Format timedelta64 string output

Question

In a similar vein to this question, I have a numpy.timedelta64 column in a pandas DataFrame. As per this answer to the aforementioned question, there is a function pandas.tslib.repr_timedelta64 which nicely displays a timedelta in days, hours:minutes:seconds. I would like to format them only in days and hours.

So what I've got is the following:

def silly_format(hours):
    (days, hours) = divmod(hours, 24)
    if days > 0 and hours > 0:
        str_time = "{0:.0f} d, {1:.0f} h".format(days, hours)
    elif days > 0:
        str_time = "{0:.0f} d".format(days)
    else:
        str_time = "{0:.0f} h".format(hours)
    return str_time

df["time"].astype("timedelta64[h]").map(silly_format)

which gets me the desired output but I was wondering whether there is a function in numpy or pandas similar to datetime.strftime that can format numpy.timedelta64 according to some format string provided?

---

I tried to adapt @Jeff's solution further but it is way slower than my answer. Here it is:

days = time_delta.astype("timedelta64[D]").astype(int)
hours = time_delta.astype("timedelta64[h]").astype(int) % 24
result = days.astype(str)
mask = (days > 0) & (hours > 0)
result[mask] = days.astype(str) + ' d, ' + hours.astype(str) + ' h'
result[(hours > 0) & ~mask] = hours.astype(str) + ' h'
result[(days > 0) & ~mask] = days.astype(str) + ' d'

Answer 1

While the answers provided by @sebix and @Jeff show a nice way of converting the timedeltas to days and hours, and @Jeff's solution in particular retains the Series' index, they lacked in flexibility of the final formatting of the string. The solution I'm using now is:

def delta_format(days, hours):
    if days > 0 and hours > 0:
        return "{0:.0f} d, {1:.0f} h".format(days, hours)
    elif days > 0:
        return "{0:.0f} d".format(days)
    else:
        return "{0:.0f} h".format(hours)

days = time_delta.astype("timedelta64[D]")
hours = time_delta.astype("timedelta64[h]") % 24
return [delta_format(d, h) for (d, h) in izip(days, hours)]

which suits me well and I get back the index by inserting that list into the original DataFrame.

Answer 2

Here's how to do it in a vectorized manner.

In [28]: s = pd.to_timedelta(range(5),unit='d') + pd.offsets.Hour(3)

In [29]: s
Out[29]: 
0   0 days, 03:00:00
1   1 days, 03:00:00
2   2 days, 03:00:00
3   3 days, 03:00:00
4   4 days, 03:00:00
dtype: timedelta64[ns]

In [30]: days = s.astype('timedelta64[D]').astype(int)

In [31]: hours = s.astype('timedelta64[h]').astype(int)-days*24

In [32]: days
Out[32]: 
0    0
1    1
2    2
3    3
4    4
dtype: int64

In [33]: hours
Out[33]: 
0    3
1    3
2    3
3    3
4    3
dtype: int64

In [34]: days.astype(str) + ' d, ' + hours.astype(str) + ' h'
Out[34]: 
0    0 d, 3 h
1    1 d, 3 h
2    2 d, 3 h
3    3 d, 3 h
4    4 d, 3 h
dtype: object

If you want exactly as the OP posed:

In [4]: result = days.astype(str) + ' d, ' + hours.astype(str) + ' h'

In [5]: result[days==0] = hours.astype(str) + ' h'

In [6]: result
Out[6]: 
0         3 h
1    1 d, 3 h
2    2 d, 3 h
3    3 d, 3 h
4    4 d, 3 h
dtype: object