   Austin    Texas       Mango      10    Gin   200.0
2   Hoover  Alabama       Mango      90    Gin    34.0
3  Houston    Texas      Orange      10  Vodka    20.0
4   Austin    Texas      Orange       8  Vodka    33.0
5   Hoover  Alabama      Orange      14  Vodka    18.0
6  Houston    Texas  Watermelon      40    NaN     NaN
7   Austin    Texas  Watermelon      99    NaN     NaN
8   Hoover  Alabama  Watermelon      43    NaN     NaN


    

      

      
    

  

  

    
最新问题


  





  

    © www.soinside.com 2019 - 2024. All rights reserved.