I have a dataframe that has segments of consecutive values appearing in column a (the value in column b does not matter):
import pandas as pd
import numpy as np
np.random.seed(150)
df = pd.DataFrame(data={'a':[1,2,3,4,5,15,16,17,18,203,204,205],'b':np.random.randint(50000,size=(12))})
>>> df
a b
0 1 27066
1 2 28155
2 3 49177
3 4 496
4 5 2354
5 15 23292
6 16 9358
7 17 19036
8 18 29946
9 203 39785
10 204 15843
11 205 21917
I would like to add a column c whose values are sequential counts according to presenting consecutive values in column a, as shown below:
a b c
1 27066 1
2 28155 2
3 49177 3
4 496 4
5 2354 5
15 23292 1
16 9358 2
17 19036 3
18 29946 4
203 39785 1
204 15843 2
205 21917 3
How to do this?
One solution:
df["c"] = (s := df["a"] - np.arange(len(df))).groupby(s).cumcount() + 1
print(df)
Output
a b c
0 1 27066 1
1 2 28155 2
2 3 49177 3
3 4 496 4
4 5 2354 5
5 15 23292 1
6 16 9358 2
7 17 19036 3
8 18 29946 4
9 203 39785 1
10 204 15843 2
11 205 21917 3
The original idea comes from ancient Python docs.
In order to use the walrus operator ((:=) or assignment expressions) you need Python 3.8+, instead you can do:
s = df["a"] - np.arange(len(df))
df["c"] = s.groupby(s).cumcount() + 1
print(df)
A simple solution is to find consecutive groups, use cumsum to get the number sequence and then remove any extra in later groups.
a = df['a'].add(1).shift(1).eq(df['a'])
df['c'] = a.cumsum() - a.cumsum().where(~a).ffill().fillna(0).astype(int) + 1
df
Result:
a b c
0 1 27066 1
1 2 28155 2
2 3 49177 3
3 4 496 4
4 5 2354 5
5 15 23292 1
6 16 9358 2
7 17 19036 3
8 18 29946 4
9 203 39785 1
10 204 15843 2
11 205 21917 3
Related
I have a dataframe with stores and its invoices numbers and I need to find the missing consecutive invoices numbers per Store, for example:
df1 = pd.DataFrame()
df1['Store'] = ['A','A','A','A','A','B','B','B','B','C','C','C','D','D']
df1['Invoice'] = ['1','2','5','6','8','20','23','24','30','200','202','203','204','206']
Store Invoice
0 A 1
1 A 2
2 A 5
3 A 6
4 A 8
5 B 20
6 B 23
7 B 24
8 B 30
9 C 200
10 C 202
11 C 203
12 D 204
13 D 206
And I want a dataframe like this:
Store MissInvoice
0 A 3
1 A 4
2 A 7
3 B 21
4 B 22
5 B 25
6 B 26
7 B 27
8 B 28
9 B 29
10 C 201
11 D 205
Thanks in advance!
You can use groupby.apply to compute a set difference with the range from the min to max value. Then explode:
(df1.astype({'Invoice': int})
.groupby('Store')['Invoice']
.apply(lambda s: set(range(s.min(), s.max())).difference(s))
.explode().reset_index()
)
NB. if you want to ensure having sorted values, use lambda s: sorted(set(range(s.min(), s.max())).difference(s)).
Output:
Store Invoice
0 A 3
1 A 4
2 A 7
3 B 21
4 B 22
5 B 25
6 B 26
7 B 27
8 B 28
9 B 29
10 C 201
11 D 205
Here's an approach:
import pandas as pd
import numpy as np
df1 = pd.DataFrame()
df1['Store'] = ['A','A','A','A','A','B','B','B','B','C','C','C']
df1['Invoice'] = ['1','2','5','6','8','20','23','24','30','200','202','203']
df1['Invoice'] = df1['Invoice'].astype(int)
df2 = df1.groupby('Store')['Invoice'].agg(['min','max'])
df2['MissInvoice'] = [[]]*len(df2)
for store,row in df2.iterrows():
df2.at[store,'MissInvoice'] = np.setdiff1d(np.arange(row['min'],row['max']+1),
df1.loc[df1['Store'] == store, 'Invoice'])
df2 = df2.explode('MissInvoice').drop(columns = ['min','max']).reset_index()
The resulting dataframe df2:
Store MissInvoice
0 A 3
1 A 4
2 A 7
3 B 21
4 B 22
5 B 25
6 B 26
7 B 27
8 B 28
9 B 29
10 C 201
Note: Store D is absent from the dataframe in my code because it is omitted from the lines in the question defining df1.
