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
Related
I have a dataframe and I want to use columns to create new rows in a new dataframe.
>>> df_1
mix_id ngs phr d mp1 mp2 mp1_wt mp2_wt mp1_phr mp2_phr
2 M01 SBR2353 100.0 NaN MES/HPD SBR2353 0.253731 0.746269 25.373134 74.626866
3 M02 SBR2054 80.0 NaN TDAE SBR2054 0.264706 0.735294 21.176471 58.823529
I would like to have a dataframe like this.
>>> df_2
mix_id ngs phr d
1 M01 MES/HPD 25.373134 NaN
2 M01 SBR2353 74.626866 NaN
3 M02 TDAE 21.176471 NaN
4 M02 SBR2054 58.823529 NaN
IIUC
you can use pd.wide_to_long, it does however needs the repeating columns to have numbers as suffix. So, the first part of solution, just renames the columns to bring the number as suffix
df.columns=[col for col in df.columns[:6]] + [re.sub(r'\d','',col) + str(re.search(r'(\d)',col).group(0)) for col in df.columns[6:] ]
# this makes mp1_wt as mp_wt1, to support pd.wide_to_long
df2=pd.wide_to_long(df, stubnames=['mp','mp_wt','mp_phr'], i=['mix_id','ngs','d'], j='val').reset_index().drop(columns='val')
df2.drop(columns=['ngs','phr','mp_wt'], inplace=True)
df2.rename(columns={'mp':'ngs','mp_phr':'phr'}, inplace=True)
df2
mix_id d ngs phr
0 M01 NaN MES/HPD 25.373134
1 M01 NaN SBR2353 74.626866
2 M02 NaN TDAE 21.176471
3 M02 NaN SBR2054 58.823529
Given a small dataset df1 as follow:
city year quarter
0 sh 2019 q4
1 bj 2020 q3
2 bj 2020 q2
3 sh 2020 q4
4 sh 2020 q1
5 bj 2021 q1
I would like to create date range in quarter from 2019-q2 to 2021-q1 as column names, then check if each row in df1's year and quarter for each city exist in df2.
If they exist, then return ys for that cell, otherwise, return NaNs.
The final result will like:
city 2019-q2 2019-q3 2019-q4 2020-q1 2020-q2 2020-q3 2020-q4 2021-q1
0 bj NaN NaN NaN NaN y y NaN y
1 sh NaN NaN y y NaN NaN y NaN
To create column names for df2:
pd.date_range('2019-04-01', '2021-04-01', freq = 'Q').to_period('Q')
How could I achieve this in Python? Thanks.
We can use crosstab on city and the string concatenation of the year and quarter columns:
new_df = pd.crosstab(df['city'], df['year'].astype(str) + '-' + df['quarter'])
new_df:
col_0 2019-q4 2020-q1 2020-q2 2020-q3 2020-q4 2021-q1
city
bj 0 0 1 1 0 1
sh 1 1 0 0 1 0
We can convert to bool, replace False and True to be the correct values, reindex to add missing columns, and cleanup axes and index to get exact output:
col_names = pd.date_range('2019-01-01', '2021-04-01', freq='Q').to_period('Q')
new_df = (
pd.crosstab(df['city'], df['year'].astype(str) + '-' + df['quarter'])
.astype(bool) # Counts to boolean
.replace({False: np.NaN, True: 'y'}) # Fill values
.reindex(columns=col_names.strftime('%Y-q%q')) # Add missing columns
.rename_axis(columns=None) # Cleanup axis name
.reset_index() # reset index
)
new_df:
city 2019-q1 2019-q2 2019-q3 2019-q4 2020-q1 2020-q2 2020-q3 2020-q4 2021-q1
0 bj NaN NaN NaN NaN NaN y y NaN y
1 sh NaN NaN NaN y y NaN NaN y NaN
DataFrame and imports:
import numpy as np
import pandas as pd
df = pd.DataFrame({
'city': ['sh', 'bj', 'bj', 'sh', 'sh', 'bj'],
'year': [2019, 2020, 2020, 2020, 2020, 2021],
'quarter': ['q4', 'q3', 'q2', 'q4', 'q1', 'q1']
})
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
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
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)