Python列表匹配需求：筛选以指定前缀开头的文件条目

现有两个Python列表：

prefixList = ["12","9"]
files = ["12-a.csv","12-b.csv","9-t.txt","8-a.txt"]

需要创建新列表fileOutput，把所有以prefixList里的元素开头的文件条目都放进去，最终要得到：

fileOutput = ["12-a.csv","12-b.csv","9-t.txt"]

几种实现方法 #

方法1：列表推导式（最常用） #

直接遍历文件列表，逐个检查是否符合前缀要求：

prefixList = ["12","9"]
files = ["12-a.csv","12-b.csv","9-t.txt","8-a.txt"]

fileOutput = [file for file in files if any(file.startswith(prefix) for prefix in prefixList)]
print(fileOutput)

运行后就能得到预期的结果。

方法2：用filter函数（函数式风格） #

如果习惯用函数式写法，可以用filter配合lambda表达式：

prefixList = ["12","9"]
files = ["12-a.csv","12-b.csv","9-t.txt","8-a.txt"]

fileOutput = list(filter(lambda f: any(f.startswith(p) for p in prefixList), files))
print(fileOutput)

方法3：优化版列表推导式 #

str.startswith()其实支持传入元组做批量检查，把前缀列表转成元组能提升一点性能，写法也更简洁：

prefixList = ["12","9"]
files = ["12-a.csv","12-b.csv","9-t.txt","8-a.txt"]

prefix_tuple = tuple(prefixList)
fileOutput = [file for file in files if file.startswith(prefix_tuple)]
print(fileOutput)

内容的提问来源于stack exchange，提问作者WelshOne