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I have very junior question in python - i have a dataframe with a column containing some IDs and separate dataframe that contains 2 columns, out of which 1 is an array:
df1 = pd.DataFrame({"some_id": [1, 2, 3, 4, 5]})
df2 = pd.DataFrame([["A", [1, 2]], ["B", [3, 4]], ["C", [5]]], columns=['letter', 'some_ids'])
I want to add do df1 new column "letter' that for a given "some_id" will look up df2, check if this id is in df2['some_ids'] and return df2['letter']
I tried this:
df1['letter'] = df2[df1[some_id].isin(df2['some_ids')].letter
and get NaNs - any suggestion where I make mistake?
Create dictionary with flatten nested lists in dict comprehension and then use Series.map:
d = {x: a for a,b in zip(df2['letter'], df2['some_ids']) for x in b}
df1['letter'] = df1['some_id'].map(d)
Or mapping by Series created by DataFrame.explode with DataFrame.set_index:
df1['letter'] = df1['some_id'].map(df2.explode('some_ids').set_index('some_ids')['letter'])
Or use left join with rename column:
df1 = df1.merge(df2.explode('some_ids').rename(columns={'some_ids':'some_id'}), how='left')
print (df1)
some_id letter
0 1 A
1 2 A
2 3 B
3 4 B
4 5 C
A similar question was asked here Pandas merge on multiple columns ignoring NaN but without answer, so I'll ask maybe someone can help.
I need to merge values from df2 into df1, but the key used in the merge differs between rows in df2, as the rows in df2 have NaNs in different columns, and in that case I want to ignore those columns, and use for each row only the columns that have values.
df1 = pd.DataFrame([[0, 1, 3], [0, 2, 4], [1, 2, 5]], columns=['level1', 'level2', 'level3'])
df1
level1 level2 level3
0 0 1 3
1 0 2 4
2 1 2 5
df2 = pd.DataFrame([[0, None, None, 10], [0, 1, None, 12], [None, 2, 5, 13]], columns=['level1', 'level2', 'level3', 'value'])
df2
level1 level2 level3 value
0 0.0 NaN NaN 10
1 0.0 1.0 NaN 12
2 NaN 2.0 5.0 13
When I do df1.merge(df2, how='left'), I get df1 with NaN in the value column, since there is no match on all the level columns as pandas is trying to match the Nan values as well.
What I do want is to get a match for any rows in df2 without trying to match the NaNs:
level1 level2 level3 value
0 0 1 3 10
1 0 1 3 12
2 0 2 4 10
3 1 2 5 13
Explanation:
Row 0 in df1 has a match on the non-NaN columns of rows 0 and 1 in df2, so it gets values 10 and 12 from there. Row 1 in df1 has a match on the non-NaN columns of row 0 in df2, so it gets value 12 from there. Row 2 in df1 has a match on the non-NaN columns of row 2 in df2, so it gets value 13 from there.
In the real data I actually have 6 level columns and the non-NaN columns for each row in df2 can be any combination or a single column from there.
What I do now is to iterrows the rows in df2, create for each one a mini-dataframe of only the non-NaN columns, and merge df1 with it. But as we know, it's not really efficient, and I wonder it there something better that can be done.
I think I figured out a vectorized solution.
Fundamentally, the idea is that you merge df1 with df2 three separate times for the number of levels and then concat the dataframes together into one.
From there, you count how many columns are null and subtract from the number of levels. This tells you how many duplicates (or matches) are required in order to prevent the data from being dropped later.
Then, you calculate how many matches or duplicates there actually are. If actual is the same as required, then that means the row is a match, and it gets kept in the dataframe.
It's not pretty, but to improve my answer you could create a merging function to cut some of the code. Most importantly, it should be highly performant compared to looping through every row. As a final note, for the duplicates_required helper column, you will need to change the 3 to a 6 since you have 6 columns in your actual dataset and you will obviously need to repeat some of my merging code:
df1 = pd.DataFrame([[0, 1, 3], [0, 2, 4], [1, 2, 5]], columns=['level1', 'level2', 'level3'])
df2 = pd.DataFrame([[0, None, None, 10], [0, 1, None, 12], [None, 2, 5, 13]], columns=['level1', 'level2', 'level3', 'value'])
df2 = df2.assign(duplicates_required = 3 - df2.isnull().sum(axis=1))
df = pd.concat([
df1.merge(df2[['level1','value', 'duplicates_required']], on='level1'),
df1.merge(df2[['level2','value', 'duplicates_required']], on='level2'),
df1.merge(df2[['level3','value', 'duplicates_required']], on='level3')
])
cols = ['level1', 'level2', 'level3', 'value']
df['actual_duplicates'] = df.groupby(cols)['value'].transform('size')
df = (df[df['duplicates_required'].eq(df['actual_duplicates'])]
.drop_duplicates(subset=cols)
.drop(['duplicates_required', 'actual_duplicates'], axis=1)
.reset_index(drop=True))
df
Out[1]:
level1 level2 level3 value
0 0 1 3 10
1 0 1 3 12
2 0 2 4 10
3 1 2 5 13
I think this works better than my previous answer using regex. Similar process, but a bit simpler to understand.
