Having trouble with time.sleep
When I run, for example:
print("[",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
prin开发者_StackOverflow社区t("=",end=" ")
time.sleep(1)
print("=",end=" ")
time.sleep(1)
print("]",end=" ")
Nothing happens for 10 seconds, then the whole [ = = = = = = = = = = ] appears. How can I prevent that so that it can act as a sort of progress bar?
Try flushing stdout after each print:
import sys
print("=",end=" ")
sys.stdout.flush()
Actually, a progress bar belongs to sys.stderr
, which is (very conveniently and not coincidentally at all) not buffered. So I suggest you:
print("=", end=" ", file=sys.stderr)
instead.
PS a synopsis of the standard input, output and error streams in POSIX-conformant operating systems can be found in Wikipedia: Standard streams. In a few words: stdin is the input to a process; stdout is the useful output of a process, the results; stderr is for warnings, errors and out-of-band (e.g. progress bars) output.
sys.stdout.flush()
You need to flush stdout
using sys.stdout.flush()
every time you want to write the updates.
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