I am having trouble applying some logic across my entire dataset. I am able to apply the logic on a small "group" but not on all of the groups (note, the groups are made by primaryFilter and secondaryFilter. Do you all mind pointing me in the right direction to go about this?
Entire Data
import pandas as pd
import numpy as np
myInput = {
'primaryFilter': [100,100,100,100,100,100,100,100,100,100,200,200,200,200,200,200,200,200,200,200],
'secondaryFilter': [1,1,1,1,2,2,2,3,3,3,1,1,2,2,2,2,3,3,3,3],
'constantValuePerGroup': [15,15,15,15,20,20,20,17,17,17,10,10,30,30,30,30,22,22,22,22],
'someValue':[3,1,4,7,9,9,2,7,3,7,6,4,7,10,10,3,4,6,7,5]
}
df_input = pd.DataFrame(data=myInput)
df_input
Test Data (First Group)
df_test = df_input[df_input.primaryFilter.isin([100])]
df_test = df_test[df_test.secondaryFilter == 1.0]
df_test['newColumn'] = np.nan
for index,row in df_test.iterrows():
if index==0:
print("start")
df_test.loc[0, 'newColumn'] = 0
elif index==df_test.shape[0]-1:
df_test.loc[index, 'newColumn'] = df_test.loc[index-1, 'newColumn'] + df_test.loc[index-1, 'someValue']
print("end")
else:
print("inter")
df_test.loc[index, 'newColumn'] = df_test.loc[index-1, 'newColumn'] + df_test.loc[index-1, 'someValue']
df_test["delta"] = df_test["constantValuePerGroup"] - df_test['newColumn']
df_test.head()
Here is the output of the test
I now would like to apply the above logic to the remaining groups 100,2 and 100,3 and 200,1 and so forth..
No need to use iterrows here, you can group the dataframe on primaryFilter and secondaryFilter columns then for each unique group take the cumulative sum of values in column someValue and shift the resulting cummulative sum by 1 position downwards to obtain newColumn. Finally subtract newColumn from constantValuePerGroup to get the delta.
df_input['newColumn'] = df_input.groupby(['primaryFilter', 'secondaryFilter'])['someValue'].apply(lambda s: s.cumsum().shift(fill_value=0))
df_input['delta'] = df_input['constantValuePerGroup'] - df_input['newColumn']
>>> df_input
primaryFilter secondaryFilter constantValuePerGroup someValue newColumn delta
0 100 1 15 3 0 15
1 100 1 15 1 3 12
2 100 1 15 4 4 11
3 100 1 15 7 8 7
4 100 2 20 9 0 20
5 100 2 20 9 9 11
6 100 2 20 2 18 2
7 100 3 17 7 0 17
8 100 3 17 3 7 10
9 100 3 17 7 10 7
10 200 1 10 6 0 10
11 200 1 10 4 6 4
12 200 2 30 7 0 30
13 200 2 30 10 7 23
14 200 2 30 10 17 13
15 200 2 30 3 27 3
16 200 3 22 4 0 22
17 200 3 22 6 4 18
18 200 3 22 7 10 12
19 200 3 22 5 17 5
I have a dataframe1 like the following:
A B C D
1 111 a 9
2 121 b 8
3 122 c 7
4 121 d 6
5 131 e 5
Also, I have another dataframe2:
Code String
111 s
12 b
13 u
What I want is to creat a dataframe like the following:
A B C D
1 111 S 9
2 121 b 8
3 122 c 7
4 121 b 6
5 131 u 5
That would be, take the first n digits (where n is the number of digits in Code column of dataframe2) and if it has the same numbers that the code, then the column C in dataframe1 would change for the string in dataframe2.
Is this what you want ? The code is not very neat but work..
import pandas as pd
DICT=df2.set_index('Code').T.to_dict('list')
Temp=[]
for key, value in DICT.items():
n=len(str(key))
D1={str(key):value[0]}
T=df1.B.astype(str).apply(lambda x: x[:n]).map(D1)
Temp2=(df1.B.astype(str).apply(lambda x: x[:n]))
Tempdf=pd.DataFrame({'Ori':df1.B,'Now':Temp2,'C':df1.C})
TorF=(Tempdf.groupby(['Now'])['Ori'].transform(min) == Tempdf['Ori'])
for n, i in enumerate(T):
if TorF[n]==False:
T[n]=Tempdf.ix[n,0]
Temp.append(T)
df1.C=pd.DataFrame(data=Temp).fillna(method='bfill').T.ix[:,0]
Out[255]:
A B C D
0 1 111 s 9
1 2 121 b 8
2 3 122 c 7
3 4 121 b 6
4 5 131 u 5
I'm trying to get all records where the mean of the last 3 rows is greater than the overall mean for all rows in a filtered set.
