如何在Python中匹配样本名与文件名并生成多值字典？

我有一个样本名列表和一个包含样本名的文件名列表：

sample_names = ['SampleA', 'SampleB', 'SampleC']

file_names = ['/path/to/group1/bin_1/group_1_bin_1_SampleA.tsv', 
              '/path/to/group1/bin_1/group_1_bin_1_SampleB.tsv',
              '/path/to/group1/bin_1/group_1_bin_1_SampleC.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleA.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleB.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleC.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleA.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleB.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleC.tsv']

需要将每个样本名与对应的所有文件名匹配，生成如下格式的多值字典：

dictionary = {'SampleA': ['/path/to/group1/bin_1/group_1_bin_1_SampleA.tsv',
                          '/path/to/group1/bin_2/group_1_bin_2_SampleA.tsv',
                          '/path/to/group1/bin_3/group_1_bin_3_SampleA.tsv'],
              'SampleB': ['/path/to/group1/bin_1/group_1_bin_1_SampleB.tsv',
                          '/path/to/group1/bin_2/group_1_bin_2_SampleB.tsv',
                          '/path/to/group1/bin_3/group_1_bin_3_SampleB.tsv'],
              'SampleC': ['/path/to/group1/bin_1/group_1_bin_1_SampleC.tsv',
                          '/path/to/group1/bin_2/group_1_bin_2_SampleC.tsv',
                          '/path/to/group1/bin_3/group_1_bin_3_SampleC.tsv']}

方案一：基础循环实现 #

先初始化每个样本名对应的空列表，再遍历所有文件名，判断当前文件名属于哪个样本后添加到对应列表：

sample_names = ['SampleA', 'SampleB', 'SampleC']
file_names = ['/path/to/group1/bin_1/group_1_bin_1_SampleA.tsv', 
              '/path/to/group1/bin_1/group_1_bin_1_SampleB.tsv',
              '/path/to/group1/bin_1/group_1_bin_1_SampleC.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleA.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleB.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleC.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleA.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleB.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleC.tsv']

# 初始化字典，每个样本对应空列表
result = {sample: [] for sample in sample_names}

for file in file_names:
    for sample in sample_names:
        if sample in file:
            result[sample].append(file)
            break  # 找到匹配样本后跳出内层循环，提升效率

print(result)

方案二：精准匹配文件名（避免路径干扰） #

如果担心样本名出现在路径目录中导致误匹配，可提取文件名后再判断：

import os

sample_names = ['SampleA', 'SampleB', 'SampleC']
file_names = ['/path/to/group1/bin_1/group_1_bin_1_SampleA.tsv', 
              '/path/to/group1/bin_1/group_1_bin_1_SampleB.tsv',
              '/path/to/group1/bin_1/group_1_bin_1_SampleC.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleA.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleB.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleC.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleA.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleB.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleC.tsv']

result = {sample: [] for sample in sample_names}

for file in file_names:
    filename = os.path.basename(file)
    for sample in sample_names:
        if sample in filename:
            result[sample].append(file)
            break

print(result)

方案三：使用collections.defaultdict简化初始化 #

无需手动为每个样本创建空列表，defaultdict会自动处理列表初始化：

from collections import defaultdict

sample_names = ['SampleA', 'SampleB', 'SampleC']
file_names = ['/path/to/group1/bin_1/group_1_bin_1_SampleA.tsv', 
              '/path/to/group1/bin_1/group_1_bin_1_SampleB.tsv',
              '/path/to/group1/bin_1/group_1_bin_1_SampleC.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleA.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleB.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleC.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleA.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleB.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleC.tsv']

result = defaultdict(list)

for file in file_names:
    for sample in sample_names:
        if sample in file:
            result[sample].append(file)
            break

# 若需要转为普通字典，执行：result = dict(result)
print(result)

方案四：字典推导式精简代码 #

用字典推导式结合列表推导式，一行代码完成逻辑：

sample_names = ['SampleA', 'SampleB', 'SampleC']
file_names = ['/path/to/group1/bin_1/group_1_bin_1_SampleA.tsv', 
              '/path/to/group1/bin_1/group_1_bin_1_SampleB.tsv',
              '/path/to/group1/bin_1/group_1_bin_1_SampleC.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleA.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleB.tsv',
              '/path/to/group1/bin_2/group_1_bin_2_SampleC.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleA.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleB.tsv',
              '/path/to/group1/bin_3/group_1_bin_3_SampleC.tsv']

result = {sample: [file for file in file_names if sample in file] for sample in sample_names}

print(result)

内容的提问来源于stack exchange，提问作者eb0906