Python中读取大文件的惰性方法？

Posted 2021-02-26

f = open('test.zip', 'rb')
 
for data in iter( lambda: f.read(1024) , ''):
    pass # do stuff
 
# or without lambda
 
f = open('really_big_file.dat')
def read1k():
    return f.read(1024)
 
for piece in iter(read1k, ''):
    process_data(piece)