I have this DataFrame (df1
) in Pandas:
df1 = pd.DataFrame(np.random.rand(10,4),columns=list('ABCD')) print df1 A B C D 0.860379 0.726956 0.394529 0.833217 0.014180 0.813828 0.559891 0.339647 0.782838 0.698993 0.551252 0.361034 0.833370 0.982056 0.741821 0.006864 0.855955 0.546562 0.270425 0.136006 0.491538 0.445024 0.971603 0.690001 0.911696 0.065338 0.796946 0.853456 0.744923 0.545661 0.492739 0.337628 0.576235 0.219831 0.946772 0.752403 0.164873 0.454862 0.745890 0.437729
I would like to check if any row (all columns) from another dataframe (df2
) are present in df1
. Here is df2
:
df2 = df1.ix[4:8] df2.reset_index(drop=True,inplace=True) df2.loc[-1] = [2, 3, 4, 5] df2.loc[-2] = [14, 15, 16, 17] df2.reset_index(drop=True,inplace=True) print df2 A B C D 0.855955 0.546562 0.270425 0.136006 0.491538 0.445024 0.971603 0.690001 0.911696 0.065338 0.796946 0.853456 0.744923 0.545661 0.492739 0.337628 0.576235 0.219831 0.946772 0.752403 2.000000 3.000000 4.000000 5.000000 14.000000 15.000000 16.000000 17.000000
I tried using df.lookup
to search for one row at a time. I did it this way:
list1 = df2.ix[0].tolist() cols = df1.columns.tolist() print df1.lookup(list1, cols)
but I got this error message:
File "C:\Users\test.py", line 19, in <module> print df1.lookup(list1, cols) File "C:\python27\lib\site-packages\pandas\core\frame.py", line 2217, in lookup raise KeyError('One or more row labels was not found') KeyError: 'One or more row labels was not found'
I also tried .all()
using:
print (df2 == df1).all(1).any()
but I got this error message:
File "C:\Users\test.py", line 12, in <module> print (df2 == df1).all(1).any() File "C:\python27\lib\site-packages\pandas\core\ops.py", line 884, in f return self._compare_frame(other, func, str_rep) File "C:\python27\lib\site-packages\pandas\core\frame.py", line 3010, in _compare_frame raise ValueError('Can only compare identically-labeled ' ValueError: Can only compare identically-labeled DataFrame objects
I also tried isin()
like this:
print df2.isin(df1)
but I got False
everywhere, which is not correct:
A B C D False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False False
Is it possible to search for a set of rows in a DataFrame, by comparing it to another dataframe's rows?
EDIT: Is is possible to drop df2
rows if those rows are also present in df1
?
To find the common rows between two DataFrames with merge(), use the parameter “how” as “inner” since it works like SQL Inner Join and this is what we want to achieve.
You can use the DataFrame. diff() function to find the difference between two rows in a pandas DataFrame. where: periods: The number of previous rows for calculating the difference.
One possible solution to your problem would be to use merge. Checking if any row (all columns) from another dataframe (df2) are present in df1 is equivalent to determining the intersection of the the two dataframes. This can be accomplished using the following function:
pd.merge(df1, df2, on=['A', 'B', 'C', 'D'], how='inner')
For example, if df1 was
A B C D 0 0.403846 0.312230 0.209882 0.397923 1 0.934957 0.731730 0.484712 0.734747 2 0.588245 0.961589 0.910292 0.382072 3 0.534226 0.276908 0.323282 0.629398 4 0.259533 0.277465 0.043652 0.925743 5 0.667415 0.051182 0.928655 0.737673 6 0.217923 0.665446 0.224268 0.772592 7 0.023578 0.561884 0.615515 0.362084 8 0.346373 0.375366 0.083003 0.663622 9 0.352584 0.103263 0.661686 0.246862
and df2 was defined as:
A B C D 0 0.259533 0.277465 0.043652 0.925743 1 0.667415 0.051182 0.928655 0.737673 2 0.217923 0.665446 0.224268 0.772592 3 0.023578 0.561884 0.615515 0.362084 4 0.346373 0.375366 0.083003 0.663622 5 2.000000 3.000000 4.000000 5.000000 6 14.000000 15.000000 16.000000 17.000000
The function pd.merge(df1, df2, on=['A', 'B', 'C', 'D'], how='inner')
produces:
A B C D 0 0.259533 0.277465 0.043652 0.925743 1 0.667415 0.051182 0.928655 0.737673 2 0.217923 0.665446 0.224268 0.772592 3 0.023578 0.561884 0.615515 0.362084 4 0.346373 0.375366 0.083003 0.663622
The results are all of the rows (all columns) that are both in df1 and df2.
