Pandas: Get all columns that have constant value

Use the pandas not-so-well-known builtin nunique():

df.columns[df.nunique() <= 1]
Index(['B', 'C'], dtype='object')

Notes:

• Use nunique(dropna=False) option if you want na's counted as a separate value
• It's the cleanest code, but not the fastest. (But in general code should prioritize clarity and readability).

try this,

print [col for col in df.columns if len(df[col].unique())==1]

Output:

['B', 'C']

You can use set and apply a filter on a series:

vals = df.apply(set, axis=0)
res = vals[vals.map(len) == 1].index

print(res)

Index(['B', 'C'], dtype='object')

Use res.tolist() if having a list output is important.

Solution 1:

c = [c for c in df.columns if len(set(df[c])) == 1]
print (c)

['B', 'C']

Solution 2:

c = df.columns[df.eq(df.iloc[0]).all()].tolist()
print (c)
['B', 'C']

Explanation for Solution 2:

First compare all rows to the first row with DataFrame.eq...

print (df.eq(df.iloc[0]))
       A     B     C      D
0   True  True  True   True
1  False  True  True  False
2  False  True  True  False

... then check each column is all Trues with DataFrame.all...

print (df.eq(df.iloc[0]).all())
A    False
B     True
C     True
D    False
dtype: bool

... finally filter columns' names for which result is True:

print (df.columns[df.eq(df.iloc[0]).all()])
Index(['B', 'C'], dtype='object')

Timings:

np.random.seed(100)
df = pd.DataFrame(np.random.randint(10, size=(1000,100)))

df[np.random.randint(100, size=20)] = 100
print (df)

# Solution 1 (second-fastest):
In [243]: %timeit ([c for c in df.columns if len(set(df[c])) == 1])
3.59 ms ± 43.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# Solution 2 (fastest):
In [244]: %timeit df.columns[df.eq(df.iloc[0]).all()].tolist()
1.62 ms ± 13.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#Mohamed Thasin ah solution
In [245]: %timeit ([col for col in df.columns if len(df[col].unique())==1])
6.8 ms ± 352 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

#jpp solution
In [246]: %%timeit
     ...: vals = df.apply(set, axis=0)
     ...: res = vals[vals.map(len) == 1].index
     ...: 
5.59 ms ± 64.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

#smci solution 1
In [275]: %timeit df.columns[ df.nunique()==1 ]
11 ms ± 105 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

#smci solution 2
In [276]: %timeit [col for col in df.columns if not df[col].is_unique]
9.25 ms ± 80 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

#smci solution 3
In [277]: %timeit df.columns[ df.apply(lambda col: not col.is_unique) ]
11.1 ms ± 511 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)