Do a full merge of the two dataframes
Compare across levels and count number of mismatches
Filter to rows where mismatch count == 0
import pandas as pd
df1 = pd.DataFrame([[0, 1, 3], [0, 2, 4], [1, 2, 5]], columns=['level1', 'level2', 'level3'])
df2 = pd.DataFrame([[0, None, None, 10], [0, 1, None, 12], [None, 2, 5, 13]], columns=['level1', 'level2', 'level3', 'value'])
levels_to_match = ['level1','level2','level3']
levels_to_match_df2 = [level + '_df2' for level in levels_to_match]
for df in [df1,df2]:
df['temp'] = 1
df1 = df1.merge(df2, on='temp', suffixes=[None,'_df2']).drop(columns=['temp'])
df1['mismatch'] = df1.apply(lambda x:
sum([(1 - (y == z or pd.isna(z))) for y, z in zip(list(x[levels_to_match]), list(x[levels_to_match_df2]))]),
axis=1)
df1 = df1.loc[df1['mismatch'] == 0, :].drop(columns=['mismatch'] + levels_to_match_df2)
print(df1)
level1 level2 level3 value
0 0 1 3 10
1 0 1 3 12
3 0 2 4 10
8 1 2 5 13
Old answer with regex
Probably not ideal, but maybe try converting your levels into strings and regex expressions, then do a full merge of all possible combinations, and finally filter using a regex search/match across two helper columns (one from df1 and the other from df2).
Assuming the data you're matching on are either int or NaN then this seems to work okay. If you have other data types in your real data then the string/regex transformations will need to be adjusted accordingly.
import pandas as pd
import re
df1 = pd.DataFrame([[0, 1, 3], [0, 2, 4], [1, 2, 5]], columns=['level1', 'level2', 'level3'])
df2 = pd.DataFrame([[0, None, None, 10], [0, 1, None, 12], [None, 2, 5, 13]], columns=['level1', 'level2', 'level3', 'value'])
levels_to_match = ['level1','level2','level3']
for df in [df1,df2]:
df['helper'] = df[levels_to_match].apply(list, axis=1)
df['helper'] = df['helper'].apply(lambda x: ','.join([str(int(item)) if pd.notna(item) else '.*' for item in x]))
df['temp'] = 1
df1 = df1.merge(df2.drop(columns=levels_to_match), on='temp', suffixes=[None,'_df2']).drop(columns=['temp'])
df1['match'] = df1.apply(lambda x: re.search(x['helper_df2'], x['helper']) is not None, axis=1)
df1 = df1.loc[df1['match'], :].drop(columns=['helper','helper_df2','match'])
print(df1)
level1 level2 level3 value
0 0 1 3 10
1 0 1 3 12
3 0 2 4 10
8 1 2 5 13
I've two pandas data frames that have some rows in common.
Suppose dataframe2 is a subset of dataframe1.
How can I get the rows of dataframe1 which are not in dataframe2?
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5], 'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3], 'col2' : [10, 11, 12]})
df1
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
df2
col1 col2
0 1 10
1 2 11
2 3 12
Expected result:
col1 col2
3 4 13
4 5 14
The currently selected solution produces incorrect results. To correctly solve this problem, we can perform a left-join from df1 to df2, making sure to first get just the unique rows for df2.