_filtered_d_all = _filtered_d.iloc[:, 0:50].loc[:, _filtered_d.mean()>0.05]
_last_n_records = _filtered_d.tail(3)
Something like this
_filtered_growing = _filtered_d.iloc[:, 0:50].loc[:, _last_n_records.mean() > _filtered_d.mean()]
However, the problem here is that the value length is incorrect. Any tips?
ValueError: Series lengths must match to compare
Sample Data
This has an index on the year and month, and 2 columns.
Col1 Col2
year month
2005 12 0.533835 0.170679
12 0.494733 0.198347
2006 3 0.440098 0.202240
6 0.410285 0.188421
9 0.502420 0.200188
12 0.522253 0.118680
2007 3 0.378120 0.171192
6 0.431989 0.145158
9 0.612036 0.178097
12 0.519766 0.252196
2008 3 0.547705 0.202163
6 0.560985 0.238591
9 0.617320 0.199537
12 0.343939 0.253855
Why not just boolean index directly on your filtered DataFrame with
df[df.tail(3).mean() > df.mean()]
Demo
>>> df
0 1 2 3 4
0 4 8 2 4 6
1 0 0 0 2 8
2 5 3 0 9 3
3 7 5 5 1 2
4 9 7 8 9 4
>>> df[df.tail(3).mean() > df.mean()]
0 1 2 3 4
0 4 8 2 4 6
1 0 0 0 2 8
2 5 3 0 9 3
3 7 5 5 1 2
Update example for MultiIndex edit
The same should work fine for your MultiIndex sample, we just have to mask a bit differently of course.
>>> df
col1 col2
2005 12 -0.340088 -0.574140
12 -0.814014 0.430580
2006 3 0.464008 0.438494
6 0.019508 -0.635128
9 0.622645 -0.824526
12 -1.674920 -1.027275
2007 3 0.397133 0.659467
6 0.026170 -0.052063
9 0.835561 0.608067
12 0.736873 -0.613877
2008 3 0.344781 -0.566392
6 -0.653290 -0.264992
9 0.080592 -0.548189
12 0.585642 1.149779
>>> df.loc[:,df.tail(3).mean() > df.mean()]
col2
2005 12 -0.574140
12 0.430580
2006 3 0.438494
6 -0.635128
9 -0.824526
12 -1.027275
2007 3 0.659467
6 -0.052063
9 0.608067
12 -0.613877
2008 3 -0.566392
6 -0.264992
9 -0.548189
12 1.149779
I am new on pandas. I try to sort a column and group them by their numbers.
df = pd.read_csv("12Patients150526 mutations-ORIGINAL.txt", sep="\t", header=0)
samp=df["SAMPLE"]
samp
Out[3]:
0 11
1 2
2 9
3 1
4 8
5 2
6 1
7 3
8 10
9 4
10 5
..
53157 12
53158 3
53159 2
53160 10
53161 2
53162 3
53163 4
53164 11
53165 12
53166 11
Name: SAMPLE, dtype: int64
#sorting
grp=df.sort(samp)
This code does not work. Can somebody help me about my problem, please.
How can I sort and group them by their numbers?
To sort df based on a particular column, use df.sort() and pass column name as parameter.
import pandas as pd
import numpy as np
# data
# ===========================
np.random.seed(0)
df = pd.DataFrame(np.random.randint(1,10,1000), columns=['SAMPLE'])
df
SAMPLE
0 6
1 1
2 4
3 4
4 8
5 4
6 6
7 3
.. ...
992 3
993 2
994 1
995 2
996 7
997 4
998 5
999 4
[1000 rows x 1 columns]
# sort
# ======================
df.sort('SAMPLE')
SAMPLE
310 1
710 1
935 1
463 1
462 1
136 1
141 1
144 1
.. ...
174 9
392 9
386 9
382 9
178 9
772 9
890 9
307 9
[1000 rows x 1 columns]