Check if all values in dataframe column are the same

Question

I want to do a quick and easy check if all column values for counts are the same in a dataframe:

In:

import pandas as pd

d = {\'names\'         


        

Answer 1

An efficient way to do this is by comparing the first value with the rest, and using all:

def is_unique(s):
    a = s.to_numpy() # s.values (pandas<0.24)
    return (a[0] == a).all()

is_unique(df['counts'])
# False

---

For an entire dataframe

In the case of wanting to perform the same task on an entire dataframe, we can extend the above by setting axis=0 in all:

def unique_cols(df):
    a = df.to_numpy() # df.values (pandas<0.24)
    return (a[0] == a).all(0)

For the shared example, we'd get:

unique_cols(df)
# array([False, False])

---

Here's a benchmark of the above methods compared with some other approaches, such as using nunique (for a pd.Series):

s_num = pd.Series(np.random.randint(0, 1_000, 1_100_000))

perfplot.show(
    setup=lambda n: s_num.iloc[:int(n)], 

    kernels=[
        lambda s: s.nunique() == 1,
        lambda s: is_unique(s)
    ],

    labels=['nunique', 'first_vs_rest'],
    n_range=[2**k for k in range(0, 20)],
    xlabel='N'
)

---

And bellow are the timings for a pd.DataFrame. Let's compare too with a numba approach, which is especially useful here since we can take advantage of short-cutting as soon as we see a repeated value in a given column (note: the numba approach will only work with numerical data):

from numba import njit

@njit
def unique_cols_nb(a):
    n_cols = a.shape[1]
    out = np.zeros(n_cols, dtype=np.int32)
    for i in range(n_cols):
        init = a[0, i]
        for j in a[1:, i]:
            if j != init:
                break
        else:
            out[i] = 1
    return out

If we compare the three methods:

df = pd.DataFrame(np.concatenate([np.random.randint(0, 1_000, (500_000, 200)), 
                                  np.zeros((500_000, 10))], axis=1))

perfplot.show(
    setup=lambda n: df.iloc[:int(n),:], 

    kernels=[
        lambda df: (df.nunique(0) == 1).values,
        lambda df: unique_cols_nb(df.values).astype(bool),
        lambda df: unique_cols(df) 
    ],

    labels=['nunique', 'unique_cols_nb', 'unique_cols'],
    n_range=[2**k for k in range(0, 20)],
    xlabel='N'
)

Answer 2

I think nunique does much more work than necessary. Iteration can stop at the first difference. This simple and generic solution uses itertools:

import itertools

def all_equal(iterable):
    "Returns True if all elements are equal to each other"
    g = itertools.groupby(iterable)
    return next(g, True) and not next(g, False)

all_equal(df.counts)

One can use this even to find all columns with constant contents in one go:

constant_columns = df.columns[df.apply(all_equal)]

---

A slightly more readable but less performant alternative:

df.counts.min() == df.counts.max()

Add skipna=False here if necessary.

Answer 3

Update using np.unique

len(np.unique(df.counts))==1
False

Or

len(set(df.counts.tolist()))==1

Or

df.counts.eq(df.counts.iloc[0]).all()
False

Or

df.counts.std()==0
False