I have the following dataframes.
import pandas as pd
d={'P':['A','B','C'],
'Q':[5,6,7]
}
df=pd.DataFrame(data=d)
print(df)
d={'P':['A','C','D'],
'Q':[5,7,8]
}
df1=pd.DataFrame(data=d)
print(df1)
d={'P':['B','E','F'],
'Q':[5,7,8]
}
df3=pd.DataFrame(data=d)
print(df3)
Code to check one dataframe column not present in other is this:
df.loc[~df['P'].isin(df1['P'])]
How to check the same in multiple columns?
How to find P column in df3 not in P column of df and df1?
Expected Output:
P Q
0 E 7
1 F 8
You can chain 2 conditions with & for bitwise AND:
cond1 = ~df3['P'].isin(df1['P'])
cond2 = ~df3['P'].isin(df['P'])
df = df3.loc[cond1 & cond2]
print (df)
P Q
1 E 7
2 F 8
Or join values of columns - by concatenate or join list by +:
df = df3.loc[~df3['P'].isin(np.concatenate([df1['P'],df['P']]))]
#another solution
#df = df3.loc[~df3['P'].isin(df1['P'].tolist() + df['P'].tolist())]
What about, However jezrael already given expert answer :)
You can simply define the conditions, and then combine them logically, like:
con1 = df3['P'].isin(df['P'])
con2 = df3['P'].isin(df1['P'])
df = df3[~ (con1 | con2)]
>>> df
P Q
1 E 7
2 F 8
Related
I want to combine two dataframes:
df1=pd.DataFrame({'A':['a','a',],'B':['b','b']})
df2=pd.DataFrame({'B':['b','b'],'A':['a','a']})
pd.concat([df1,df2],ignore_index=True)
result:
But I want the output to be like this (I want the same code as SQL's union/union all):
Another way is to use numpy to stack the two dataframes and then use pd.DataFrame constructor:
pd.DataFrame(np.vstack([df1.values,df2.values]), columns = df1.columns)
Output:
A B
0 a b
1 a b
2 b a
3 b a
Here is a proposition to do an SQL UNION ALL with pandas by using pandas.concat :
list_dfs = [df1, df2]
out = (
pd.concat([pd.DataFrame(sub_df.to_numpy()) for sub_df in list_dfs],
ignore_index=True)
.set_axis(df1.columns, axis=1)
)
# Output :
print(out)
A B
0 a b
1 a b
2 b a
3 b a
Say I have two DataFrames
df1 = pd.DataFrame({'A':[1,2], 'B':[3,4]}, index = [0,1])
df2 = pd.DataFrame({'B':[8,9], 'C':[10,11]}, index = [1,2])
I want to merge so that any values in df1 are overwritten in there is a value in df2 at that location and any new values in df2 are added including the new rows and columns.
The result should be:
A B C
0 1 3 nan
1 2 8 10
2 nan 9 11
I've tried combine_first but that causes only nan values to be overwritten
updated has the issue where new rows are created rather than overwritten
merge has many issues.
I've tried writing my own function
def take_right(df1, df2, j, i):
print (df1)
print (df2)
try:
s1 = df1[j][i]
except:
s1 = np.NaN
try:
s2 = df2[j][i]
except:
s2 = np.NaN
if math.isnan(s2):
#print(s1)
return s1
else:
# print(s2)
return s2
def combine_df(df1, df2):
rows = (set(df1.index.values.tolist()) | set(df2.index.values.tolist()))
#print(rows)
columns = (set(df1.columns.values.tolist()) | set(df2.columns.values.tolist()))
#print(columns)
df = pd.DataFrame()
#df.columns = columns
for i in rows:
#df[:][i]=[]
for j in columns:
df = df.insert(int(i), j, take_right(df1,df2,j,i), allow_duplicates=False)
# print(df)
return df
This won't add new columns or rows to an empty DataFrame.
Thank you!!
One approach is to create an empty output dataframe with the union of columns and indices from df1 and df2 and then use the df.update method to assign their values into the out_df
import pandas as pd
df1 = pd.DataFrame({'A':[1,2], 'B':[3,4]}, index = [0,1])
df2 = pd.DataFrame({'B':[8,9], 'C':[10,11]}, index = [1,2])
out_df = pd.DataFrame(
columns = df1.columns.union(df2.columns),
index = df1.index.union(df2.index),
)
out_df.update(df1)
out_df.update(df2)
out_df
Why does combine_first not work?
df = df2.combine_first(df1)
print(df)
Output:
A B C
0 1.0 3 NaN
1 2.0 8 10.0
2 NaN 9 11.0
I am stuck with a seemingly easy problem: dropping unique rows in a pandas dataframe. Basically, the opposite of drop_duplicates().
Let's say this is my data:
A B C
0 foo 0 A
1 foo 1 A
2 foo 1 B
3 bar 1 A
I would like to drop the rows when A, and B are unique, i.e. I would like to keep only the rows 1 and 2.
I tried the following:
# Load Dataframe
df = pd.DataFrame({"A":["foo", "foo", "foo", "bar"], "B":[0,1,1,1], "C":["A","A","B","A"]})
uniques = df[['A', 'B']].drop_duplicates()
duplicates = df[~df.index.isin(uniques.index)]
But I only get the row 2, as 0, 1, and 3 are in the uniques!