First, we need to modify the original DataFrame to add the row with data [3, 10].
df1 = pd.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3],
'col2' : [10, 11, 12, 13, 14, 10]})
df2 = pd.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
df1
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
5 3 10
df2
col1 col2
0 1 10
1 2 11
2 3 12
Perform a left-join, eliminating duplicates in df2 so that each row of df1 joins with exactly 1 row of df2. Use the parameter indicator to return an extra column indicating which table the row was from.
df_all = df1.merge(df2.drop_duplicates(), on=['col1','col2'],
how='left', indicator=True)
df_all
col1 col2 _merge
0 1 10 both
1 2 11 both
2 3 12 both
3 4 13 left_only
4 5 14 left_only
5 3 10 left_only
Create a boolean condition:
df_all['_merge'] == 'left_only'
0 False
1 False
2 False
3 True
4 True
5 True
Name: _merge, dtype: bool
Why other solutions are wrong
A few solutions make the same mistake - they only check that each value is independently in each column, not together in the same row. Adding the last row, which is unique but has the values from both columns from df2 exposes the mistake:
common = df1.merge(df2,on=['col1','col2'])
(~df1.col1.isin(common.col1))&(~df1.col2.isin(common.col2))
0 False
1 False
2 False
3 True
4 True
5 False
dtype: bool
This solution gets the same wrong result:
df1.isin(df2.to_dict('l')).all(1)
One method would be to store the result of an inner merge form both dfs, then we can simply select the rows when one column's values are not in this common:
In [119]:
common = df1.merge(df2,on=['col1','col2'])
print(common)
df1[(~df1.col1.isin(common.col1))&(~df1.col2.isin(common.col2))]
col1 col2
0 1 10
1 2 11
2 3 12
Out[119]:
col1 col2
3 4 13
4 5 14
EDIT
Another method as you've found is to use isin which will produce NaN rows which you can drop:
In [138]:
df1[~df1.isin(df2)].dropna()
Out[138]:
col1 col2
3 4 13
4 5 14
However if df2 does not start rows in the same manner then this won't work:
df2 = pd.DataFrame(data = {'col1' : [2, 3,4], 'col2' : [11, 12,13]})
will produce the entire df:
In [140]:
df1[~df1.isin(df2)].dropna()
Out[140]:
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
Assuming that the indexes are consistent in the dataframes (not taking into account the actual col values):
df1[~df1.index.isin(df2.index)]
As already hinted at, isin requires columns and indices to be the same for a match. If match should only be on row contents, one way to get the mask for filtering the rows present is to convert the rows to a (Multi)Index:
In [77]: df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3], 'col2' : [10, 11, 12, 13, 14, 10]})
In [78]: df2 = pandas.DataFrame(data = {'col1' : [1, 3, 4], 'col2' : [10, 12, 13]})
In [79]: df1.loc[~df1.set_index(list(df1.columns)).index.isin(df2.set_index(list(df2.columns)).index)]
Out[79]:
col1 col2
1 2 11
4 5 14
5 3 10
If index should be taken into account, set_index has keyword argument append to append columns to existing index. If columns do not line up, list(df.columns) can be replaced with column specifications to align the data.
pandas.MultiIndex.from_tuples(df<N>.to_records(index = False).tolist())
could alternatively be used to create the indices, though I doubt this is more efficient.
Suppose you have two dataframes, df_1 and df_2 having multiple fields(column_names) and you want to find the only those entries in df_1 that are not in df_2 on the basis of some fields(e.g. fields_x, fields_y), follow the following steps.
Step1.Add a column key1 and key2 to df_1 and df_2 respectively.
Step2.Merge the dataframes as shown below. field_x and field_y are our desired columns.
Step3.Select only those rows from df_1 where key1 is not equal to key2.
Step4.Drop key1 and key2.
This method will solve your problem and works fast even with big data sets. I have tried it for dataframes with more than 1,000,000 rows.
df_1['key1'] = 1
df_2['key2'] = 1
df_1 = pd.merge(df_1, df_2, on=['field_x', 'field_y'], how = 'left')
df_1 = df_1[~(df_1.key2 == df_1.key1)]
df_1 = df_1.drop(['key1','key2'], axis=1)
a bit late, but it might be worth checking the "indicator" parameter of pd.merge.
See this other question for an example:
Compare PandaS DataFrames and return rows that are missing from the first one
This is the best way to do it:
df = df1.drop_duplicates().merge(df2.drop_duplicates(), on=df2.columns.to_list(),
how='left', indicator=True)
df.loc[df._merge=='left_only',df.columns!='_merge']
Note that drop duplicated is used to minimize the comparisons. It would work without them as well. The best way is to compare the row contents themselves and not the index or one/two columns and same code can be used for other filters like 'both' and 'right_only' as well to achieve similar results. For this syntax dataframes can have any number of columns and even different indices. Only the columns should occur in both the dataframes.
Why this is the best way?
index.difference only works for unique index based comparisons
pandas.concat() coupled with drop_duplicated() is not ideal because it will also get rid of the rows which may be only in the dataframe you want to keep and are duplicated for valid reasons.
