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I am using Pandas 1.51 and I'm trying to get the rank of each row in a dataframe in a rolling window that looks ahead by employing FixedForwardWindowIndexer. But I can't make sense of the results. My code:
df = pd.DataFrame({"X":[9,3,4,5,1,2,8,7,6,10,11]})
window_size = 5
indexer = pd.api.indexers.FixedForwardWindowIndexer(window_size=window_size)
df.rolling(window=indexer).rank(ascending=False)
results:
X
0 5.0
1 4.0
2 1.0
3 2.0
4 3.0
5 1.0
6 1.0
7 NaN
8 NaN
9 NaN
10 NaN
By my reckoning, it should look like:
X
0 1.0 # based on the window [9,3,4,5,1], 9 is ranked 1st w/ascending = False
1 3.0 # based on the window [3,4,5,1,2], 3 is ranked 3rd
2 3.0 # based on the window [4,5,1,2,8], 4 is ranked 3rd
3 3.0 # etc
4 5.0
5 5.0
6 3.0
7 NaN
8 NaN
9 NaN
10 NaN
I am basing this on a backward-looking window, which works fine:
>>> df.rolling(window_size).rank(ascending=False)
X
0 NaN
1 NaN
2 NaN
3 NaN
4 5.0
5 4.0
6 1.0
7 2.0
8 3.0
9 1.0
10 1.0
Any assistance is most welcome.
Here is another way to do it:
df["rank"] = [
x.rank(ascending=False).iloc[0].values[0]
for x in df.rolling(window_size)
if len(x) == window_size
] + [pd.NA] * (window_size - 1)
Then:
print(df)
# Output
X rank
0 9 1.0
1 3 3.0
2 4 3.0
3 5 3.0
4 1 5.0
5 2 5.0
6 8 3.0
7 7 <NA>
8 6 <NA>
9 10 <NA>
10 11 <NA>
I am looking for a method to create an array of numbers to label groups, based on the value of the 'number' column. If it's possible?
With this abbreviated example DF:
number = [nan,nan,1,nan,nan,nan,2,nan,nan,3,nan,nan,nan,nan,nan,4,nan,nan]
df = pd.DataFrame(columns=['number'])
df = pd.DataFrame.assign(df, number=number)
Ideally I would like to make a new column, 'group', based on the int in column 'number' - so there would be effectively be array's of 1, ,2, 3, etc. FWIW, the DF is 1000's lines long, with sporadically placed int's.
The result would be a new column, something like this:
number group
0 NaN 0
1 NaN 0
2 1.0 1
3 NaN 1
4 NaN 1
5 NaN 1
6 2.0 2
7 NaN 2
8 NaN 2
9 3.0 3
10 NaN 3
11 NaN 3
12 NaN 3
13 NaN 3
14 NaN 3
15 4.0 4
16 NaN 4
17 NaN 4
All advice much appreciated!
You can use notna combined with cumsum:
df['group'] = df['number'].notna().cumsum()
NB. if you had zeros: df['group'] = df['number'].ne(0).cumsum().
output:
number group
0 NaN 0
1 NaN 0
2 1.0 1
3 NaN 1
4 NaN 1
5 NaN 1
6 2.0 2
7 NaN 2
8 NaN 2
9 3.0 3
10 NaN 3
11 NaN 3
12 NaN 3
13 NaN 3
14 NaN 3
15 4.0 4
16 NaN 4
17 NaN 4
You can use forward fill:
df['number'].ffill().fillna(0)
Output:
0 0.0
1 0.0
2 1.0
3 1.0
4 1.0
5 1.0
6 2.0
7 2.0
8 2.0
9 3.0
10 3.0
11 3.0
12 3.0
13 3.0
14 3.0
15 4.0
16 4.0
17 4.0
Name: number, dtype: float64
The objective is to fill NaN with respect to two columns (i.e., a, b) .
a b c d
2,0,1,4
5,0,5,6
6,0,1,1
1,1,1,4
4,1,5,6
5,1,5,6
6,1,1,1
1,2,2,3
6,2,5,6
Such that, there should be continous value of between 1 to 6 for the column a for a fixed value in column b. Then, the other rows assigned to nan.
The code snippet does the trick
import numpy as np
import pandas as pd
maxval_col_a=6
lowval_col_a=1
maxval_col_b=2
lowval_col_b=0
r=list(range(lowval_col_b,maxval_col_b+1))
df=pd.DataFrame(np.column_stack([[2,5,6,1,4,5,6,1,6,],
[0,0,0,1,1,1,1,2,2,], [1,5,1,1,5,5,1,2,5,],[4,6,1,4,6,6,1,3,6,]]),columns=['a','b','c','d'])
all_df=[]
for idx in r:
k=df.loc[df['b']==idx].set_index('a').reindex(range(lowval_col_a, maxval_col_a+1, 1)).reset_index()
k['b']=idx
all_df.append(k)
df=pd.concat(all_df)
But, I am curious whether there are more efficient and better way of doing this with Pandas.
