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

Although the most intuitive idea could possibly be to count the amount of unique values and check if there is only one, this would have a needlessly high complexity for what we're trying to do. Numpy's' np.unique, called by pandas' nunique, implements a sorting of the underlying arrays, which has an evarage complexity of O(n·log(n)) using quicksort (default). The above approach is O(n).

The difference in performance becomes more obvious when we're applying this to an entire dataframe (see below).

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 below 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'
)

I think the cleanest way is to check all columns against the first column using eq:

CopyIn [11]: df
Out[11]: 
   a  b  c  d
0  C  C  C  C
1  C  C  A  A
2  A  A  A  A

In [12]: df.iloc[:, 0]
Out[12]: 
0    C
1    C
2    A
Name: a, dtype: object

In [13]: df.eq(df.iloc[:, 0], axis=0)
Out[13]: 
      a     b      c      d
0  True  True   True   True
1  True  True  False  False
2  True  True   True   True

Now you can use all (if they are all equal to the first item, they are all equal):

CopyIn [14]: df.eq(df.iloc[:, 0], axis=0).all(1)
Out[14]: 
0     True
1    False
2     True
dtype: bool

You also need to be careful to create a copy of the DataFrame, otherwise the csvdata_old will be updated with csvdata (since it points to the same object):

csvdata_old = csvdata.copy()

To check whether they are equal, you can use assert_frame_equal as in this answer:

from pandas.util.testing import assert_frame_equal
assert_frame_equal(csvdata, csvdata_old)

You can wrap this in a function with something like:

try:
    assert_frame_equal(csvdata, csvdata_old)
    return True
except:  # appeantly AssertionError doesn't catch all
    return False

There was discussion of a better way...

You can use assert_frame_equal and not check the dtype of the columns.

# Pre v. 0.20.3
# from pandas.util.testing import assert_frame_equal

from pandas.testing import assert_frame_equal

assert_frame_equal(df1, df2, check_dtype=False)

Use DataFrame.all with compare by Series.eq:

if df.B.eq(9).all():
    print ('all values are 9')

EDIT: Like mentioned @mozway use:

assert df.B.eq(9).all()
#AssertionError

With custom error:

if df.B.ne(9).any():
    raise Exception('All values in column are not equal 9')
    
Exception: All values in column are not equal 9