How do I delete a column that contains only zeros in Pandas?

Question

I currently have a dataframe consisting of columns with 1\'s and 0\'s as values, I would like to iterate through the columns and delete the ones that are made up of only 0\'

Answer 1

In case you'd like a more expressive way of getting the zero-column names so you can print / log them, and drop them, in-place, by their names:

zero_cols = [ col for col, is_zero in ((df == 0).sum() == df.shape[0]).items() if is_zero ]
df.drop(zero_cols, axis=1, inplace=True)

Some break down:

# a pandas Series with {col: is_zero} items
# is_zero is True when the number of zero items in that column == num_all_rows
(df == 0).sum() == df.shape[0])

# a list comprehension of zero_col_names is built from the_series
[ col for col, is_zero in the_series.items() if is_zero ]

Answer 2

df.loc[:, (df != 0).any(axis=0)]

---

Here is a break-down of how it works:

In [74]: import pandas as pd

In [75]: df = pd.DataFrame([[1,0,0,0], [0,0,1,0]])

In [76]: df
Out[76]: 
   0  1  2  3
0  1  0  0  0
1  0  0  1  0

[2 rows x 4 columns]

df != 0 creates a boolean DataFrame which is True where df is nonzero:

In [77]: df != 0
Out[77]: 
       0      1      2      3
0   True  False  False  False
1  False  False   True  False

[2 rows x 4 columns]

(df != 0).any(axis=0) returns a boolean Series indicating which columns have nonzero entries. (The any operation aggregates values along the 0-axis -- i.e. along the rows -- into a single boolean value. Hence the result is one boolean value for each column.)

In [78]: (df != 0).any(axis=0)
Out[78]: 
0     True
1    False
2     True
3    False
dtype: bool

And df.loc can be used to select those columns:

In [79]: df.loc[:, (df != 0).any(axis=0)]
Out[79]: 
   0  2
0  1  0
1  0  1

[2 rows x 2 columns]

---

To "delete" the zero-columns, reassign df:

df = df.loc[:, (df != 0).any(axis=0)]

Answer 3

Here is an alternative way to use is

df.replace(0,np.nan).dropna(axis=1,how="all")

Compared with the solution of unutbu, this way is obviously slower:

%timeit df.loc[:, (df != 0).any(axis=0)]
652 µs ± 5.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df.replace(0,np.nan).dropna(axis=1,how="all")
1.75 ms ± 9.49 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)