首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >如何根据颜色值创建chord图的矩阵:

如何根据颜色值创建chord图的矩阵:
EN

Stack Overflow用户
提问于 2019-09-09 07:33:41
回答 1查看 343关注 0票数 0

假设我有一个数据框架,它具有以下格式的数据。

代码语言:javascript
复制
UID | Name | ID
----------------
1 | ABC | IM-1
2 | XYZ | IM-2
3 | XYZ | IM-2
4 | PQR | IM-3
5 | PQR | IM-4
6 | PQR | IM-5
7 | XYZ | IM-5
8 | ABC | IM-5

我需要创建一个矩阵,输入到chord图代码中。这需要以下列格式输出:

代码语言:javascript
复制
(array([[0,1,1,1],
        [1,1,1,0],
        [1,1,0,2]]),['ABC','XYZ','PQR'])

注:在本例中,“名称”在列表中是有限的(即ABC、XYZ或PQR) - " ID“在记录之间共享-第四列是独立的记录数(例如,ABC是单个记录IM-1的一部分,PQR在IM-4和IM-5中出现两次)矩阵的其他成员是基于ID的名称之间的联系(例如IM-5,增加PQR-XYZ,XYZ-PQR,PQR-ABC,ABC-PQR,ABC-PQR,-目标是为"Name“字段之间的连接创建一个chord图

我知道这是个不错的读物。提前谢谢你的帮助。

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2019-09-09 16:49:34

更新了我的答案,但方法基本上是一样的。将数据解析为数据框架,在ID上执行内连接以获得通过共享公共ID链接的名称对。然后将这个边表转换成一个邻接矩阵。最后,一些人四处走动以获得“悬空”边缘,即只有一次出现的ID (添加在更新的答案中),并根据相应的Name对其计数进行分组。

代码语言:javascript
复制
#!/usr/bin/env python
"""
Create adjacency matrix from a dataframe, where edges are implicitly defined by shared attributes.

Answer to:
https://stackoverflow.com/questions/57849602/how-to-create-the-matrix-for-chord-diagram-based-on-coloumn-value
"""
import numpy as np
import pandas as pd
from collections import Counter

def parse_data_format(file_path):
    # read data skipping second line
    df = pd.read_csv(file_path, sep='|', skiprows=[1])

    # strip whitespace from column names
    df = df.rename(columns=lambda x: x.strip())

    # strip whitespace from values
    df_obj = df.select_dtypes(['object'])
    df[df_obj.columns] = df_obj.apply(lambda x: x.str.strip())

    return df


def get_edges(df):
    """Get all combinations of 'Name' that share a 'ID' value (using an inner join)."""
    inner_self_join = df.merge(df, how='inner', on='ID')
    excluding_self_pairs = inner_self_join[inner_self_join['UID_x']!=inner_self_join['UID_y']]
    edges = excluding_self_pairs[['Name_x', 'Name_y']].values
    return edges


def get_adjacency(edges):
    "Convert a list of 2-tuples specifying source and target of a connection into an adjacency matrix."
    order = np.unique(edges)
    total_names = len(order)
    name_to_idx = dict(list(zip(order, range(total_names))))
    adjacency = np.zeros((total_names, total_names))
    for (source, target) in edges:
        adjacency[name_to_idx[source], name_to_idx[target]] += 1
    return adjacency, order


def get_dangling_edge_counts(df):
    # get IDs with count 1
    counts = Counter(df['ID'].values)
    singles = [ID for (ID, count) in counts.items() if count == 1]
    # get corresponding names
    names = [df[df['ID']==ID]['Name'].values[0] for ID in singles]
    # convert into counts
    return Counter(names)


if __name__ == '__main__':

    # here we read in the data as a file buffer;
    # however, normally we would hand a file path to parse_data_format instead
    import sys
    if sys.version_info[0] < 3:
        from StringIO import StringIO
    else:
        from io import StringIO

    data = StringIO(
        """UID | Name | ID
        ----------------
        1 | ABC | IM-1
        2 | XYZ | IM-2
        3 | XYZ | IM-2
        4 | PQR | IM-3
        5 | PQR | IM-4
        6 | PQR | IM-5
        7 | XYZ | IM-5
        8 | ABC | IM-5
        """
    )

    df = parse_data_format(data)
    edges = get_edges(df)
    adjacency, order = get_adjacency(edges)
    print(adjacency)
    # [[0. 1. 1.]
    #  [1. 0. 1.]
    #  [1. 1. 0.]]
    print(order)
    # ['ABC' 'PQR' 'XYZ']

    dangling_edge_counts = get_dangling_edge_counts(df)
    print(dangling_edge_counts)
    # Counter({'PQR': 2, 'ABC': 1})

    last_column = np.zeros_like(order, dtype=np.int)
    for ii, name in enumerate(order):
        if name in dangling_edge_counts:
            last_column[ii] = dangling_edge_counts[name]
    combined = np.concatenate([adjacency, last_column[:, np.newaxis]], axis=-1)
    print(combined)
    #[[0. 1. 1. 1.]
    # [1. 0. 1. 2.]
    # [1. 1. 2. 0.]]
票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/57849602

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档