Floor or ceiling of a pandas series in python?

Question

I have a pandas series series. If I want to get the element-wise floor or ceiling, is there a built in method or do I have to write the function and use apply? I ask because the data is big so I appreciate efficiency. Also this question has not been asked with respect to the Pandas package.

score 146 · Accepted Answer · answered Dec 21 '14 at 18:37

146

You can use NumPy's built in methods to do this: np.ceil(series) or np.floor(series).

Both return a Series object (not an array) so the index information is preserved.

answered Dec 21 '14 at 18:37

Alex Riley

169,130
45
262
238

5

how can I chain this like using `round()` on a pandas `Series`? – iamyojimbo Aug 31 '20 at 14:22
This works: dataframe["new_series"] = np.ceil(dataframe[series].round(0)) – Dave Sottimano Dec 09 '22 at 02:58
2

@iamyojimbo, if you want to chain the method, use `pipe()`, `df['column'].pipe(np.ceil)` – renan-eccel Jun 09 '23 at 17:41

score 37 · Answer 2 · answered Dec 21 '14 at 18:37

37

I am the OP, but I tried this and it worked:

np.floor(series)

answered Dec 21 '14 at 18:37

wolfsatthedoor

7,163
18
46
90

Jivan · Answer 3 · 2020-07-08T12:05:13.610

23

UPDATE: THIS ANSWER IS WRONG, DO NOT DO THIS

Explanation: using Series.apply() with a native vectorized Numpy function makes no sense in most cases as it will run the Numpy function in a Python loop, leading to much worse performance. You'd be much better off using np.floor(series) directly, as suggested by several other answers.

You could do something like this using NumPy's floor, for instance, with a dataframe:

floored_data = data.apply(np.floor)

Can't test it right now but an actual and working solution might not be far from it.

edited Jul 08 '20 at 12:05

answered Dec 21 '14 at 18:36

Jivan

21,522
15
80
131

If you use the argument `raw=True` then the [`df.apply`](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.apply.html) function will achieve much better performance. – Raisin Jan 26 '23 at 21:20
The apply function is not a vectorized implementation therefore it will be super slow. You can call np.floor directly to the dataframe. – Péter Szilvási Feb 02 '23 at 12:50

score 20 · Answer 4 · edited Jul 29 '21 at 08:28

20

With pd.Series.clip, you can set a floor via clip(lower=x) or ceiling via clip(upper=x):

s = pd.Series([-1, 0, -5, 3])
    
print(s.clip(lower=0))
# 0    0
# 1    0
# 2    0
# 3    3
# dtype: int64
    
print(s.clip(upper=0))
# 0   -1
# 1    0
# 2   -5
# 3    0
# dtype: int64

pd.Series.clip allows generalised functionality, e.g. applying and flooring a ceiling simultaneously, e.g. s.clip(-1, 1)

NOTE: Answer originally referred to clip_lower / clip_upper which were removed in pandas 1.0.0.

edited Jul 29 '21 at 08:28

pbarber

139
2
8

answered Jan 14 '19 at 00:05

jpp

159,742
34
281
339

8

I can see that it's worth mentioning `clip` et al. as *somewhat* related functions, but clipping a value is a very different operation to finding the floor/ceiling of that value... – Alex Riley Jan 14 '19 at 13:14
@AlexRiley, Point taken; the reason I mention it is there are situations where the upper and lower bounds are variables from an external input, and you may want (for example) to use `s.clip(-1, np.inf)` or `s.clip(-np.inf, 1)` to handle those situations seamlessly. – jpp Jan 14 '19 at 13:15
It was more that I was puzzled by what OP's wants to do given the wording of their question (and indeed OP's own answer). Unless I'm missing something, if you have `s = pd.Series([3.1, 2.2, 5.6])` there does not seem to be a way to use `clip` alone to compute the same result as `np.floor(s)` for example. (Not to detract from your answer which is well-written and useful information - I was just confused when I reread the question.) – Alex Riley Jan 14 '19 at 13:24
3

@AlexRiley, Ah, now rereading the question is confusing me also. I came by the question when googling "how to floor a Pandas series". So I'll leave this answer for others who reach the somewhat ambiguous question via the same route! – jpp Jan 14 '19 at 13:27

score 7 · Answer 5 · answered Sep 24 '21 at 02:19

The pinned answer already the fastest. Here's I provide some alternative to do ceiling and floor using pure pandas and compare it with the numpy approach.

series = pd.Series(np.random.normal(100,20,1000000))

Floor

%timeit np.floor(series) # 1.65 ms ± 18.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit series.astype(int) # 2.2 ms ± 131 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit (series-0.5).round(0) # 3.1 ms ± 47 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit round(series-0.5,0) # 2.83 ms ± 60.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Why astype int works? Because in Python, when converting to integer, that it always get floored.

Ceil

%timeit np.ceil(series) # 1.67 ms ± 21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit (series+0.5).round(0) # 3.15 ms ± 46.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit round(series+0.5,0) # 2.99 ms ± 103 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

So yeah, just use the numpy function.

Floor or ceiling of a pandas series in python?

5 Answers5

Linked

Related