I have a long form dataframe that contains multiple samples and time points for each subject. The number of samples and timepoint can vary, and the days between time points can also vary:
test_df = pd.DataFrame({"subject_id":[1,1,1,2,2,3],
"sample":["A", "B", "C", "D", "E", "F"],
"timepoint":[19,11,8,6,2,12],
"time_order":[3,2,1,2,1,1]
})
subject_id sample timepoint time_order
0 1 A 19 3
1 1 B 11 2
2 1 C 8 1
3 2 D 6 2
4 2 E 2 1
5 3 F 12 1
I need to figure out a way to generalize grouping this dataframe by subject_id and putting all samples and time points on the same row, in time order.
DESIRED OUTPUT:
subject_id sample1 timepoint1 sample2 timepoint2 sample3 timepoint3
0 1 C 8 B 11 A 19
1 2 E 2 D 6 null null
5 3 F 12 null null null null
Pivot gets me close, but I'm stuck on how to proceed from there:
test_df = test_df.pivot(index=['subject_id', 'sample'],
columns='time_order', values='timepoint')
Use DataFrame.set_index with DataFrame.unstack for pivoting, sorting MultiIndex in columns, flatten it and last convert subject_id to column:
df = (test_df.set_index(['subject_id', 'time_order'])
.unstack()
.sort_index(level=[1,0], axis=1))
df.columns = df.columns.map(lambda x: f'{x[0]}{x[1]}')
df = df.reset_index()
print (df)
subject_id sample1 timepoint1 sample2 timepoint2 sample3 timepoint3
0 1 C 8.0 B 11.0 A 19.0
1 2 E 2.0 D 6.0 NaN NaN
2 3 F 12.0 NaN NaN NaN NaN
a=test_df.iloc[:,:3].groupby('subject_id').last().add_suffix('1')
b=test_df.iloc[:,:3].groupby('subject_id').nth(-2).add_suffix('2')
c=test_df.iloc[:,:3].groupby('subject_id').nth(-3).add_suffix('3')
pd.concat([a, b,c], axis=1)
sample1 timepoint1 sample2 timepoint2 sample3 timepoint3
subject_id
1 C 8 B 11.0 A 19.0
2 E 2 D 6.0 NaN NaN
3 F 12 NaN NaN NaN NaN
Related
I have a margin table
item margin
0 a 3
1 b 4
2 c 5
and an item table
item sequence
0 a 1
1 a 2
2 a 3
3 b 1
4 b 2
5 c 1
6 c 2
7 c 3
I want to join the two table so that the margin will only be joined to the product with maximum sequence number, the desired outcome is
item sequence margin
0 a 1 NaN
1 a 2 NaN
2 a 3 3.0
3 b 1 NaN
4 b 2 4.0
5 c 1 NaN
6 c 2 NaN
7 c 3 5.0
How to achieve this?
Below is the code for margin and item table
import pandas as pd
df_margin=pd.DataFrame({"item":["a","b","c"],"margin":[3,4,5]})
df_item=pd.DataFrame({"item":["a","a","a","b","b","c","c","c"],"sequence":[1,2,3,1,2,1,2,3]})
One option would be to merge then replace extra values with NaN via Series.where:
new_df = df_item.merge(df_margin)
new_df['margin'] = new_df['margin'].where(
new_df.groupby('item')['sequence'].transform('max').eq(new_df['sequence'])
)
Or with loc:
new_df = df_item.merge(df_margin)
new_df.loc[new_df.groupby('item')['sequence']
.transform('max').ne(new_df['sequence']), 'margin'] = np.NAN
Another option would be to assign a temp column to both frames df_item with True where the value is maximal, and df_margin is True everywhere then merge outer and drop the temp column:
new_df = (
df_item.assign(
t=df_item
.groupby('item')['sequence']
.transform('max')
.eq(df_item['sequence'])
).merge(df_margin.assign(t=True), how='outer').drop('t', 1)
)
Both produce new_df:
item sequence margin
0 a 1 NaN
1 a 2 NaN
2 a 3 3.0
3 b 1 NaN
4 b 2 4.0
5 c 1 NaN
6 c 2 NaN
7 c 3 5.0
You could do:
df_item.merge(df_item.groupby('item')['sequence'].max().\
reset_index().merge(df_margin), 'left')
item sequence margin
0 a 1 NaN
1 a 2 NaN
2 a 3 3.0
3 b 1 NaN
4 b 2 4.0
5 c 1 NaN
6 c 2 NaN
7 c 3 5.0
Breakdown:
df_new = df_item.groupby('item')['sequence'].max().reset_index().merge(df_margin)
df_item.merge(df_new, 'left')
I have a dataframe as
col 1 col 2
A 2020-07-13
A 2020-07-15
A 2020-07-18
A 2020-07-19
B 2020-07-13
B 2020-07-19
C 2020-07-13
C 2020-07-18
I want it to become the following in a new dataframe
col_3 diff_btw_1st_2nd_date diff_btw_2nd_3rd_date diff_btw_3rd_4th_date
A 2 3 1
B 6 NaN NaN
C 5 NaN NaN
I tried getting the groupby at Col 1 level , but not getting the intended result. Can anyone help?