We can also modify this example if the columns are not the same in df1 and df2 and just compare the row values that are the same for a subset of the columns. If we modify the original example:
df1 = pd.DataFrame(np.random.rand(10,4),columns=list('ABCD')) df2 = df1.ix[4:8] df2.reset_index(drop=True,inplace=True) df2.loc[-1] = [2, 3, 4, 5] df2.loc[-2] = [14, 15, 16, 17] df2.reset_index(drop=True,inplace=True) df2 = df2[['A', 'B', 'C']] # df2 has only columns A B C
Then we can look at the common columns using common_cols = list(set(df1.columns) & set(df2.columns))
between the two dataframes then merge:
pd.merge(df1, df2, on=common_cols, how='inner')
EDIT: New question (comments), having identified the rows from df2 that were also present in the first dataframe (df1), is it possible to take the result of the pd.merge() and to then drop the rows from df2 that are also present in df1
I do not know of a straightforward way to accomplish the task of dropping the rows from df2 that are also present in df1. That said, you could use the following:
ds1 = set(tuple(line) for line in df1.values) ds2 = set(tuple(line) for line in df2.values) df = pd.DataFrame(list(ds2.difference(ds1)), columns=df2.columns)
There probably exists a better way to accomplish that task but i am unaware of such a method / function.
EDIT 2: How to drop the rows from df2 that are also present in df1 as shown in @WR answer.
The method provided df2[~df2['A'].isin(df12['A'])]
does not account for all types of situations. Consider the following DataFrames:
df1:
A B C D 0 6 4 1 6 1 7 6 6 8 2 1 6 2 7 3 8 0 4 1 4 1 0 2 3 5 8 4 7 5 6 4 7 1 1 7 3 7 3 4 8 5 2 8 8 9 3 2 8 4
df2:
A B C D 0 1 0 2 3 1 8 4 7 5 2 4 7 1 1 3 3 7 3 4 4 5 2 8 8 5 1 1 1 1 6 2 2 2 2
df12:
A B C D 0 1 0 2 3 1 8 4 7 5 2 4 7 1 1 3 3 7 3 4 4 5 2 8 8
Using the above DataFrames with the goal of dropping rows from df2 that are also present in df1 would result in the following:
A B C D 0 1 1 1 1 1 2 2 2 2
Rows (1, 1, 1, 1) and (2, 2, 2, 2) are in df2 and not in df1. Unfortunately, using the provided method (df2[~df2['A'].isin(df12['A'])]
) results in:
A B C D 6 2 2 2 2
This occurs because the value of 1 in column A is found in both the intersection DataFrame (i.e. (1, 0, 2, 3)) and df2 and thus removes both (1, 0, 2, 3) and (1, 1, 1, 1). This is unintended since the row (1, 1, 1, 1) is not in df1 and should not be removed.
I think the following will provide a solution. It creates a dummy column that is later used to subset the DataFrame to the desired results:
df12['key'] = 'x' temp_df = pd.merge(df2, df12, on=df2.columns.tolist(), how='left') temp_df[temp_df['key'].isnull()].drop('key', axis=1)
@Andrew: I believe I found a way to drop the rows of one dataframe that are already present in another (i.e. to answer my EDIT) without using loops - let me know if you disagree and/or if my OP + EDIT did not clearly state this:
THIS WORKS
The columns for both dataframes are always the same - A
, B
, C
and D
. With this in mind, based heavily on Andrew's approach, here is how to drop the rows from df2
that are also present in df1
:
common_cols = df1.columns.tolist() #generate list of column names df12 = pd.merge(df1, df2, on=common_cols, how='inner') #extract common rows with merge df2 = df2[~df2['A'].isin(df12['A'])]
Line 3 does the following:
df2
that do not match rows in df1
:A
to make this comparison - it isNOTE: this method is essentially the equivalent of the SQL NOT IN()
.
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