I have a problem with getting nearest values for some rows in pandas dataframe and fill another column with values from those rows.
data sample I have:
id su_id r_value match_v
A A1 0 1
A A2 0 1
A A3 70 2
A A4 120 100
A A5 250 3
A A6 250 100
B B1 0 1
B B2 30 2
The thing is, wherever match_v is equal to 100, I need to replace that 100 with a value from the row where r_value is the closest to r_value from origin row(where match_v is equal to 100), but just withing group (grouped by id)
Expected output
id su_id r_value match_v
A A1 0 1
A A2 0 1
A A3 70 2
A A4 120 2
A A5 250 3
A A6 250 3
B B1 0 1
B B2 30 2
I have tried with creating lead and leg with shift and then finding differences. But doesn't work well and it somehow messed up already good values.
I haven't tried anything else cause I really don't have any idea.
Any help or hint is welcomed and I if you need any additional info, I'm here.
Thanks in advance.
More like merge_asof
s=df.loc[df.match_v!=100]
s=pd.merge_asof(df.sort_values('r_value'),s.sort_values('r_value'),on='r_value',by='id',direction='nearest')
df['match_v']=df['su_id'].map(s.set_index('su_id_x')['match_v_y'])
df
Out[231]:
id su_id r_value match_v
0 A A1 0 1
1 A A2 0 1
2 A A3 70 2
3 A A4 120 2
4 A A5 250 3
5 A A6 250 3
6 B B1 0 1
7 B B2 30 2
Here is another way using numpy broadcast , build for speed up calculation
l=[]
for x , y in df.groupby('id'):
s1=y.r_value.values
s=abs((s1-s1[:,None])).astype(float)
s[np.tril_indices(s.shape[0], 0)] = 999999
s=s.argmin(0)
s2=y.match_v.values
l.append(s2[s][s2==100])
df.loc[df.match_v==100,'match_v']=np.concatenate(l)
df
Out[264]:
id su_id r_value match_v
0 A A1 0 1
1 A A2 0 1
2 A A3 70 2
3 A A4 120 2
4 A A5 250 3
5 A A6 250 3
6 B B1 0 1
7 B B2 30 2
You could define a custom function which does the calculation and substitution, and then use it with groupby and apply.
def mysubstitution(x):
for i in x.index[x['match_v'] == 100]:
diff = (x['r_value'] - (x['r_value'].iloc[i])).abs()
exclude = x.index.isin([i])
closer_idx = diff[~exclude].idxmin()
x['match_v'].iloc[i] = x['match_v'].iloc[closer_idx]
return x
ddf = df.groupby('id').apply(mysubstitution)
ddf is:
id su_id r_value match_v
0 A A1 0 1
1 A A2 0 1
2 A A3 70 2
3 A A4 120 2
4 A A5 250 3
5 A A6 250 3
6 B B1 0 1
7 B B2 30 2
Assuming there is always at least one valid value within the group when a 100 is first encountered.
m = dict()
for i in range(len(df)):
if df.loc[i, "match_v"] == 100:
df.loc[i, "match_v"] = m[df.loc[i, "id"]]
else:
m[df.loc[i, "id"]] = df.loc[i, "match_v"]
Related
I have a dataframe:
id life_day value
a1 1 10
a1 2 20
a1 3 10
a1 4 5
a1 5 5
a1 6 1
b2 1 7
b2 3 11
b2 4 10
b2 5 20
I want to sum values for each id till life_day 4. So desired result is:
id life_day value
a1 4 45
b2 4 28
How to do that? I tried df[df["life_day"] == 90].groupby("id).sum() but brings wrong results
Your approach almost works, but I don't know why you wrote == 90 in df["life_day"] == 90, and it looks like you want the max of life_day, not the sum.
df[df['life_day'] <= 4].groupby('id').agg({'life_day': 'max', 'value': 'sum'})
life_day value
id
a1 4 45
b2 4 28
Use the pandas where condition to mask and then groupby agg
df.where(df['life_day'].le(4)).groupby('id').agg({'life_day':'last','value':'sum'}).reset_index()
id life_day value
0 a1 4.0 45.0
1 b2 4.0 28.0
I have a dataframe:
id value
a1 0
a1 1
a1 2
a1 3
a2 0
a2 1
a3 0
a3 1
a3 2
a3 3
I want to filter id's and leave only those which have value higher than 3. So in this example id a2 must be removed since it only has values 0 and 1. So desired result is:
id value
a1 0
a1 1
a1 2
a1 3
a3 0
a3 1
a3 2
a3 3
a3 4
a3 5
How to to that in pandas?
Updated.