Use GroupBy.cumcount for counter pre column col 1 and reshape by DataFrame.set_index with Series.unstack, then use DataFrame.diff, remove first only NaNs columns by DataFrame.iloc, convert timedeltas to days by Series.dt.days per all columns and change columns names by DataFrame.add_prefix:
df['col 2'] = pd.to_datetime(df['col 2'])
df = (df.set_index(['col 1',df.groupby('col 1').cumcount()])['col 2']
.unstack()
.diff(axis=1)
.iloc[:, 1:]
.apply(lambda x: x.dt.days)
.add_prefix('diff_')
.reset_index())
print (df)
col 1 diff_1 diff_2 diff_3
0 A 2 3.0 1.0
1 B 6 NaN NaN
2 C 5 NaN NaN
Or use DataFrameGroupBy.diff with counter for new columns by DataFrame.assign, reshape by DataFrame.pivot and remove NaNs by c2 with DataFrame.dropna:
df['col 2'] = pd.to_datetime(df['col 2'])
df = (df.assign(g = df.groupby('col 1').cumcount(),
c1 = df.groupby('col 1')['col 2'].diff().dt.days)
.dropna(subset=['c1'])
.pivot('col 1','g','c1')
.add_prefix('diff_')
.rename_axis(None, axis=1)
.reset_index())
print (df)
col 1 diff_1 diff_2 diff_3
0 A 2.0 3.0 1.0
1 B 6.0 NaN NaN
2 C 5.0 NaN NaN
You can assign a cumcount number grouped by col 1, and pivot the table using that cumcount number.
Solution
df["col 2"] = pd.to_datetime(df["col 2"])
# 1. compute date difference in days using diff() and dt accessor
df["diff"] = df.groupby(["col 1"])["col 2"].diff().dt.days
# 2. assign cumcount for pivoting
df["cumcount"] = df.groupby("col 1").cumcount()
# 3. partial transpose, discarding the first difference in nan
df2 = df[["col 1", "diff", "cumcount"]]\
.pivot(index="col 1", columns="cumcount")\
.drop(columns=[("diff", 0)])
Result
# replace column names for readability
df2.columns = [f"d{i+2}-d{i+1}" for i in range(len(df2.columns))]
print(df2)
d2-d1 d3-d2 d4-d3
col 1
A 2.0 3.0 1.0
B 6.0 NaN NaN
C 5.0 NaN NaN
df after assing cumcount is like this
print(df)
col 1 col 2 diff cumcount
0 A 2020-07-13 NaN 0
1 A 2020-07-15 2.0 1
2 A 2020-07-18 3.0 2
3 A 2020-07-19 1.0 3
4 B 2020-07-13 NaN 0
5 B 2020-07-19 6.0 1
6 C 2020-07-13 NaN 0
7 C 2020-07-18 5.0 1
Cant be this hard. I Have
df=pd.DataFrame({'id':[1,2,3],'name':['j','l','m'], 'mnt':['f','p','p'],'nt':['b','w','e'],'cost':[20,30,80],'paid':[12,23,45]})
I need
import numpy as np
df1=pd.DataFrame({'id':[1,2,3,1,2,3],'name':['j','l','m','j','l','m'], 't':['f','p','p','b','w','e'],'paid':[12,23,45,np.nan,np.nan,np.nan],'cost':[20,30,80,np.nan,np.nan,np.nan]})
I have 45 columns to invert.
I tried
(df.set_index(['id', 'name'])
.rename_axis(['paid'], axis=1)
.stack().reset_index())
EDIT: I think simpliest here is set missing values by variable column in DataFrame.melt:
df2 = df.melt(['id', 'name','cost','paid'], value_name='t')
df2.loc[df2.pop('variable').eq('nt'), ['cost','paid']] = np.nan
print (df2)
id name cost paid t
0 1 j 20.0 12.0 f
1 2 l 30.0 23.0 p
2 3 m 80.0 45.0 p
3 1 j NaN NaN b
4 2 l NaN NaN w
5 3 m NaN NaN e
Use lreshape working with dictionary of lists for specified which columns are 'grouped' together:
df2 = pd.lreshape(df, {'t':['mnt','nt'], 'mon':['cost','paid']})
print (df2)
id name t mon
0 1 j f 20
1 2 l p 30
2 3 m p 80
3 1 j b 12
4 2 l w 23
5 3 m e 45
I am trying to populate column 'C' with values from column 'A' based on conditions in column 'B'. Example: If column 'B' equals 'nan', then row under column 'C' equals the row in column 'A'. If column 'B' does NOT equal 'nan', then leave column 'C' as is (ie 'nan'). Next, the values in column 'A' to be removed (only the values that were copied from column A to C).
