如何将数据框转换为字典

Question

我有一个如下所述的数据框。

我想为上面的数据框中的所有非零列提供一个字典，就像下面的一样。

{
(0, 'aan'): 1,
 (0, 'abcc'): 1,
 (1, 'acd'): 1,
 (3, 'access'): 5,
 (3, 'acd'): 3,
 (4, 'aao'): 2,
 (4, 'access’): 4
}

Answer 1

也许，有两个步骤可以解决该问题：

1. 获取布尔值，是否元素为0
2. 总结价值

#include <stdio.h>
#include <string.h>
#include <math.h>

unsigned long long int bin2dec(const char *string, const size_t size)
{
    unsigned long long int bit, value = 0;
    for(size_t index=0;index<size;index++)
    {
        // moving from the end to the beginning, get a character from the string
        // and convert it from a character containing a digit to a number
        bit = string[size-index-1]-'0';

        // in the original question this was: value += bit*pow(2,index);
        // but we can just do this and get the same effect
        // without multiplication or library function
        value += bit<<index;
    }
    return value;
}

int main()
{
    const char * binary = "111111111111111111111111111111";
    unsigned long long int decimal = bin2dec(binary, strlen(binary));
    printf("%llu\n",decimal);
    return 0;
}

Answer 2

通过稀疏矩阵对其进行管道传输，然后将DataFrame作为dict返回。 不幸的是，熊猫的稀疏矩阵功能有限，因此我们需要使用scipy。以下代码应适用于您的应用程序。

import scipy as sp
import pandas as pd
import numpy as np #for the random dataframe as example.

# Example dataframe
df = pd.DataFrame(np.random.randint(0,10,size=(1000, 10)))

# Use scipy to create sparse matrix
coo = sp.sparse.csc_matrix(df).tocoo(copy=False)

# Parse sparse matrix back into dataframe without zeroes.
df = pd.DataFrame({'index': coo.row, 'col': coo.col, 'data': coo.data})[
    ['index', 'col', 'data']].sort_values(['index', 'col']).reset_index(drop=True)

# Create index to split (row, column) from value.
ix = pd.MultiIndex.from_frame(df[['index','col']])
df = df['data'].copy(True)
df.index = ix

# Output as dict
df.to_dict()

df
   0  1  2  3  4  5  6  7  8  9
0  4  7  0  3  4  8  6  0  5  3
1  3  3  9  2  1  2  8  2  7  2
2  0  1  5  5  4  3  2  0  4  1
3  6  7  7  7  2  1  3  7  1  1
4  2  5  9  8  9  7  5  4  0  3

{(0, 0): 4,
 (0, 1): 7,
 (0, 3): 3, # Notice (0,2) is gone.
 (0, 4): 4,
 (0, 5): 8,
 (0, 6): 6,
 (0, 8): 5,
 (0, 9): 3,
 (1, 0): 3,
 (1, 1): 3,
 (1, 2): 9,
 (1, 3): 2,
 (1, 4): 1,
 (1, 5): 2,
 (1, 6): 8,
 (1, 7): 2,
 (1, 8): 7,
 (1, 9): 2,
 (2, 1): 1, # Notice (2,0) is gone.
 (2, 2): 5,
 (2, 3): 5,
 (2, 4): 4,
 (2, 5): 3,
 (2, 6): 2,
 (2, 8): 4,
 (2, 9): 1,

Answer 3

这是一种非常基本的暴力手段。 不可扩展。

data = {'aan': [1, 2,0], 'aao': [0,3, 4], 'access':[0,0,1]}
df = pandas.pandas.DataFrame(data=data)
master= {}
for t in df.itertuples():
    _ = {(t.Index, col):getattr(t, col) for col in df.columns if getattr(t, col)}
    if not _:continue
    master.update(_)

打印

{(0, 'aan'): 1, (1, 'aan'): 2, (1, 'aao'): 3, (2, 'aao'): 4, (2, 'access'): 1}