I think those answers containing merging are extremely slow. Therefore I would suggest another way of getting those rows which are different between the two dataframes:
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5], 'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3], 'col2' : [10, 11, 12]})
DISCLAIMER: My solution works if you're interested in one specific column where the two dataframes differ. If you are interested only in those rows, where all columns are equal do not use this approach.
Let's say, col1 is a kind of ID, and you only want to get those rows, which are not contained in both dataframes:
ids_in_df2 = df2.col1.unique()
not_found_ids = df[~df['col1'].isin(ids_in_df2 )]
And that's it. You get a dataframe containing only those rows where col1 isn't appearent in both dataframes.
You can also concat df1, df2:
x = pd.concat([df1, df2])
and then remove all duplicates:
y = x.drop_duplicates(keep=False, inplace=False)
I have an easier way in 2 simple steps:
As the OP mentioned Suppose dataframe2 is a subset of dataframe1, columns in the 2 dataframes are the same,
df1 = pd.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3],
'col2' : [10, 11, 12, 13, 14, 10]})
df2 = pd.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
### Step 1: just append the 2nd df at the end of the 1st df
df_both = df1.append(df2)
### Step 2: drop rows which contain duplicates, Drop all duplicates.
df_dif = df_both.drop_duplicates(keep=False)
## mission accompliched!
df_dif
Out[20]:
col1 col2
3 4 13
4 5 14
5 3 10
you can do it using isin(dict) method:
In [74]: df1[~df1.isin(df2.to_dict('l')).all(1)]
Out[74]:
col1 col2
3 4 13
4 5 14
Explanation:
In [75]: df2.to_dict('l')
Out[75]: {'col1': [1, 2, 3], 'col2': [10, 11, 12]}
In [76]: df1.isin(df2.to_dict('l'))
Out[76]:
col1 col2
0 True True
1 True True
2 True True
3 False False
4 False False
In [77]: df1.isin(df2.to_dict('l')).all(1)
Out[77]:
0 True
1 True
2 True
3 False
4 False
dtype: bool
Here is another way of solving this:
df1[~df1.index.isin(df1.merge(df2, how='inner', on=['col1', 'col2']).index)]
Or:
df1.loc[df1.index.difference(df1.merge(df2, how='inner', on=['col1', 'col2']).index)]
extract the dissimilar rows using the merge function
df = df1.merge(df2.drop_duplicates(), on=['col1','col2'],
how='left', indicator=True)
save the dissimilar rows in CSV
df[df['_merge'] == 'left_only'].to_csv('output.csv')
My way of doing this involves adding a new column that is unique to one dataframe and using this to choose whether to keep an entry
df2[col3] = 1
df1 = pd.merge(df_1, df_2, on=['field_x', 'field_y'], how = 'outer')
df1['Empt'].fillna(0, inplace=True)
This makes it so every entry in df1 has a code - 0 if it is unique to df1, 1 if it is in both dataFrames. You then use this to restrict to what you want
answer = nonuni[nonuni['Empt'] == 0]
How about this:
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5],
'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
records_df2 = set([tuple(row) for row in df2.values])
in_df2_mask = np.array([tuple(row) in records_df2 for row in df1.values])
result = df1[~in_df2_mask]
Easier, simpler and elegant
uncommon_indices = np.setdiff1d(df1.index.values, df2.index.values)
new_df = df1.loc[uncommon_indices,:]
pd.concat([df1, df2]).drop_duplicates(keep=False) will concatenate the two DataFrames together, and then drop all the duplicates, keeping only the unique rows. By default it will keep the first occurrence of the duplicate, but setting keep=False will drop all the duplicates.
Keep in mind that if you need to compare the DataFrames with columns with different names, you will have to make sure the columns have the same name before concatenating the dataframes.
Also, if the dataframes have a different order of columns, it will also affect the final result.
I've two pandas data frames that have some rows in common.
Suppose dataframe2 is a subset of dataframe1.
How can I get the rows of dataframe1 which are not in dataframe2?
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5], 'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3], 'col2' : [10, 11, 12]})
df1
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
df2
col1 col2
0 1 10
1 2 11
2 3 12
Expected result:
col1 col2
3 4 13
4 5 14
The currently selected solution produces incorrect results. To correctly solve this problem, we can perform a left-join from df1 to df2, making sure to first get just the unique rows for df2.
