Question

我有一个带有多个标头的pandas数据框。我想知道如何将其转换为嵌套目录列表。熊猫数据框中的每一行将是列表中的嵌套字典。

这是一个例子

#Creaet an example multiheader dataframe

col =['id','x, single room','x, double room','y, single room','y, double room' ]
df = pd.DataFrame([[1,2,3,4,5], [3,4,7,5,3]], columns=col)
a = df.columns.str.split(', ', expand=True).values
#swap values in NaN and replace NAN to ''
df.columns = pd.MultiIndex.from_tuples([('', x[0]) if pd.isnull(x[1]) else x for x in a])
df

结果

        x   y
id  single room double room single room double room
0   1   2   3   4   5
1   3   4   7   5   3

这是我想转换为嵌套字典列表的数据框。所以这是理想的结果

[{'id': 1,
  'x': {'double room': 3, 'single room': 2},
  'y': {'double room': 5, 'single room': 4}},
 {'id': 3,
  'x': {'double room': 7, 'single room': 4},
  'y': {'double room': 3, 'single room': 5}}]

在下面的代码中，我直接创建此列表。

firstDict = { 'id':1, 'x':{'single room':2, 'double room':3}, 'y':{'single room':4, 'double room':5} }
secondDict = { 'id':3, 'x':{'single room':4, 'double room':7}, 'y':{'single room':5, 'double room':3} }
dictList = []
dictList.append( firstDict )
dictList.append( secondDict )
dictList

[{'id': 1,
  'x': {'double room': 3, 'single room': 2},
  'y': {'double room': 5, 'single room': 4}},
 {'id': 3,
  'x': {'double room': 7, 'single room': 4},
  'y': {'double room': 3, 'single room': 5}}]

因此，总而言之，我如何将数据帧df转换为dictList。

编辑：

这是一个最小的示例，我正在寻找的解决方案应该推广到更长的标头。

Answer 1

我认为这样做没有直接的方法，也就是说，您可以使用stack + to_dict以及随后的一些后处理：

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class MapGenerator : MonoBehaviour
{

    public Transform tilePrefab; //This is the part which I want to be random
    public Vector2 mapSize;

    private void Start()
    {
        GenerateMap();
    }
    public void GenerateMap()
    {
        for (int x = 0; x < mapSize.x; x++)
        {
            for (int y = 0; y < mapSize.y; y++)
            {
                Vector3 tilePosition = new Vector3(-mapSize.x / 2 + 0.5f + x, 0,-mapSize.y / 2 + 0.5f + y);
                Transform newTile = Instantiate(tilePrefab, tilePosition, Quaternion.Euler(Vector3.right * 360)) as Transform;
            }
        }
    }

}

输出

# prepare the DataFrame
df = df.set_index(('', 'id')).stack(level=0)
df.index.names = ['id', None]

# convert to a dicts of dicts
d = {}
for (idi, key), values in df.to_dict('index').items():
    d.setdefault(idi, {}).update({key: values})

# convert d to list of dicts
result = [{'id': k, **values} for k, values in d.items()]

Answer 2

不确定标头的数量可以多长时间，目前它处于易于手动编码的状态，如下所示-

dct = []

for x in df.values:
  nd = {
          "id": x[0],
          "x": {
                  "single room": x[1],
                  "double room": x[2]
               },
          "y": {
                  "single room": x[3],
                  "double room": x[4]
               }
       }
  dct.append(nd)

请让我知道是否有大量的标头，并且代码需要在不显式键入的情况下处理它们。

Answer 3

像这样吗？

import pandas as pd
col =['id','x, single room','x, double room','y, single room','y, double room' ]
df = pd.DataFrame([[1,2,3,4,5], [3,4,7,5,3]], columns=col)
a = df.columns.str.split(', ', expand=True).values
#swap values in NaN and replace NAN to ''
df.columns = pd.MultiIndex.from_tuples([('', x[0]) if pd.isnull(x[1]) else x for x in a])
print(df)

dict_list = []
for index, row in df.iterrows():
    d = {}
#    _dict [row["id"]]
    print(type(row), row)#, row.select(1, axis = 0) )
    d["id"] = row[0]
    d["x"] = {'single room':row[1], 'double room':row[1]}
    d["y"] = {'single room':row[3], 'double room':row[4]}
    dict_list.append(d)

print(dict_list)

输出：

[{'id': 1, 
  'x': {'single room': 2, 'double room': 2}, 
  'y': {'single room': 4, 'double room': 5}
  }, 
{'id': 3, 
  'x': {'single room': 4, 'double room': 4}, 
  'y': {'single room': 5, 'double room': 3}
}
]

Answer 4

您可以使用

l = []
d = None
for i, row in df.iterrows():
    for (i1,i2),v in row.iteritems():
        if i2 == 'id':
            d = {i2:v}
            l.append(d)
            continue
        try:
            d[i1][i2]=v
        except KeyError:
            d[i1] = {i2:v}

或者您可以对预期结果稍作修改：

from collections import defaultdict
l =[]
for i, row in df.iterrows():
    d = defaultdict(dict)
    for (i1,i2),v in row.iteritems():
        if i2 == 'id':
            d[i2][v]=v
        else:
            d[i1][i2]=v
    l.append(dict(d))

输出：

[{'id': {1: 1},
  'x': {'single room': 2, 'double room': 3},
  'y': {'single room': 4, 'double room': 5}},
 {'id': {3: 3},
  'x': {'single room': 4, 'double room': 7},
  'y': {'single room': 5, 'double room': 3}}]

Answer 5

我喜欢接受的解决方案，但是在这里我的两个选择都没有堆叠。

此解决方案简单明了，但随着列数的增加，重复和容易出错：

lst = [{'id': d[('', 'id')], 
        'x': {'single room': d[('x', 'single room')], 'double room': d[('x', 'double room')]},
        'y': {'single room': d[('y', 'single room')], 'double room': d[('y', 'double room')]},}
        for d in df.to_dict('records')
]

让我们尝试使其更具可扩展性，您可以从Arbitrarily nested dictionary from tuples获得nest函数：

def nest(d: dict) -> dict:
    result = {}
    for key, value in d.items():
        target = result
        for k in key[:-1]:  
            target = target.setdefault(k, {})
        target[key[-1]] = value
    return result

但是对于('', id)，我们需要不同的行为：

def nest_m(d: dict) -> dict:
    result = {}
    for key, value in d.items():
        if key == ('', 'id'):
            result['id'] = value
        else:
            target = result
            for k in key[:-1]:  
                target = target.setdefault(k, {})
            target[key[-1]] = value    
    return result

最后一行：

lst = [nest_m(d) for d in df.to_dict('records')]

输出：

[{'id': 1,
  'x': {'single room': 2, 'double room': 3},
  'y': {'single room': 4, 'double room': 5}},
 {'id': 3,
  'x': {'single room': 4, 'double room': 7},
  'y': {'single room': 5, 'double room': 3}}]

如何将多标头熊猫数据框转换为嵌套词典列表

5 个答案: