I have two dataframes (df1, df2).
df1
df2
I would like to have the final dataframe like this:
Final dataframe
How do I do this using Pandas?
EDIT The following solution is suggested by #sophocles
df1 = pd.DataFrame({'name':['a','b','c'],
'val1':[1,None,3],
'val2':[4,5,6] })
df2 = pd.DataFrame({'name':['b'],
'val1':2})
df1 and df2:
name val1 val2
0 a 1.0 4
1 b NaN 5
2 c 3.0 6
name val1
0 b 2
Simply using fillna
df1.set_index('name').fillna(df2.set_index('name')).reset_index()
This one is much faster than using merge method
Related
EDITED*
I have a large df with many rows that share the same value in some of the columns.
I want to do the following:
new df = identify the rows in df that have a value in a certain column (not empty).
'''
df = pd.DataFrame({"a": [1, 2,2,2, 3, 4],
"b":['A','B','B', 'B','C','D'],
"c":[NaN, 2,NaN,NaN,NaN,NaN]})
'''
df1=df[~df['c'].isnull()]
'''
add to 'new_df' the rows from df that share 2 keys.
I tried to use merge:
df2 = pd.merge(df1,df,on=['a','b'], how='left')
But the result was that It added the same row a few times and not the unique rows
a b c_x c_y
0 2 B 2.0 2.0
1 2 B 2.0 NaN
2 2 B 2.0 NaN
I want to keep only one 'c' column with all the values. Not sure what approach to use.
Hope I made it clear...
Thanks!
As far as I understand, you would like to group by 'a' and 'b' and return only those groups where at least one row does not have a NaN in column 'c'. If that's the case. here you go
Load the df:
df = pd.DataFrame({"a": [1,1,1, 2,2,2, 3, 4], "b":['A','A','A','B','B', 'B','C','D'], "c":[None, None,None,2,None,None,None,None]})
filter for any non-NaNs:
df.groupby(['a','b']).filter(lambda g: any(~g['c'].isna()))
output:
a b c
3 2 B 2.0
4 2 B NaN
5 2 B NaN
I have 3 data frame:
df1
id,k,a,b,c
1,2,1,5,1
2,3,0,1,0
3,6,1,1,0
4,1,0,5,0
5,1,1,5,0
df2
name,a,b,c
p,4,6,8
q,1,2,3
df3
type,w_ave,vac,yak
n,3,5,6
v,2,1,4
from the multiplication, using pandas and numpy, I want to the output in df1:
id,k,a,b,c,w_ave,vac,yak
1,2,1,5,1,16,15,18
2,3,0,1,0,0,3,6
3,6,1,1,0,5,4,7
4,1,0,5,0,0,11,14
5,1,1,5,0,13,12,15
the conditions are:
The value of the new column will be =
#its not a code
df1["w_ave"][1] = df3["w_ave"]["v"]+ df1["a"][1]*df2["a"]["q"]+df1["b"][1]*df2["b"]["q"]+df1["c"][1]*df2["c"]["q"]
for output["w_ave"][1]= 2 +(1*1)+(5*2)+(1*3)
df3["w_ave"]["v"]=2
df1["a"][1]=1, df2["a"]["q"]=1 ;
df1["b"][1]=5, df2["b"]["q"]=2 ;
df1["c"][1]=1, df2["c"]["q"]=3 ;
Which means:
- a new column will be added in df1, from the name of the column from df3.
- for each row of the df1, the value of a, b, c will be multiplied with the same-named q value from df2. and summed together with the corresponding value of df3.
-the column name of df1 , matched will column name of df2 will be multiplied. The other not matched column will not be multiplied, like df1[k].
- However, if there is any 0 in df1["a"], the corresponding output will be zero.
I am struggling with this. It was tough to explain also. My attempts are very silly. I know this attempt will not work. However, I have added this:
import pandas as pd, numpy as np
data1 = "Sample_data1.csv"
data2 = "Sample_data2.csv"
data3 = "Sample_data3.csv"
folder = '~Sample_data/'
df1 =pd.read_csv(folder + data1)
df2 =pd.read_csv(folder + data2)
df3 =pd.read_csv(folder + data3)
df1= df2 * df1
Ok, so this will in no way resemble your desired output, but vectorizing the formula you provided:
df2=df2.set_index("name")
df3=df3.set_index("type")
df1["w_ave"] = df3.loc["v", "w_ave"]+ df1["a"].mul(df2.loc["q", "a"])+df1["b"].mul(df2.loc["q", "b"])+df1["c"].mul(df2.loc["q", "c"])
Outputs:
id k a b c w_ave
0 1 2 1 5 1 16
1 2 3 0 1 0 4
2 3 6 1 1 0 5
3 4 1 0 5 0 12
4 5 1 1 5 0 13
I am trying to map a column to a dataframe from another dataframe where all words exist from the target dataframe
multiple matches are fine as I can filter them out after.
Thanks in advance!
df1
ColA
this is a sentence
with some words
in a column
and another
for fun
df2
ColB ColC
this a 123
in column 456
fun times 789
Some attempts
dfResult = df1.apply(lambda x: np.all([word in x.df1['ColA'].split(' ') for word in x.df2['ColB'].split(' ')]),axis = 1)
dfResult = df1.ColA.apply(lambda sentence: all(word in sentence for word in df2.ColB))
desired output
dfResult
ColA ColC
this is a sentence 123
with some words NaN
in a column 456
and another NaN
for fun NaN
Turn to set and look for subsets with Numpy broadcasting
Disclaimer: No assurances that this will be fast.
A = df1.ColA.str.split().apply(set).to_numpy() # If pandas version is < 0.24 use `.values`
B = df2.ColB.str.split().apply(set).to_numpy() # instead of `.to_numpy()`
C = df2.ColC.to_numpy()
# When `dtype` is `object` Numpy falls back on performing
# the operation on each pair of values. Since these are `set` objects
# `<=` tests for subset.
i, j = np.where(B <= A[:, None])
out = pd.array([np.nan] * len(A), pd.Int64Dtype()) # Empty nullable integers
# Use `out = np.empty(len(A), dtype=object)` if pandas version is < 0.24
out[i] = C[j]
df1.assign(ColC=out)
ColA ColC
0 this is a sentence 123
1 with some words NaN
2 in a column 456
3 and another NaN
4 for fun NaN
By using loop and set.issubset
pd.DataFrame([[y if set(z.split()).issubset(set(x.split())) else np.nan for z,y in zip(df2.ColB,df2.ColC)] for x in df1.ColA ]).max(1)
Out[34]:
0 123.0
1 NaN
2 456.0
3 NaN
4 NaN
dtype: float64
df = pd.DataFrame(np.arange(4*3).reshape(4,3), index=[['a','a','b','b'],[1,2,1,2]], columns=list('xyz'))
where df looks like:
Now I add a new row by:
df.loc['new',:]=[0,0,0]
Then df becomes:
Now I want to do the same but with a different df that has non-unique multi-index:
df = pd.DataFrame(np.arange(4*3).reshape(4,3), index=[['a','a','b','b'],[1,1,2,2]], columns=list('xyz'))
,which looks like:
and call
df.loc['new',:]=[0,0,0]
The result is "Exception: cannot handle a non-unique multi-index!"
How could I achieve the goal?
Use append or concat with helper DataFrame:
df1 = pd.DataFrame([[0,0,0]],
columns=df.columns,
index=pd.MultiIndex.from_arrays([['new'], ['']]))
df2 = df.append(df1)
df2 = pd.concat([df, df1])
print (df2)
x y z
a 1 0 1 2
1 3 4 5
b 2 6 7 8
2 9 10 11
new 0 0 0
I have a DataFrame like this
DataFrame({"key":["a","b","c","d","e"], "value": [5,4,3,2,1]})
I am mainly interested in row "a", "b" and "c". I want to merge everything else into an "others" row like this
key value
0 a 5
1 b 4
2 c 3
3 others 3
I wonder how can this be done.
First create a dataframe without d and e:
df2 = df[df.key.isin(["a","b","c"])]
Then find the value that you want the other column to have (using the sum function in this example):
val = df[~df["key"].isin(["a","b","c"])].sum()["value"]
Finally, append this column to the second df:
df2.append({"key":"others", "value":val},ignore_index=True)
df2 is now:
key value
0 a 5
1 b 4
2 c 3
3 others 3
I have found a way to do it. Not sure if it is the best way.
In [3]: key_map = {"a":"a", "b":"b", "c":"c"}
In [4]: data['key1'] = data['key'].map(lambda k: key_map.get(k, "others"))
In [5]: data.groupby("key1").sum()
Out[5]:
value
key1
a 5
b 4
c 3
others 3