Parse hours without leading zeroes by strptime in Python

Question

Suppose you have time in this format:

a = [..., 800.0, 830.0, 900.0, 930.0, 1000.0, 1030.0, ...]

The problem is that leading zeroes for hou

Answer 1

If those are really float literals in there vs strings, you can do this:

a=[800., 830., 900., 930., 1000., 1030.]
hours=[time.strptime('{:04.0f}'.format(f), '%H%M') for f in a]

That will round the decimal if any (1033.66666 would be 1034 hence becoming 10:34 AM)

You can also truncate like so:

[800.0, 830.0, 900.0, 930.0, 1000.0, 1030.0, 1033.3333333, 1033.66666]
hours=[time.strptime(str(f).split('.')[0], '%H%M') for f in a]

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edit from comments

If you have values outside the range, you do this:

a=[800., 830., 900., 930., 1000., 1030., 2400.]
hours=[time.strptime(s,'%H%M') for s in ['{:04.0f}'.format(f) if f <2400 else '0000' for f in a]]

or, you can make your original code work that way as well:

[time.strptime(i,'%H%M') for i in[str(int(f)) if f<2400 else '0000' for f in a]]