First, we need to modify the original DataFrame to add the row with data [3, 10].
df1 = pd.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3],
'col2' : [10, 11, 12, 13, 14, 10]})
df2 = pd.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
df1
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
5 3 10
df2
col1 col2
0 1 10
1 2 11
2 3 12
Perform a left-join, eliminating duplicates in df2 so that each row of df1 joins with exactly 1 row of df2. Use the parameter indicator to return an extra column indicating which table the row was from.
df_all = df1.merge(df2.drop_duplicates(), on=['col1','col2'],
how='left', indicator=True)
df_all
col1 col2 _merge
0 1 10 both
1 2 11 both
2 3 12 both
3 4 13 left_only
4 5 14 left_only
5 3 10 left_only
Create a boolean condition:
df_all['_merge'] == 'left_only'
0 False
1 False
2 False
3 True
4 True
5 True
Name: _merge, dtype: bool
Why other solutions are wrong
A few solutions make the same mistake - they only check that each value is independently in each column, not together in the same row. Adding the last row, which is unique but has the values from both columns from df2 exposes the mistake:
common = df1.merge(df2,on=['col1','col2'])
(~df1.col1.isin(common.col1))&(~df1.col2.isin(common.col2))
0 False
1 False
2 False
3 True
4 True
5 False
dtype: bool
This solution gets the same wrong result:
df1.isin(df2.to_dict('l')).all(1)
One method would be to store the result of an inner merge form both dfs, then we can simply select the rows when one column's values are not in this common:
In [119]:
common = df1.merge(df2,on=['col1','col2'])
print(common)
df1[(~df1.col1.isin(common.col1))&(~df1.col2.isin(common.col2))]
col1 col2
0 1 10
1 2 11
2 3 12
Out[119]:
col1 col2
3 4 13
4 5 14
EDIT
Another method as you've found is to use isin which will produce NaN rows which you can drop:
In [138]:
df1[~df1.isin(df2)].dropna()
Out[138]:
col1 col2
3 4 13
4 5 14
However if df2 does not start rows in the same manner then this won't work:
df2 = pd.DataFrame(data = {'col1' : [2, 3,4], 'col2' : [11, 12,13]})
will produce the entire df:
In [140]:
df1[~df1.isin(df2)].dropna()
Out[140]:
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
Assuming that the indexes are consistent in the dataframes (not taking into account the actual col values):
df1[~df1.index.isin(df2.index)]
As already hinted at, isin requires columns and indices to be the same for a match. If match should only be on row contents, one way to get the mask for filtering the rows present is to convert the rows to a (Multi)Index:
In [77]: df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3], 'col2' : [10, 11, 12, 13, 14, 10]})
In [78]: df2 = pandas.DataFrame(data = {'col1' : [1, 3, 4], 'col2' : [10, 12, 13]})
In [79]: df1.loc[~df1.set_index(list(df1.columns)).index.isin(df2.set_index(list(df2.columns)).index)]
Out[79]:
col1 col2
1 2 11
4 5 14
5 3 10
If index should be taken into account, set_index has keyword argument append to append columns to existing index. If columns do not line up, list(df.columns) can be replaced with column specifications to align the data.
pandas.MultiIndex.from_tuples(df<N>.to_records(index = False).tolist())
could alternatively be used to create the indices, though I doubt this is more efficient.
Suppose you have two dataframes, df_1 and df_2 having multiple fields(column_names) and you want to find the only those entries in df_1 that are not in df_2 on the basis of some fields(e.g. fields_x, fields_y), follow the following steps.
Step1.Add a column key1 and key2 to df_1 and df_2 respectively.
Step2.Merge the dataframes as shown below. field_x and field_y are our desired columns.
Step3.Select only those rows from df_1 where key1 is not equal to key2.
Step4.Drop key1 and key2.
This method will solve your problem and works fast even with big data sets. I have tried it for dataframes with more than 1,000,000 rows.
df_1['key1'] = 1
df_2['key2'] = 1
df_1 = pd.merge(df_1, df_2, on=['field_x', 'field_y'], how = 'left')
df_1 = df_1[~(df_1.key2 == df_1.key1)]
df_1 = df_1.drop(['key1','key2'], axis=1)
a bit late, but it might be worth checking the "indicator" parameter of pd.merge.
See this other question for an example:
Compare PandaS DataFrames and return rows that are missing from the first one
This is the best way to do it:
df = df1.drop_duplicates().merge(df2.drop_duplicates(), on=df2.columns.to_list(),
how='left', indicator=True)
df.loc[df._merge=='left_only',df.columns!='_merge']
Note that drop duplicated is used to minimize the comparisons. It would work without them as well. The best way is to compare the row contents themselves and not the index or one/two columns and same code can be used for other filters like 'both' and 'right_only' as well to achieve similar results. For this syntax dataframes can have any number of columns and even different indices. Only the columns should occur in both the dataframes.