The expected output
a b c d
0 1 0 NaN NaN
1 2 0 1.0 4.0
2 3 0 NaN NaN
3 4 0 NaN NaN
4 5 0 5.0 6.0
5 6 0 1.0 1.0
0 1 1 1.0 4.0
1 2 1 NaN NaN
2 3 1 NaN NaN
3 4 1 5.0 6.0
4 5 1 5.0 6.0
5 6 1 1.0 1.0
0 1 2 2.0 3.0
1 2 2 NaN NaN
2 3 2 NaN NaN
3 4 2 NaN NaN
4 5 2 NaN NaN
5 6 2 5.0 6.0
Create the cartesian product of combinations:
mi = pd.MultiIndex.from_product([df['b'].unique(), range(1, 7)],
names=['b', 'a']).swaplevel()
out = df.set_index(['a', 'b']).reindex(mi).reset_index()
print(out)
# Output
a b c d
0 1 0 NaN NaN
1 2 0 1.0 4.0
2 3 0 NaN NaN
3 4 0 NaN NaN
4 5 0 5.0 6.0
5 6 0 1.0 1.0
6 1 1 1.0 4.0
7 2 1 NaN NaN
8 3 1 NaN NaN
9 4 1 5.0 6.0
10 5 1 5.0 6.0
11 6 1 1.0 1.0
12 1 2 2.0 3.0
13 2 2 NaN NaN
14 3 2 NaN NaN
15 4 2 NaN NaN
16 5 2 NaN NaN
17 6 2 5.0 6.0
First create a multindex with cols [a,b] then a new multindex with all the combinations and then you reindex with the new multindex:
(showing all steps)
# set both a and b as index (it's a multiindex)
df.set_index(['a','b'],drop=True,inplace=True)
# create the new multindex
new_idx_a=np.tile(np.arange(0,6+1),3)
new_idx_b=np.repeat([0,1,2],6+1)
new_multidx=pd.MultiIndex.from_arrays([new_idx_a,
new_idx_b])
# reindex
df=df.reindex(new_multidx)
# convert the multindex back to columns
df.index.names=['a','b']
df.reset_index()
results:
a b c d
0 0 0 NaN NaN
1 1 0 NaN NaN
2 2 0 1.0 4.0
3 3 0 NaN NaN
4 4 0 NaN NaN
5 5 0 5.0 6.0
6 6 0 1.0 1.0
7 0 1 NaN NaN
8 1 1 1.0 4.0
9 2 1 NaN NaN
10 3 1 NaN NaN
11 4 1 5.0 6.0
12 5 1 5.0 6.0
13 6 1 1.0 1.0
14 0 2 NaN NaN
15 1 2 2.0 3.0
16 2 2 NaN NaN
17 3 2 NaN NaN
18 4 2 NaN NaN
19 5 2 NaN NaN
20 6 2 5.0 6.0
We can do it by using a groupby on the column b, then set a as index and add the missing values of a using numpy.arange.
To finish, reset the index to get the expected result :
import numpy as np
df.groupby('b').apply(lambda x : x.set_index('a').reindex(np.arange(1, 7))).drop('b', 1).reset_index()
Output :
b a c d
0 0 1 NaN NaN
1 0 2 1.0 4.0
2 0 3 NaN NaN
3 0 4 NaN NaN
4 0 5 5.0 6.0
5 0 6 1.0 1.0
6 1 1 1.0 4.0
7 1 2 NaN NaN
8 1 3 NaN NaN
9 1 4 5.0 6.0
10 1 5 5.0 6.0
11 1 6 1.0 1.0
12 2 1 2.0 3.0
13 2 2 NaN NaN
14 2 3 NaN NaN
15 2 4 NaN NaN
16 2 5 NaN NaN
17 2 6 5.0 6.0
i'm in this situation,
my df is like that
A B
0 0.0 2.0
1 3.0 4.0
2 NaN 1.0
3 2.0 NaN
4 NaN 1.0
5 4.8 NaN
6 NaN 1.0
and i want to apply this line of code:
df['A'] = df['B'].fillna(df['A'])
and I expect a workflow and final output like that:
A B
0 2.0 2.0
1 4.0 4.0
2 1.0 1.0
3 NaN NaN
4 1.0 1.0
5 NaN NaN
6 1.0 1.0
A B
0 2.0 2.0
1 4.0 4.0
2 1.0 1.0
3 2.0 NaN
4 1.0 1.0
5 4.8 NaN
6 1.0 1.0
but I receive this error:
TypeError: Unsupported type Series
probably because each time there is an NA it tries to fill it with the whole series and not with the single element with the same index of the B column.
I receive the same error with a syntax like that:
df['C'] = df['B'].fillna(df['A'])
so the problem seems not to be the fact that I'm first changing the values of A with the ones of B and then trying to fill the "B" NA with the values of a column that is technically the same as B
I'm in a databricks environment and I'm working with koalas data frames but they work as the pandas ones.
can you help me?
