I have 2 dataframes which I need to join using left join. In sql I have the query as
SELECT A.* INTO NewTable FROM A LEFT JOIN B ON A.id=B.id WHERE B.id IS NULL;
I have the 2 dataframes as:
df1:
id
name
1
one
2
two
3
three
4
four
df2:
id
2
3
What I am expecting is:
id
name
1
one
4
four
What I have tried?
common = df1.merge(df2, on=['id', 'id'])
result = df1[~df1.id.isin(common.id)]
I get more results in this then what the query returns. Any help is appreciated.
you have the right solution,only you do interpret the results wrong.
This will give you the result without index
import pandas as pd
d = {'id': [1, 2,3,4], 'col2': ['one','two','three','four']}
d1 = {'id': [2,3]}
df1 = pd.DataFrame(data=d)
df2 = pd.DataFrame(data=d1)
result = df1[~df1.id.isin(df2.id)]
print(result.to_string(index=False))
You can use left join with .merge() with indicator= parameter turned on. Then, filter the indicator values equal to "left_only" with .query(), as follows:
df1.merge(df2, on='id', how='left', indicator='ind').query('ind == "left_only"')
Result:
id name ind
0 1 one left_only
3 4 four left_only
Optionally, you can remove the indicator column, as follows:
df1.merge(df2, on='id', how='left', indicator='ind').query('ind == "left_only"').drop('ind', axis=1)
Result:
id name
0 1 one
3 4 four
Try:
print(df1[~df1["id"].isin(df2["id"])])
Prints:
id name
0 1 one
3 4 four
import pandas as pd
import numpy as np
df1 = pd.DataFrame(np.array([[1, "one"], [2, "two"], [3, "three"], [4, "four"]]),
columns=['id', 'name '])
df2 = pd.DataFrame(np.array([[1], [2]]),
columns=['id'])
df1.drop(df2['id'], axis=0,inplace=True)
df1
Related
I have the following data frames:
DF1
df1 = pd.DataFrame(columns = ['Key', 'Value'])
df1['Key'] = ['A', 'B', 'C', 'D']
DF2
df2 = pd.DataFrame(columns = ['Key', 'Value'])
df2['Key'] = ['A', 'C']
df2['Value'] = [1,7]
I would like to merge these two data frames such that the data from DF2 under the column 'Value' is filled in DF1, where the remaining letters 'B' and 'D' have zero.
I tried this:
df3 = pd.merge(df1,df2,how='outer', on = 'Key')
However, this creates an additional column Value_x and Value_y which is not what I want.
Thanks
I think the shortest way to accomplish this is
df1[['Key']].merge(df2, on='Key', how='outer')
by not including Value from the left frame, you don't have 2 columns in the resulting data frame.
You could remove the Value column from df1 and use your existing merge.
Or only use the Key column from df1 when merging.
df3 = pd.merge(df1['Key'],df2,how='outer', on = 'Key').fillna(value=0)
Key Value
0 A 1.0
1 B 0.0
2 C 7.0
3 D 0.0
Another way to do it is by concatenating the dataframes and then grouping the Key value like this:
df3 = pd.concat([df1, df2]).fillna(0).groupby('Key').sum().reset_index()
Output:
Key Value
0 A 1
1 B 0
2 C 7
3 D 0
This way is a little verbose but easier to read and extensible to more than 2 DFs.
Say I have two data frames:
df1:
A
0 a
1 b
df2:
A
0 a
1 c
I want the result to be the union of the two frames with an extra column showing the source data frame that the row belongs to. In case of duplicates, duplicates should be removed and the respective extra column should show both sources:
A B
0 a df1, df2
1 b df1
2 c df2
I can get the concatenated data frame (df3) without duplicates as follows:
import pandas as pd
df3=pd.concat([df1,df2],ignore_index=True).drop_duplicates().reset_index(drop=True)
I can't think of/find a method to have control over what element goes where. How can I add the extra column?
Thank you very much for any tips.
Merge with an indicator argument, and remap the result:
m = {'left_only': 'df1', 'right_only': 'df2', 'both': 'df1, df2'}
result = df1.merge(df2, on=['A'], how='outer', indicator='B')
result['B'] = result['B'].map(m)
result
A B
0 a df1, df2
1 b df1
2 c df2
Use the command below:
df3 = pd.concat([df1.assign(source='df1'), df2.assign(source='df2')]) \
.groupby('A') \
.aggregate(list) \
.reset_index()
The result will be:
A source
0 a [df1, df2]
1 b [df1]
2 c [df2]
The assign will add a column named source with value df1 and df2 to your dataframes. groupby command groups rows with same A value to single row. aggregate command describes how to aggregate other columns (source) for each group of rows with same A. I have used list aggregate function so that the source column be the list of values with same A.