Why this is the best way?
index.difference only works for unique index based comparisons
pandas.concat() coupled with drop_duplicated() is not ideal because it will also get rid of the rows which may be only in the dataframe you want to keep and are duplicated for valid reasons.
I think those answers containing merging are extremely slow. Therefore I would suggest another way of getting those rows which are different between the two dataframes:
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5], 'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3], 'col2' : [10, 11, 12]})
DISCLAIMER: My solution works if you're interested in one specific column where the two dataframes differ. If you are interested only in those rows, where all columns are equal do not use this approach.
Let's say, col1 is a kind of ID, and you only want to get those rows, which are not contained in both dataframes:
ids_in_df2 = df2.col1.unique()
not_found_ids = df[~df['col1'].isin(ids_in_df2 )]
And that's it. You get a dataframe containing only those rows where col1 isn't appearent in both dataframes.
You can also concat df1, df2:
x = pd.concat([df1, df2])
and then remove all duplicates:
y = x.drop_duplicates(keep=False, inplace=False)
I have an easier way in 2 simple steps:
As the OP mentioned Suppose dataframe2 is a subset of dataframe1, columns in the 2 dataframes are the same,
df1 = pd.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3],
'col2' : [10, 11, 12, 13, 14, 10]})
df2 = pd.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
### Step 1: just append the 2nd df at the end of the 1st df
df_both = df1.append(df2)
### Step 2: drop rows which contain duplicates, Drop all duplicates.
df_dif = df_both.drop_duplicates(keep=False)
## mission accompliched!
df_dif
Out[20]:
col1 col2
3 4 13
4 5 14
5 3 10
you can do it using isin(dict) method:
In [74]: df1[~df1.isin(df2.to_dict('l')).all(1)]
Out[74]:
col1 col2
3 4 13
4 5 14
Explanation:
In [75]: df2.to_dict('l')
Out[75]: {'col1': [1, 2, 3], 'col2': [10, 11, 12]}
In [76]: df1.isin(df2.to_dict('l'))
Out[76]:
col1 col2
0 True True
1 True True
2 True True
3 False False
4 False False
In [77]: df1.isin(df2.to_dict('l')).all(1)
Out[77]:
0 True
1 True
2 True
3 False
4 False
dtype: bool
Here is another way of solving this:
df1[~df1.index.isin(df1.merge(df2, how='inner', on=['col1', 'col2']).index)]
Or:
df1.loc[df1.index.difference(df1.merge(df2, how='inner', on=['col1', 'col2']).index)]
extract the dissimilar rows using the merge function
df = df1.merge(df2.drop_duplicates(), on=['col1','col2'],
how='left', indicator=True)
save the dissimilar rows in CSV
df[df['_merge'] == 'left_only'].to_csv('output.csv')
My way of doing this involves adding a new column that is unique to one dataframe and using this to choose whether to keep an entry
df2[col3] = 1
df1 = pd.merge(df_1, df_2, on=['field_x', 'field_y'], how = 'outer')
df1['Empt'].fillna(0, inplace=True)
This makes it so every entry in df1 has a code - 0 if it is unique to df1, 1 if it is in both dataFrames. You then use this to restrict to what you want
answer = nonuni[nonuni['Empt'] == 0]
How about this:
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5],
'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
records_df2 = set([tuple(row) for row in df2.values])
in_df2_mask = np.array([tuple(row) in records_df2 for row in df1.values])
result = df1[~in_df2_mask]
Easier, simpler and elegant
uncommon_indices = np.setdiff1d(df1.index.values, df2.index.values)
new_df = df1.loc[uncommon_indices,:]
pd.concat([df1, df2]).drop_duplicates(keep=False) will concatenate the two DataFrames together, and then drop all the duplicates, keeping only the unique rows. By default it will keep the first occurrence of the duplicate, but setting keep=False will drop all the duplicates.
Keep in mind that if you need to compare the DataFrames with columns with different names, you will have to make sure the columns have the same name before concatenating the dataframes.
Also, if the dataframes have a different order of columns, it will also affect the final result.
I've two pandas data frames that have some rows in common.
Suppose dataframe2 is a subset of dataframe1.
How can I get the rows of dataframe1 which are not in dataframe2?
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5], 'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3], 'col2' : [10, 11, 12]})
df1
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
df2
col1 col2
0 1 10
1 2 11
2 3 12
Expected result:
col1 col2
3 4 13
4 5 14
The currently selected solution produces incorrect results. To correctly solve this problem, we can perform a left-join from df1 to df2, making sure to first get just the unique rows for df2.