Original Dataset:
index A B C
0 6 nan nan
1 6 nan nan
2 9 3 nan
3 9 3 nan
4 2 8 nan
5 2 8 nan
6 3 4 nan
7 3 nan nan
8 4 nan nan
Output:
index A B C
0 nan nan 6
1 nan nan 6
2 9 3 nan
3 9 3 nan
4 2 8 nan
5 2 8 nan
6 3 4 nan
7 nan nan 3
8 nan nan 4
Below is what I have tried so far, but its not working.
def impute_unit(cols):
Legal_Block = cols[0]
Legal_Lot = cols[1]
Legal_Unit = cols[2]
if pd.isnull(Legal_Lot):
return 3
else:
return Legal_Unit
bk_Final_tax['Legal_Unit'] = bk_Final_tax[['Legal_Block', 'Legal_Lot',
'Legal_Unit']].apply(impute_unit, axis = 1)
Seems like you need
df['C'] = np.where(df.B.isna(), df.A, df.C)
df['A'] = np.where(df.B.isna(), np.nan, df.A)
A different, maybe fancy way to do it would be to swap A and C values only when B is np.nan
m = df.B.isna()
df.loc[m, ['A', 'C']] = df.loc[m, ['C', 'A']].values
In other words, change
bk_Final_tax['Legal_Unit'] = bk_Final_tax[['Legal_Block', 'Legal_Lot',
'Legal_Unit']].apply(impute_unit, axis = 1)
for
bk_Final_tax['Legal_Unit'] = np.where(df.Legal_Lot.isna(), df.Legal_Block, df.Legal_Unit)
bk_Final_tax['Legal_Block'] = np.where(df.Legal_Lot.isna(), np.nan, df.Legal_Block)
I am trying to calculate the difference between rows based on multiple columns. The data set is very large and I am pasting dummy data below that describes the problem:
if I want to calculate the daily difference in weight at a pet+name level. So far I have only come up with the solution of concatenating these columns and creating multiindex based on the new column and the date column. But I think there should be a better way. In the real dataset I have more than 3 columns I am using calculate row difference.
df['pet_name']=df.pet + df.name
df.set_index(['pet_name','date'],inplace = True)
df.sort_index(inplace=True)
df['diffs']=np.nan
for idx in t.index.levels[0]:
df.diffs[idx] = df.weight[idx].diff()
Base on your description , you can try groupby
df['pet_name']=df.pet + df.name
df.groupby('pet_name')['weight'].diff()
Use groupby by 2 columns:
df.groupby(['pet', 'name'])['weight'].diff()
All together:
#convert dates to datetimes
df['date'] = pd.to_datetime(df['date'])
#sorting
df = df.sort_values(['pet', 'name','date'])
#get differences per groups
df['diffs'] = df.groupby(['pet', 'name', 'date'])['weight'].diff()
Sample:
np.random.seed(123)
N = 100
L = list('abc')
df = pd.DataFrame({'pet': np.random.choice(L, N),
'name': np.random.choice(L, N),
'date': pd.Series(pd.date_range('2015-01-01', periods=int(N/10)))
.sample(N, replace=True),
'weight':np.random.rand(N)})
df['date'] = pd.to_datetime(df['date'])
df = df.sort_values(['pet', 'name','date'])
df['diffs'] = df.groupby(['pet', 'name', 'date'])['weight'].diff()
df['pet_name'] = df.pet + df.name
df = df.sort_values(['pet_name','date'])
df['diffs1'] = df.groupby(['pet_name', 'date'])['weight'].diff()
print (df.head(20))
date name pet weight diffs pet_name diffs1
1 2015-01-02 a a 0.105446 NaN aa NaN
2 2015-01-03 a a 0.845533 NaN aa NaN
2 2015-01-03 a a 0.980582 0.135049 aa 0.135049
2 2015-01-03 a a 0.443368 -0.537214 aa -0.537214
3 2015-01-04 a a 0.375186 NaN aa NaN
6 2015-01-07 a a 0.715601 NaN aa NaN
7 2015-01-08 a a 0.047340 NaN aa NaN
9 2015-01-10 a a 0.236600 NaN aa NaN
0 2015-01-01 b a 0.777162 NaN ab NaN
2 2015-01-03 b a 0.871683 NaN ab NaN
3 2015-01-04 b a 0.988329 NaN ab NaN
4 2015-01-05 b a 0.918397 NaN ab NaN
4 2015-01-05 b a 0.016119 -0.902279 ab -0.902279
5 2015-01-06 b a 0.095530 NaN ab NaN
5 2015-01-06 b a 0.894978 0.799449 ab 0.799449
5 2015-01-06 b a 0.365719 -0.529259 ab -0.529259
5 2015-01-06 b a 0.887593 0.521874 ab 0.521874
7 2015-01-08 b a 0.792299 NaN ab NaN
7 2015-01-08 b a 0.313669 -0.478630 ab -0.478630
7 2015-01-08 b a 0.281235 -0.032434 ab -0.032434