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's. Here's what I have tried so far:
ones = []
zeros = []
for year in years:
for i in range(0,599):
if year[str(i)].values.any() == 1:
ones.append(i)
if year[str(i)].values.all() == 0:
zeros.append(i)
for j in ones:
if j in zeros:
zeros.remove(j)
for q in zeros:
del year[str(q)]
In which years is a list of dataframes for the various years I am analyzing, ones consists of columns with a one in them and zeros is a list of columns containing all zeros. Is there a better way to delete a column based on a condition? For some reason I have to check whether the ones columns are in the zeros list as well and remove them from the zeros list to obtain a list of all the zero columns.
ベストアンサー1
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)]