Em Qua, 2006-03-15 às 13:49 -0800, vinjvinj escreveu: > f = open(someFilePath, "rb") > content = [] > for data in content.read() > content.append(data) > fullContent = "".join(content) > > Is there a more efficient way of doing this? I'll be running this > operation on 10,000+ files where each file is an image file with size > 50k-100k
If you really need everything in memory, why not just... fullContent = open(someFilePath, "rb").read() ...? -- http://mail.python.org/mailman/listinfo/python-list
