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我有一个数据框

sample1  0 0 0 0 0 1 1 1 1 1 1 1 1 L1
sample2  0 0 0 0 0 1 1 1 1 1 0 0 0 L1-1
sample3  0 0 0 0 0 1 1 0 0 0 0 0 0 L1-1-1
sample4  0 0 0 0 0 1 0 0 0 0 0 0 0 L1-1-1-1
sample5  0 0 0 0 0 0 0 1 1 0 0 0 0 L1-1-2
sample6  0 0 0 0 0 0 0 1 0 0 0 0 0 L1-1-2-1
sample7  0 0 0 0 0 0 0 0 0 1 0 0 0 L1-1-3
sample8  0 0 0 0 0 0 0 0 0 0 1 1 1 L1-2
sample9  0 0 0 0 0 0 0 0 0 0 1 1 0 L1-2-1
sample10 0 0 0 0 0 0 0 0 0 0 0 0 1 L1-2-2
sample11 1 1 1 1 1 0 0 0 0 0 0 0 0 L2
sample12 1 1 1 0 0 0 0 0 0 0 0 0 0 L2-1
sample13 1 1 0 0 0 0 0 0 0 0 0 0 0 L2-1-1
sample14 1 0 0 0 0 0 0 0 0 0 0 0 0 L2-1-1-1
sample15 0 0 0 1 0 0 0 0 0 0 0 0 0 L2-2
sample16 0 0 0 0 1 0 0 0 0 0 0 0 0 L2-3

如您所见，每一行都是聚集的。

我想为每个样本命名“基于谱系”的标签。

例如，sample1 将是 lin1，因为它首先出现，sample2 将是 lin1-1。

Sample3 将是 lin1-1-1，sample4 将是 lin1-1-1-1。

接下来，sample5 将是 lin1-2，sample6 将是 lin1-2-1...

Sample11 将是血统 lin2 的新起点。

我最初的命名想法是。

"sample1 是 lin1，如果下一个样本包含在前一个样本中，则 lin1 + "-1" 如果没有，则 lin(1+1)"

样本1-> lin1

sample2 -> lin1-1（sample2 包含在 sample1 中）

sample3 -> lin1-1-1（sample3 包含在 sample2 中）

sample4 -> lin1-1-1-1（sample4 包含在 sample3 中）

sample5 -> lin1-1-2 (sample5 不包含在 sample4 中) .... 这样的逻辑。

我无法将这个逻辑变成 python 脚本。

4

1 回答 1

1

这可以通过几个步骤来完成。

步骤 1. 数据预处理

将数据按降序排序并删除重复的，否则可能无法正常工作。假设完成。

import numpy as np

data = '''sample1  0 0 0 0 0 1 1 1 1 1 1 1 1
sample2  0 0 0 0 0 1 1 1 1 1 0 0 0
sample3  0 0 0 0 0 1 1 0 0 0 0 0 0
sample4  0 0 0 0 0 1 0 0 0 0 0 0 0
sample5  0 0 0 0 0 0 0 1 1 0 0 0 0
sample6  0 0 0 0 0 0 0 1 0 0 0 0 0
sample7  0 0 0 0 0 0 0 0 0 1 0 0 0
sample8  0 0 0 0 0 0 0 0 0 0 1 1 1
sample9  0 0 0 0 0 0 0 0 0 0 1 1 0
sample10 0 0 0 0 0 0 0 0 0 0 0 0 1
sample11 1 1 1 1 1 0 0 0 0 0 0 0 0
sample12 1 1 1 0 0 0 0 0 0 0 0 0 0
sample13 1 1 0 0 0 0 0 0 0 0 0 0 0
sample14 1 0 0 0 0 0 0 0 0 0 0 0 0
sample15 0 0 0 1 0 0 0 0 0 0 0 0 0
sample16 0 0 0 0 1 0 0 0 0 0 0 0 0'''

data = [x.split() for x in data.split('\n')]
data = [x[1:] for x in data]
data = np.array(data, dtype=int)  
data

array([[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1],
       [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
       [1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0]])

步骤 2. 将样本编码到位置。每个元素都是一个frozenset。

nrow, ncol = data.shape

def to_position(sample):
    ncol = len(sample)
    return frozenset(i for i in range(ncol) if sample[i] == 1)

position = [to_position(data[i]) for i in range(nrow)]
# print(position)

步骤 3. 将每个样本位置分配给一个集群，该集群目前表示为一个元组。

def assign_cluster(sample, clusters, parent):
    if parent not in clusters:
        clusters[parent] = sample
    elif sample < clusters[parent]:
        # Find child
        parent = parent + (0,)
        assign_cluster(sample, clusters, parent)
    else:
        # Find siblings
        parent = parent[:-1] + (parent[-1] + 1, )
        assign_cluster(sample, clusters, parent)


clusters = {}
root = (0,)
clusters[root] = position[0]
for i in range(1, nrow):
    sample = position[i]
    assign_cluster(sample, clusters, parent=root)

# print(clusters)

步骤 4. 将簇转换为字符串并显示结果。

def cluster_to_string(c):
    c = [str(_ + 1) for _ in c]
    return 'L' + '-'.join(c)

position_dict = {v: k for k, v in clusters.items()}

for sample in data:
    sample = to_position(sample)
    c = position_dict[sample]
    print(cluster_to_string(c))

L1
L1-1
L1-1-1
L1-1-1-1
L1-1-2
L1-1-2-1
L1-1-3
L1-2
L1-2-1
L1-2-2
L2
L2-1
L2-1-1
L2-1-1-1
L2-2
L2-3
于 2021-11-19T13:51:07.780 回答