quickly drop dataframe columns with only one distinct value

One step:

df = df[[c for c
        in list(df)
        if len(df[c].unique()) > 1]]

Two steps:

Create a list of column names that have more than 1 distinct value.

keep = [c for c
        in list(df)
        if len(df[c].unique()) > 1]

Drop the columns that are not in 'keep'

df = df[keep]

Note: this step can also be done using a list of columns to drop:

drop_cols = [c for c
             in list(df)
             if df[c].nunique() <= 1]
df = df.drop(columns=drop_cols)

You can use Series.unique() method to find out all the unique elements in a column, and for columns whose .unique() returns only 1 element, you can drop that. Example -

for col in df.columns:
    if len(df[col].unique()) == 1:
        df.drop(col,inplace=True,axis=1)

A method that does not do inplace dropping -

res = df
for col in df.columns:
    if len(df[col].unique()) == 1:
        res = res.drop(col,axis=1)

---

Demo -

In [154]: df = pd.DataFrame([[1,2,3],[1,3,3],[1,2,3]])

In [155]: for col in df.columns:
   .....:     if len(df[col].unique()) == 1:
   .....:         df.drop(col,inplace=True,axis=1)
   .....:

In [156]: df
Out[156]:
   1
0  2
1  3
2  2

---

Timing results -

In [166]: %paste
def func1(df):
        res = df
        for col in df.columns:
                if len(df[col].unique()) == 1:
                        res = res.drop(col,axis=1)
        return res

## -- End pasted text --

In [172]: df = pd.DataFrame({'a':1, 'b':np.arange(5), 'c':[0,0,2,2,2]})

In [178]: %timeit func1(df)
1000 loops, best of 3: 1.05 ms per loop

In [180]: %timeit df[df.apply(pd.Series.value_counts).dropna(thresh=2, axis=1).columns]
100 loops, best of 3: 8.81 ms per loop

In [181]: %timeit df.apply(pd.Series.value_counts).dropna(thresh=2, axis=1)
100 loops, best of 3: 5.81 ms per loop

The fastest method still seems to be the method using unique and looping through the columns.