Solutions for select all duplicated rows:
You can use duplicated with subset and parameter keep=False for select all duplicates:
df = df[df.duplicated(subset=['A','B'], keep=False)]
print (df)
A B C
1 foo 1 A
2 foo 1 B
Solution with transform:
df = df[df.groupby(['A', 'B'])['A'].transform('size') > 1]
print (df)
A B C
1 foo 1 A
2 foo 1 B
A bit modified solutions for select all unique rows:
#invert boolean mask by ~
df = df[~df.duplicated(subset=['A','B'], keep=False)]
print (df)
A B C
0 foo 0 A
3 bar 1 A
df = df[df.groupby(['A', 'B'])['A'].transform('size') == 1]
print (df)
A B C
0 foo 0 A
3 bar 1 A
I came up with a solution using groupby:
groupped = df.groupby(['A', 'B']).size().reset_index().rename(columns={0: 'count'})
uniques = groupped[groupped['count'] == 1]
duplicates = df[~df.index.isin(uniques.index)]
Duplicates now has the proper result:
A B C
2 foo 1 B
3 bar 1 A
Also, my original attempt in the question can be fixed by simply adding keep=False in the drop_duplicates method:
# Load Dataframe
df = pd.DataFrame({"A":["foo", "foo", "foo", "bar"], "B":[0,1,1,1], "C":["A","A","B","A"]})
uniques = df[['A', 'B']].drop_duplicates(keep=False)
duplicates = df[~df.index.isin(uniques.index)]
Please #jezrael answer, I think it is safest(?), as I am using pandas indexes here.
df1 = df.drop_duplicates(['A', 'B'],keep=False)
df1 = pd.concat([df, df1])
df1 = df1.drop_duplicates(keep=False)
This technique is more suitable when you have two datasets dfX and dfY with millions of records. You may first concatenate dfX and dfY and follow the same steps.
Let's say I have a data frame with such column names:
['a','b','c','d','e','f','g']
And I would like to change names from 'c' to 'f' (actually add string to the name of column), so the whole data frame column names would look like this:
['a','b','var_c_equal','var_d_equal','var_e_equal','var_f_equal','g']
Well, firstly I made a function that changes column names with the string i want:
df.rename(columns=lambda x: 'or_'+x+'_no', inplace=True)
But now I really want to understand how to implement something like this:
df.loc[:,'c':'f'].rename(columns=lambda x: 'var_'+x+'_equal', inplace=True)
You can a use a list comprehension for that like:
Code:
new_columns = ['var_{}_equal'.format(c) if c in 'cdef' else c for c in columns]
Test Code:
import pandas as pd
df = pd.DataFrame({'a':(1,2), 'b':(1,2), 'c':(1,2), 'd':(1,2)})
print(df)
df.columns = ['var_{}_equal'.format(c) if c in 'cdef' else c
for c in df.columns]
print(df)
Results:
a b c d
0 1 1 1 1
1 2 2 2 2
a b var_c_equal var_d_equal
0 1 1 1 1
1 2 2 2 2
One way is to use a dictionary instead of an anonymous function. Both the below variations assume the columns you need to rename are contiguous.
Contiguous columns by position
d = {k: 'var_'+k+'_equal' for k in df.columns[2:6]}
df = df.rename(columns=d)
Contiguous columns by name
If you need to calculate the numerical indices:
cols = df.columns.get_loc
d = {k: 'var_'+k+'_equal' for k in df.columns[cols('c'):cols('f')+1]}
df = df.rename(columns=d)
Specifically identified columns
If you want to provide the columns explicitly:
d = {k: 'var_'+k+'_equal' for k in 'cdef'}
df = df.rename(columns=d)
I have a dataframe such as:
label column1
a 1
a 2
b 6
b 4
I would like to make a dataframe with a new column, with the opposite value from column1 where the labels match. Such as:
label column1 column2
a 1 2
a 2 1
b 6 4
b 4 6
I know this is probably very simple to do with a groupby command but I've been searching and can't find anything.
The following uses groupby and apply and seems to work okay:
x = pd.DataFrame({ 'label': ['a','a','b','b'],
'column1': [1,2,6,4] })
y = x.groupby('label').apply(
lambda g: g.assign(column2 = np.asarray(g.column1[::-1])))
y = y.reset_index(drop=True) # optional: drop weird index
print(y)
you can try the code block below:
#create the Dataframe
df = pd.DataFrame({'label':['a','a','b','b'],
'column1':[1,2,6,4]})
#Group by label
a = df.groupby('label').first().reset_index()
b = df.groupby('label').last().reset_index()
#Concat those groups to create columns2
df2 = (pd.concat([b,a])
.sort_values(by='label')
.rename(columns={'column1':'column2'})
.reset_index()
.drop('index',axis=1))
#Merge with the original Dataframe
df = df.merge(df2,left_index=True,right_index=True,on='label')[['label','column1','column2']]
Hope this helps
Assuming their are only pairs of labels, you could use the following as well:
# Create dataframe
df = pd.DataFrame(data = {'label' :['a', 'a', 'b', 'b'],
'column1' :[1,2, 6,4]})
# iterate over dataframe, identify matching label and opposite value
for index, row in df.iterrows():
newvalue = int(df[(df.label == row.label) & (df.column1 != row.column1)].column1.values[0])
# set value to new column
df.set_value(index, 'column2', newvalue)
df.head()
You can use groupby with apply where create new Series with back order:
df['column2'] = df.groupby('label')["column1"] \
.apply(lambda x: pd.Series(x[::-1].values)).reset_index(drop=True)
print (df)
column1 label column2
0 1 a 2
1 2 a 1
2 6 b 4
3 4 b 6