We use outer join to solve this -
df1 = pd.DataFrame({'A':['a','b']})
df2 = pd.DataFrame({'A':['a','c']})
df1['col1']='df1'
df2['col2']='df2'
df=pd.merge(df1, df2, on=['A'], how="outer").fillna('')
df['B']=df['col1']+','+df['col2']
df['B'] = df['B'].str.strip(',')
df=df[['A','B']]
df
A B
0 a df1,df2
1 b df1
2 c df2
I have two dataframes with a common column called 'upc' as such:
df1:
upc
23456793749
78907809834
35894796324
67382808404
93743008374
df2:
upc
4567937
9078098
8947963
3828084
7430083
Notice that df2 'upc' values are the innermost 7 values of df1 'upc' values.
Note that both df1 and df2 have other columns not shown above.
What I want to do is do an inner merge on 'upc' but only on the innermost 7 values. How can I achieve this?
1) Create both dataframes and convert to string type.
2) pd.merge the two frames, but using the left_on keyword to access the inner 7 characters of your 'upc' series
df1 = pd.DataFrame(data=[
23456793749,
78907809834,
35894796324,
67382808404,
93743008374,], columns = ['upc1'])
df1 = df1.astype(str)
df2 = pd.DataFrame(data=[
4567937,
9078098,
8947963,
3828084,
7430083,], columns = ['upc2'])
df2 = df2.astype(str)
pd.merge(df1, df2, left_on=df1['upc1'].astype(str).str[2:-2], right_on='upc2', how='inner')
Out[5]:
upc1 upc2
0 23456793749 4567937
1 78907809834 9078098
2 35894796324 8947963
3 67382808404 3828084
4 93743008374 7430083
Using str.extact, match all items in df1 with df2, then we using the result as merge key merge with df2
df1['keyfordf2']=df1.astype(str).upc.str.extract(r'({})'.format('|'.join(df2.upc.astype(str).tolist())),expand=True).fillna(False)
df1.merge(df2.astype(str),left_on='keyfordf2',right_on='upc')
Out[273]:
upc_x keyfordf2 upc_y
0 23456793749 4567937 4567937
1 78907809834 9078098 9078098
2 35894796324 8947963 8947963
3 67382808404 3828084 3828084
4 93743008374 7430083 7430083
You could make a new column in df1 and merge on that.
import pandas as pd
df1= pd.DataFrame({'upc': [ 23456793749, 78907809834, 35894796324, 67382808404, 93743008374]})
df2= pd.DataFrame({'upc': [ 4567937, 9078098, 8947963, 3828084, 7430083]})
df1['upc_old'] = df1['upc'] #in case you still need the old (longer) upc column
df1['upc'] = df1['upc'].astype(str).str[2:-2].astype(int)
merged_df = pd.merge(df1, df2, on='upc')
I'm having trouble using pd.merge after groupby. Here's my hypothetical:
import pandas as pd
from pandas import DataFrame
import numpy as np
df1 = DataFrame({'key': [1,1,2,2,3,3],
'var11': np.random.randn(6),
'var12': np.random.randn(6)})
df2 = DataFrame({'key': [1,2,3],
'var21': np.random.randn(3),
'var22': np.random.randn(3)})
#group var11 in df1 by key
grouped = df1['var11'].groupby(df1['key'])
# calculate the mean of var11 by key
grouped = grouped.mean()
print grouped
key
1 1.399430
2 0.568216
3 -0.612843
dtype: float64
print grouped.index
Int64Index([1, 2, 3], dtype='int64')
print df2
key var21 var22
0 1 -0.381078 0.224325
1 2 0.836719 -0.565498
2 3 0.323412 -1.616901
df2 = pd.merge(df2, grouped, left_on = 'key', right_index = True)
At this point, I get IndexError: list index out of range.
When using groupby, the grouping variable ('key' in this example) becomes the index for the resultant series, which is why I specify 'right_index = True'. I've tried other syntax without success. Any advice?