First, we need to modify the original DataFrame to add the row with data [3, 10].
df1 = pd.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3],
'col2' : [10, 11, 12, 13, 14, 10]})
df2 = pd.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
df1
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
5 3 10
df2
col1 col2
0 1 10
1 2 11
2 3 12
Perform a left-join, eliminating duplicates in df2 so that each row of df1 joins with exactly 1 row of df2. Use the parameter indicator to return an extra column indicating which table the row was from.
df_all = df1.merge(df2.drop_duplicates(), on=['col1','col2'],
how='left', indicator=True)
df_all
col1 col2 _merge
0 1 10 both
1 2 11 both
2 3 12 both
3 4 13 left_only
4 5 14 left_only
5 3 10 left_only
Create a boolean condition:
df_all['_merge'] == 'left_only'
0 False
1 False
2 False
3 True
4 True
5 True
Name: _merge, dtype: bool
Why other solutions are wrong
A few solutions make the same mistake - they only check that each value is independently in each column, not together in the same row. Adding the last row, which is unique but has the values from both columns from df2 exposes the mistake:
common = df1.merge(df2,on=['col1','col2'])
(~df1.col1.isin(common.col1))&(~df1.col2.isin(common.col2))
0 False
1 False
2 False
3 True
4 True
5 False
dtype: bool
This solution gets the same wrong result:
df1.isin(df2.to_dict('l')).all(1)
One method would be to store the result of an inner merge form both dfs, then we can simply select the rows when one column's values are not in this common:
In [119]:
common = df1.merge(df2,on=['col1','col2'])
print(common)
df1[(~df1.col1.isin(common.col1))&(~df1.col2.isin(common.col2))]
col1 col2
0 1 10
1 2 11
2 3 12
Out[119]:
col1 col2
3 4 13
4 5 14
EDIT
Another method as you've found is to use isin which will produce NaN rows which you can drop:
In [138]:
df1[~df1.isin(df2)].dropna()
Out[138]:
col1 col2
3 4 13
4 5 14
However if df2 does not start rows in the same manner then this won't work:
df2 = pd.DataFrame(data = {'col1' : [2, 3,4], 'col2' : [11, 12,13]})
will produce the entire df:
In [140]:
df1[~df1.isin(df2)].dropna()
Out[140]:
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
Assuming that the indexes are consistent in the dataframes (not taking into account the actual col values):
df1[~df1.index.isin(df2.index)]
As already hinted at, isin requires columns and indices to be the same for a match. If match should only be on row contents, one way to get the mask for filtering the rows present is to convert the rows to a (Multi)Index:
In [77]: df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3], 'col2' : [10, 11, 12, 13, 14, 10]})
In [78]: df2 = pandas.DataFrame(data = {'col1' : [1, 3, 4], 'col2' : [10, 12, 13]})
In [79]: df1.loc[~df1.set_index(list(df1.columns)).index.isin(df2.set_index(list(df2.columns)).index)]
Out[79]:
col1 col2
1 2 11
4 5 14
5 3 10
If index should be taken into account, set_index has keyword argument append to append columns to existing index. If columns do not line up, list(df.columns) can be replaced with column specifications to align the data.
pandas.MultiIndex.from_tuples(df<N>.to_records(index = False).tolist())
could alternatively be used to create the indices, though I doubt this is more efficient.
Suppose you have two dataframes, df_1 and df_2 having multiple fields(column_names) and you want to find the only those entries in df_1 that are not in df_2 on the basis of some fields(e.g. fields_x, fields_y), follow the following steps.
Step1.Add a column key1 and key2 to df_1 and df_2 respectively.
Step2.Merge the dataframes as shown below. field_x and field_y are our desired columns.
Step3.Select only those rows from df_1 where key1 is not equal to key2.
Step4.Drop key1 and key2.
This method will solve your problem and works fast even with big data sets. I have tried it for dataframes with more than 1,000,000 rows.
df_1['key1'] = 1
df_2['key2'] = 1
df_1 = pd.merge(df_1, df_2, on=['field_x', 'field_y'], how = 'left')
df_1 = df_1[~(df_1.key2 == df_1.key1)]
df_1 = df_1.drop(['key1','key2'], axis=1)
a bit late, but it might be worth checking the "indicator" parameter of pd.merge.
