I have a dataframe df1
which looks like:
c k l 0 A 1 a 1 A 2 b 2 B 2 a 3 C 2 a 4 C 2 d
and another called df2
like:
c l 0 A b 1 C a
I would like to filter df1
keeping only the values that ARE NOT in df2
. Values to filter are expected to be as (A,b)
and (C,a)
tuples. So far I tried to apply the isin
method:
d = df[~(df['l'].isin(dfc['l']) & df['c'].isin(dfc['c']))]
That seems to me too complicated, it returns:
c k l 2 B 2 a 4 C 2 d
but I'm expecting:
c k l 0 A 1 a 2 B 2 a 4 C 2 d
Create another data frame using the random() function and randomly selecting the rows of the first dataset. Now we will use dataframe. loc[] function to select the row values of the first data frame using the indexes of the second data frame.
Filter Rows by Condition You can use df[df["Courses"] == 'Spark'] to filter rows by a condition in pandas DataFrame. Not that this expression returns a new DataFrame with selected rows.
You can do this efficiently using isin
on a multiindex constructed from the desired columns:
df1 = pd.DataFrame({'c': ['A', 'A', 'B', 'C', 'C'], 'k': [1, 2, 2, 2, 2], 'l': ['a', 'b', 'a', 'a', 'd']}) df2 = pd.DataFrame({'c': ['A', 'C'], 'l': ['b', 'a']}) keys = list(df2.columns.values) i1 = df1.set_index(keys).index i2 = df2.set_index(keys).index df1[~i1.isin(i2)]
I think this improves on @IanS's similar solution because it doesn't assume any column type (i.e. it will work with numbers as well as strings).
(Above answer is an edit. Following was my initial answer)
Interesting! This is something I haven't come across before... I would probably solve it by merging the two arrays, then dropping rows where df2
is defined. Here is an example, which makes use of a temporary array:
df1 = pd.DataFrame({'c': ['A', 'A', 'B', 'C', 'C'], 'k': [1, 2, 2, 2, 2], 'l': ['a', 'b', 'a', 'a', 'd']}) df2 = pd.DataFrame({'c': ['A', 'C'], 'l': ['b', 'a']}) # create a column marking df2 values df2['marker'] = 1 # join the two, keeping all of df1's indices joined = pd.merge(df1, df2, on=['c', 'l'], how='left') joined
# extract desired columns where marker is NaN joined[pd.isnull(joined['marker'])][df1.columns]
There may be a way to do this without using the temporary array, but I can't think of one. As long as your data isn't huge the above method should be a fast and sufficient answer.
This is pretty succinct and works well:
df1 = df1[~df1.index.isin(df2.index)]
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