pandas - 按行元素通过另一个数据框过滤数据框

Question

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

Answer 1

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 (ie 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.

Answer 2

这非常简洁并且效果很好：

df1 = df1[~df1.index.isin(df2.index)]

Answer 3

Using DataFrame.merge & DataFrame.query :

A more elegant method would be to do left join with the argument indicator=True , then filter all the rows which are left_only with query :

d = (
    df1.merge(df2, 
              on=['c', 'l'],
              how='left', 
              indicator=True)
    .query('_merge == "left_only"')
    .drop(columns='_merge')
)

print(d)
   c  k  l
0  A  1  a
2  B  2  a
4  C  2  d

indicator=True returns a dataframe with an extra column _merge which marks each row left_only, both, right_only :

df1.merge(df2, on=['c', 'l'], how='left', indicator=True)

   c  k  l     _merge
0  A  1  a  left_only
1  A  2  b       both
2  B  2  a  left_only
3  C  2  a       both
4  C  2  d  left_only

Answer 4

I think this is a quite simple approach when you want to filter a dataframe based on multiple columns from another dataframe or even based on a custom list.

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']})

#values of df2 columns 'c' and 'l' that will be used to filter df1
idxs = list(zip(df2.c.values, df2.l.values)) #[('A', 'b'), ('C', 'a')]

#so df1 is filtered based on the values present in columns c and l of df2 (idxs)
df1 = df1[pd.Series(list(zip(df1.c, df1.l)), index=df1.index).isin(idxs)]

Answer 5

How about:

df1['key'] = df1['c'] + df1['l']
d = df1[~df1['key'].isin(df2['c'] + df2['l'])].drop(['key'], axis=1)

Answer 6

Another option that avoids creating an extra column or doing a merge would be to do a groupby on df2 to get the distinct (c, l) pairs and then just filter df1 using that.

gb = df2.groupby(("c", "l")).groups
df1[[p not in gb for p in zip(df1['c'], df1['l'])]]]

For this small example, it actually seems to run a bit faster than the pandas-based approach (666 µs vs. 1.76 ms on my machine), but I suspect it could be slower on larger examples since it's dropping into pure Python.

Answer 7

You can concatenate both DataFrames and drop all duplicates:

df1.append(df2).drop_duplicates(subset=['c', 'l'], keep=False)

Output:

   c    k  l
0  A  1.0  a
2  B  2.0  a
4  C  2.0  d

This method doesn't work if you have duplicates subset=['c', 'l'] in df1 .