See this other question for an example:
Compare PandaS DataFrames and return rows that are missing from the first one
This is the best way to do it:
df = df1.drop_duplicates().merge(df2.drop_duplicates(), on=df2.columns.to_list(),
how='left', indicator=True)
df.loc[df._merge=='left_only',df.columns!='_merge']
Note that drop duplicated is used to minimize the comparisons. It would work without them as well. The best way is to compare the row contents themselves and not the index or one/two columns and same code can be used for other filters like 'both' and 'right_only' as well to achieve similar results. For this syntax dataframes can have any number of columns and even different indices. Only the columns should occur in both the dataframes.
Why this is the best way?
index.difference only works for unique index based comparisons
pandas.concat() coupled with drop_duplicated() is not ideal because it will also get rid of the rows which may be only in the dataframe you want to keep and are duplicated for valid reasons.
I think those answers containing merging are extremely slow. Therefore I would suggest another way of getting those rows which are different between the two dataframes:
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5], 'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3], 'col2' : [10, 11, 12]})
DISCLAIMER: My solution works if you're interested in one specific column where the two dataframes differ. If you are interested only in those rows, where all columns are equal do not use this approach.
Let's say, col1 is a kind of ID, and you only want to get those rows, which are not contained in both dataframes:
ids_in_df2 = df2.col1.unique()
not_found_ids = df[~df['col1'].isin(ids_in_df2 )]
And that's it. You get a dataframe containing only those rows where col1 isn't appearent in both dataframes.
You can also concat df1, df2:
x = pd.concat([df1, df2])
and then remove all duplicates:
y = x.drop_duplicates(keep=False, inplace=False)
I have an easier way in 2 simple steps:
As the OP mentioned Suppose dataframe2 is a subset of dataframe1, columns in the 2 dataframes are the same,
df1 = pd.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3],
'col2' : [10, 11, 12, 13, 14, 10]})
df2 = pd.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
### Step 1: just append the 2nd df at the end of the 1st df
df_both = df1.append(df2)
### Step 2: drop rows which contain duplicates, Drop all duplicates.
df_dif = df_both.drop_duplicates(keep=False)
## mission accompliched!
df_dif
Out[20]:
col1 col2
3 4 13
4 5 14
5 3 10
you can do it using isin(dict) method:
In [74]: df1[~df1.isin(df2.to_dict('l')).all(1)]
Out[74]:
col1 col2
3 4 13
4 5 14
Explanation:
In [75]: df2.to_dict('l')
Out[75]: {'col1': [1, 2, 3], 'col2': [10, 11, 12]}
In [76]: df1.isin(df2.to_dict('l'))
Out[76]:
col1 col2
0 True True
1 True True
2 True True
3 False False
4 False False
In [77]: df1.isin(df2.to_dict('l')).all(1)
Out[77]:
0 True
1 True
2 True
3 False
4 False
dtype: bool
Here is another way of solving this:
df1[~df1.index.isin(df1.merge(df2, how='inner', on=['col1', 'col2']).index)]
Or:
df1.loc[df1.index.difference(df1.merge(df2, how='inner', on=['col1', 'col2']).index)]
extract the dissimilar rows using the merge function
df = df1.merge(df2.drop_duplicates(), on=['col1','col2'],
how='left', indicator=True)
save the dissimilar rows in CSV
df[df['_merge'] == 'left_only'].to_csv('output.csv')
My way of doing this involves adding a new column that is unique to one dataframe and using this to choose whether to keep an entry
df2[col3] = 1
df1 = pd.merge(df_1, df_2, on=['field_x', 'field_y'], how = 'outer')
df1['Empt'].fillna(0, inplace=True)
This makes it so every entry in df1 has a code - 0 if it is unique to df1, 1 if it is in both dataFrames. You then use this to restrict to what you want
answer = nonuni[nonuni['Empt'] == 0]
How about this:
df1 = pandas.DataFrame(data = {'col1' : [1, 2, 3, 4, 5],
'col2' : [10, 11, 12, 13, 14]})
df2 = pandas.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
records_df2 = set([tuple(row) for row in df2.values])
in_df2_mask = np.array([tuple(row) in records_df2 for row in df1.values])
result = df1[~in_df2_mask]
Easier, simpler and elegant
uncommon_indices = np.setdiff1d(df1.index.values, df2.index.values)
new_df = df1.loc[uncommon_indices,:]
pd.concat([df1, df2]).drop_duplicates(keep=False) will concatenate the two DataFrames together, and then drop all the duplicates, keeping only the unique rows. By default it will keep the first occurrence of the duplicate, but setting keep=False will drop all the duplicates.
Keep in mind that if you need to compare the DataFrames with columns with different names, you will have to make sure the columns have the same name before concatenating the dataframes.
Also, if the dataframes have a different order of columns, it will also affect the final result.