First of all, next time please give an example dataset, not an image or copy of one. It's best to give a minimal example that reproduces your problem (it's also a good way to investigate your problem). This df, for example, will do the trick:
df = pd.DataFrame.from_dict({
'a':[1,0,0,0,0,0,0,0,0,0,0],
'b':[1,1,1,0,1,0,0,0,0,0,0]})
Now, the previous answers help, but if you can avoid a loop, it's preferable. You can write something simpler and more concise that will do the trick:
df.drop(columns=df.columns[df.eq(0).mean()>0.9])
Let's go through it step by step:
The df.eq(0)
returns True
\\ False
in each cell.
The .mean()
method treats True as 1 and False as 0, so comparing that mean to 0.9 is what you want.
Calling df.columns[...]
at these places will return only those where the >0.9
holds, and drop
just drops them.
The following should do the trick for you:
row_count = df.shape[0]
columns_to_drop = []
for column, count in df.apply(lambda column: (column == 0).sum()).iteritems():
if count / row_count >= 0.9:
columns_to_drop.append(column)
df = df.drop(columns_to_drop, axis=1, inplace=True)
bad_col = []
for i, x in enumerate(df.columns):
if sorted(list(df[x].value_counts(normalize = True).values))[-1] >= 0.9 :
bad_col.append(x)
Explanation inline the code .
#Suppose df is your DataFrame then execute the following code.
df_float=df.loc[:, df.dtypes == np.float64] #checks if the column contains numbers
for i in df_float.columns:
if ((len(df_float[i].loc[df_float[i]==0])/len(df_float))>0.9): #checking if 90% data is zero
df_float.drop(i,axis=1,inplace=True) #delete the column
#Your results are stored in df_float
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