Group by IDs and find their max values. Find the IDs whose max value is at or above 3:
keep = df.groupby('id')['value'].max() >= 3
Select the rows with the IDs that match:
df[df['id'].isin(keep[keep].index)]
Use boolean mask to keep rows that match condition then replace bad id (a2) by the next id (a3). Finally, group again by id an apply a cumulative sum.
mask = df.groupby('id')['value'] \
.transform(lambda x: sorted(x.tolist()) == [0, 1, 2, 3])
df1 = df[mask].reindex(df.index).bfill()
df1['value'] = df1.groupby('id').agg('cumcount')
Output:
>>> df1
id value
0 a1 0
1 a1 1
2 a1 2
3 a1 3
4 a3 0
5 a3 1
6 a3 2
7 a3 3
8 a3 4
9 a3 5
Still new into pandas but is there a way to sort df by subtotal of each group.
Area Unit Count
A A1 5
A A2 2
B B1 10
B B2 1
B B3 3
C C1 10
So I want to sort them by subtotal of each Area which results to A subtotal = 7, B subtotal=14, C subtotal = 10
The sort should be like
Area Unit Count
B B1 10
B B2 1
B B3 3
C C1 10
A A1 5
A A2 2
*Note that despite the value of B3 > B1 it should not be affected by the sort.
create a helper column 'sorter', which is the sum of the count variable, and sort ur dataframe with it
df['sorter'] = df.groupby("Area").Count.transform('sum')
df.sort_values('sorter',ascending=False).reset_index(drop=True).drop('sorter',axis=1)
Area Unit Count
0 B B1 10
1 B B2 1
2 B B3 3
3 C C1 10
4 A A1 5
5 A A2 2
I have a pandas.core.groupby.DataFrameGroupBy object where I am trying to count the number of rows where a value for TOTAL_FLOOR_AREA is > 30. I can count the number of rows for each dataframe in the groupby object using:
import numpy as np
grouped = master_lsoa.groupby('lsoa11')
grouped.aggregate(np.count_nonzero).TOTAL_FLOOR_AREA
But how do I conditionally count rows where the value for TOTAL_FLOOR_AREA is greater than 30?
Sam
I think you need:
np.random.seed(6)
N = 15
master_lso = pd.DataFrame({'lsoa11': np.random.randint(4, size=N),
'TOTAL_FLOOR_AREA': np.random.choice([0,30,40,50], size=N)})
master_lso['lsoa11'] = 'a' + master_lso['lsoa11'].astype(str)
print (master_lso)
TOTAL_FLOOR_AREA lsoa11
0 40 a2
1 50 a1
2 30 a3
3 0 a0
4 40 a2
5 0 a1
6 30 a3
7 0 a2
8 40 a0
9 0 a2
10 0 a1
11 50 a1
12 50 a3
13 40 a1
14 30 a1
First filter rows by condition by boolean indexing - it is faster before grouping, because less rows.
df = master_lso[master_lso['TOTAL_FLOOR_AREA'] > 30]
print (df)
TOTAL_FLOOR_AREA lsoa11
0 40 a2
1 50 a1
4 40 a2
8 40 a0
11 50 a1
12 50 a3
13 40 a1
Then groupby and aggregate size:
df1 = df.groupby('lsoa11')['TOTAL_FLOOR_AREA'].size().reset_index(name='Count')
print (df1)
lsoa11 Count
0 a0 1
1 a1 3
2 a2 2
3 a3 1
you could also construct a new column indicating where the condition is met and sum up like (stealing #jezrael's dataframe):
master_lso.assign(Large_Enough= lambda x:x["TOTAL_FLOOR_AREA"]>30)\
.groupby('lsoa11')["Large_Enough"].sum().reset_index()
Note that Truevalues are interpreted as 1. So the sum provides the corresponding count here.
The advantage over #jezrael's solution is that you can still sum up the total area per group
Relate to the question below,I would like to count the number of following rows.
Thanks to the answer,I could handle data.
But I met some trouble and exception.
How to count the number of following rows in pandas
A B
1 a0
2 a1
3 b1
4 a0
5 b2
6 a2
7 a2
First,I would like to cut df.with startswith("a")
df1
A B
1 a0
df2
A B
2 a1
3 b1
df3
A B
4 a0
5 b2
df4
A B
6 a2
df5
A B
7 a2
I would like to count each df's rows
"a" number
a0 1
a1 2
a0 2
a2 1
a2 1
How could be this done?
I am happy someone tell me how to handle this kind of problem.
You can use aggregate by custom Series created with cumsum:
print (df.B.str.startswith("a").cumsum())
0 1
1 2
2 2
3 3
4 3
5 4
6 5
Name: B, dtype: int32
df1 = df.B.groupby(df.B.str.startswith("a").cumsum()).agg(['first', 'size'])
df1.columns =['"A"','number']
df1.index.name = None
print (df1)
"A" number
1 a0 1
2 a1 2
3 a0 2
4 a2 1
5 a2 1