Get highest duration from a list of strings

Here's an absolute hack which solves the problem in a bad but clever way: Python's min and max functions can be used with a key function which is used to compare elements, so that it returns the element minimising or maximising that function. If the key function returns a tuple, then the order is determined by the first component of the tuple, using the second component as a tie-breaker.

We can exploit the fact that the last characters 'd', 'h' and 'm' can be compared in alphabetical order; a day is longer than an hour is longer than a minute. This means the longest duration has the minimum character in alphabetical order, with the maximum integer as a tie-breaker. Maximising that integer is the same as minimising its negation:

>>> durations = ['5d', '20h', '1h', '7m', '14d', '1m']
>>> min(durations, key=lambda d: (d[-1], -int(d[:-1])))
'14d'

Pure python solution. We could store mapping between our time extensions (m, h, d) and minutes (here time_map), to find highest duration. Here we're using max() with key argument to apply our mapping.

inp = ['5d', '20h', '1h', '7m', '14d', '1m']
time_map = {'m': 1, 'h': 60, 'd': 24*60}

print(max(inp, key=lambda x:int(x[:-1])*time_map[x[-1]]))  # -> 14d

np.argmax on pd.to_timedelta:

import numpy as np
import pandas as pd

durations = ['5d', '20h', '1h', '7m', '14d', '1m']

durations[np.argmax(pd.to_timedelta(durations))]

Out[24]: '14d'

pd.to_timedelta turns a string into a duration (source), and np.argmax returns the index of the highest element.