I think you should just do this:
In [140]:
df2 = pd.merge(df2,
pd.DataFrame(grouped, columns=['mean']),
left_on='key',
right_index=True)
print df2
key var21 var22 mean
0 1 0.324476 0.701254 0.400313
1 2 -1.270500 0.055383 -0.293691
2 3 0.804864 0.566747 0.628787
[3 rows x 4 columns]
The reason it didn't work is that grouped is a Series not a DataFrame
I would like to merge two DataFrames, and keep the index from the first frame as the index on the merged dataset. However, when I do the merge, the resulting DataFrame has integer index. How can I specify that I want to keep the index from the left data frame?
In [4]: a = pd.DataFrame({'col1': {'a': 1, 'b': 2, 'c': 3},
'to_merge_on': {'a': 1, 'b': 3, 'c': 4}})
In [5]: b = pd.DataFrame({'col2': {0: 1, 1: 2, 2: 3},
'to_merge_on': {0: 1, 1: 3, 2: 5}})
In [6]: a
Out[6]:
col1 to_merge_on
a 1 1
b 2 3
c 3 4
In [7]: b
Out[7]:
col2 to_merge_on
0 1 1
1 2 3
2 3 5
In [8]: a.merge(b, how='left')
Out[8]:
col1 to_merge_on col2
0 1 1 1.0
1 2 3 2.0
2 3 4 NaN
In [9]: _.index
Out[9]: Int64Index([0, 1, 2], dtype='int64')
EDIT: Switched to example code that can be easily reproduced
In [5]: a.reset_index().merge(b, how="left").set_index('index')
Out[5]:
col1 to_merge_on col2
index
a 1 1 1
b 2 3 2
c 3 4 NaN
Note that for some left merge operations, you may end up with more rows than in a when there are multiple matches between a and b. In this case, you may need to drop duplicates.
You can make a copy of index on left dataframe and do merge.
a['copy_index'] = a.index
a.merge(b, how='left')
I found this simple method very useful while working with large dataframe and using pd.merge_asof() (or dd.merge_asof()).
This approach would be superior when resetting index is expensive (large dataframe).
There is a non-pd.merge solution using Series.map and DataFrame.set_index.
a['col2'] = a['to_merge_on'].map(b.set_index('to_merge_on')['col2']))
col1 to_merge_on col2
a 1 1 1.0
b 2 3 2.0
c 3 4 NaN
This doesn't introduce a dummy index name for the index.
Note however that there is no DataFrame.map method, and so this approach is not for multiple columns.
df1 = df1.merge(df2, how="inner", left_index=True, right_index=True)
This allows to preserve the index of df1
Assuming that the resulting df has the same number of rows and order as your first df, you can do this:
c = pd.merge(a, b, on='to_merge_on')
c.set_index(a.index,inplace=True)
another simple option is to rename the index to what was before:
a.merge(b, how="left").set_axis(a.index)
merge preserves the order at dataframe 'a', but just resets the index so it's safe to use set_axis
You can also use DataFrame.join() method to achieve the same thing. The join method will persist the original index. The column to join can be specified with on parameter.
In [17]: a.join(b.set_index("to_merge_on"), on="to_merge_on")
Out[17]:
col1 to_merge_on col2
a 1 1 1.0
b 2 3 2.0
c 3 4 NaN
Think I've come up with a different solution. I was joining the left table on index value and the right table on a column value based off index of left table. What I did was a normal merge:
First10ReviewsJoined = pd.merge(First10Reviews, df, left_index=True, right_on='Line Number')
Then I retrieved the new index numbers from the merged table and put them in a new column named Sentiment Line Number:
First10ReviewsJoined['Sentiment Line Number']= First10ReviewsJoined.index.tolist()
Then I manually set the index back to the original, left table index based off pre-existing column called Line Number (the column value I joined on from left table index):
First10ReviewsJoined.set_index('Line Number', inplace=True)
Then removed the index name of Line Number so that it remains blank:
First10ReviewsJoined.index.name = None
Maybe a bit of a hack but seems to work well and relatively simple. Also, guess it reduces risk of duplicates/messing up your data. Hopefully that all makes sense.
For the people that wants to maintain the left index as it was before the left join:
def left_join(
a: pandas.DataFrame, b: pandas.DataFrame, on: list[str], b_columns: list[str] = None
) -> pandas.DataFrame:
if b_columns:
b_columns = set(on + b_columns)
b = b[b_columns]
df = (
a.reset_index()
.merge(
b,
how="left",
on=on,
)
.set_index(keys=[x or "index" for x in a.index.names])
)
df.index.names = a.index.names
return df