使用列表理解如果嵌套

如何解决使用列表理解如果嵌套

给出一个文件名列表，我们想将所有扩展名为hpp的文件重命名为扩展名h。为此，我们要生成一个名为newfilenames的新列表，其中包括新文件名。使用列表理解＃

filenames = ["program.c","stdio.hpp","sample.hpp","a.out","math.hpp","hpp.out"]
newfilenames = [filename[:len(filename) - 3] + "h" for filename in filenames if filename.endswith("p")]
print(newfilenames)

输出来自["stdio.h","sample.h","math.h"] 应该是["program.c","stdio.h","math.h","hpp.out"]

解决方法

您可以简单地将.hpp替换为.h

filenames = ["program.c","stdio.hpp","sample.hpp","a.out","math.hpp","hpp.out"]
x=[x.replace(".hpp",".h") for x in filenames]
print (x)

输出：

['program.c','stdio.h','sample.h','a.out','math.h','hpp.out']

,

我认为您需要

filenames = [i[:-3]+"h" if i.split(".")[-1]=="hpp" else i for i in filenames]

print(filenames)