Another option
Suppose the following dataset
import pandas as pd
import numpy as np
df = pd.DataFrame(data={'State':[1,2,3,4,5,6, 7, 8, 9, 10],
'Sno Center': ["Guntur", "Nellore", "Visakhapatnam", "Biswanath", "Doom-Dooma", "Guntur", "Labac-Silchar", "Numaligarh", "Sibsagar", "Munger-Jamalpu"],
'Mar-21': [121, 118.8, 131.6, 123.7, 127.8, 125.9, 114.2, 114.2, 117.7, 117.7],
'Apr-21': [121.1, 118.3, 131.5, np.NaN, 128.2, 128.2, 115.4, 115.1, np.NaN, 118.3]})
df
State Sno Center Mar-21 Apr-21
0 1 Guntur 121.0 121.1
1 2 Nellore 118.8 118.3
2 3 Visakhapatnam 131.6 131.5
3 4 Biswanath 123.7 NaN
4 5 Doom-Dooma 127.8 128.2
5 6 Guntur 125.9 128.2
6 7 Labac-Silchar 114.2 115.4
7 8 Numaligarh 114.2 115.1
8 9 Sibsagar 117.7 NaN
9 10 Munger-Jamalpu 117.7 118.3
Then
df.loc[(df["Mar-21"].notnull()) & (df["Apr-21"].isna()), "Apr-21"] = df["Mar-21"]
df
State Sno Center Mar-21 Apr-21
0 1 Guntur 121.0 121.1
1 2 Nellore 118.8 118.3
2 3 Visakhapatnam 131.6 131.5
3 4 Biswanath 123.7 123.7
4 5 Doom-Dooma 127.8 128.2
5 6 Guntur 125.9 128.2
6 7 Labac-Silchar 114.2 115.4
7 8 Numaligarh 114.2 115.1
8 9 Sibsagar 117.7 117.7
9 10 Munger-Jamalpu 117.7 118.3
IIUC:
try with max():
df['A']=df[['A','B']].max(axis=1)
output of df:
A B
0 2.0 2.0
1 4.0 4.0
2 1.0 1.0
3 2.0 NaN
4 1.0 1.0
5 4.8 NaN
6 1.0 1.0
I want to select rows with groupby conditions.
import pandas as pd
import numpy as np
dftest = pd.DataFrame({'A':['Feb',np.nan,'Air','Flow','Feb',
'Beta','Cat','Feb','Beta','Air'],
'B':['s','s','t','s','t','s','t','t','t','t'],
'C':[5,4,3,2,1,7,6,5,4,3],
'D':[4,np.nan,3,np.nan,2,
np.nan,2,3,np.nan,7]})
def filcols3(df,dd):
if df.iloc[0]['D']==dd:
return df
dd=4
grp=dftest.groupby('B').apply(filcols3,dd)
the result of grp is:
A B C D
B
s 0 Feb s 5 4.0
1 NaN s 4 NaN
3 Flow s 2 NaN
5 Beta s 7 NaN
this is what I want.
while if I use the following code(part 2)
def filcols3(df,dd):
if df.iloc[0]['D']<=dd:
return df
dd=3
the result is:
A B C D
0 NaN NaN NaN NaN
1 NaN NaN NaN NaN
2 Air t 3.0 3.0
3 NaN NaN NaN NaN
4 Feb t 1.0 2.0
5 NaN NaN NaN NaN
6 Cat t 6.0 2.0
7 Feb t 5.0 3.0
8 Beta t 4.0 NaN
9 Air t 3.0 7.0
I'm surprise for this result, I mean to get
A B C D
2 Air t 3 3.0
4 Feb t 1 2.0
6 Cat t 6 2.0
7 Feb t 5 3.0
8 Beta t 4 NaN
9 Air t 3 7.0
what's wrong with the code of part 2? how to get the final result I want?
apply's behaviour is a little non-intuitive here, but if the idea is to filter out entire groups based on a specific condition per group, you can use GroupBy.transform and get a mask to filter df:
df[df.groupby('B')['D'].transform('first') <= 3]
A B C D
2 Air t 3 3.0
4 Feb t 1 2.0
6 Cat t 6 2.0
7 Feb t 5 3.0
8 Beta t 4 NaN
9 Air t 3 7.0
Or, fixing your code,
df[df.groupby('B')['D'].transform(lambda x: x.values[0] <= 3)]
A B C D
2 Air t 3 3.0
4 Feb t 1 2.0
6 Cat t 6 2.0
7 Feb t 5 3.0
8 Beta t 4 NaN
9 Air t 3 7.0
May Check with filter
dftest.groupby('B').filter(lambda x : any(x['D'].head(1)<=3))
Out[538]:
A B C D
2 Air t 3 3.0
4 Feb t 1 2.0
6 Cat t 6 2.0
7 Feb t 5 3.0
8 Beta t 4 NaN
9 Air t 3 7.0
Or without groupby drop_duplicates
s=df.drop_duplicates('B').D<=3
df[df.B.isin(df.loc[s.index,'B'][s])]
Out[550]:
A B C D
2 Air t 3 3.0
4 Feb t 1 2.0
6 Cat t 6 2.0
7 Feb t 5 3.0
8 Beta t 4 NaN
9 Air t 3 7.0