As a part of model requirement, I am stuck on weird spot where I need to replace actual value with Nan for extra quarters.
In the below example,
Id 1 should have nan in column Q4, 2 should have no nan and 3 should have Q3 and Q4 both as nan.
d = {'ID': [1, 2,3], 'QTR_req': [3,4,2],'Q1':[1,1,1],'Q2':[2,2,2],'Q3':[3,3,3],'Q4':[4,4,4]}
df2 = pd.DataFrame(data=d)
I have reached till the part of accessing QTR_req using df.loc but then stuck on how to make specific quarter nan. Could you suggest what am I looking for here?
May be this:
df2[cols_needed] = (df2[ cols_needed ]
.where(df2['QTR_req'].values[:,None] >np.arange(len(cols_needed )) )
)
Output:
ID QTR_req Q1 Q2 Q3 Q4
0 1 3 1 2 3.0 NaN
1 2 4 1 2 3.0 4.0
2 3 2 1 2 NaN NaN
Related
I have a dataframe of the following scheme in pyspark:
user_id datadate page_1.A page_1.B page_1.C page_2.A page_2.B \
0 111 20220203 NaN NaN NaN NaN NaN
1 222 20220203 5 5 5 5.0 5.0
2 333 20220203 3 3 3 3.0 3.0
page_2.C page_3.A page_3.B page_3.C
0 NaN 1.0 1.0 2.0
1 5.0 NaN NaN NaN
2 4.0 NaN NaN NaN
So it contains columns like user_id, datadate, and few columns for each page (got 3 pages), which are the result of 2 joins. In this example, i have page_1, page_2, page_3, and each has 3 columns: A,B,C. Additionally, for each page columns, for each row, they will either be all null or all full, like in my example.
I don't care about the values of each of the columns per page, I just want to get for each row, the [A,B,C] values that are not null.
example for a wanted result table:
user_id datadate A B C
0 111 20220203 1 1 2
1 222 20220203 5 5 5
2 333 20220203 3 3 3
so the logic will be something like:
df[A] = page_1.A or page_2.A or page_3.A, whichever is not null
df[B] = page_1.B or page_2.B or page_3.B, whichever is not null
df[C] = page_1.C or page_2.C or page_3.C, whichever is not null
for all of the rows..
and of course, I would like to do it in an efficient way.
Thanks a lot.
You can use the sql functions greatest to extract the greatest values in a list of columns.
You can find the documentation here: https://spark.apache.org/docs/3.1.1/api/python/reference/api/pyspark.sql.functions.greatest.html
from pyspark.sql import functions as F
(df.withColumn('A', F.greates(F.col('page_1.A'), F.col('page_2.A), F.col('page_3.A'))
.withColumn('B', F.greates(F.col('page_1.B'), F.col('page_2.B), F.col('page_3.B'))
.select('userid', 'datadate', 'A', 'B'))
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 have a dataframe that contains nan values in particular column. while iterating through the rows, if it come across nan(using isnan() method) then I need to change it to some other value(since I have some conditions). I tried using replace() and fillna() with limit parameter also but they are modifying whole column when they come across the first nan value? Is there any method that I can assign value to specific nan rather than changing all the values of a column?
Example: the dataframe looks like it:
points sundar cate king varun vicky john charlie target_class
1 x2 5 'cat' 4 10 3 2 1 NaN
2 x3 3 'cat' 1 2 3 1 1 NaN
3 x4 6 'lion' 8 4 3 7 1 NaN
4 x5 4 'lion' 1 1 3 1 1 NaN
5 x6 8 'cat' 10 10 9 7 1 0.0
an I have a list like
a = [1.0, 0.0]
and I expect to be like
points sundar cate king varun vicky john charlie target_class
1 x2 5 'cat' 4 10 3 2 1 1.0
2 x3 3 'cat' 1 2 3 1 1 1.0
3 x4 6 'lion' 8 4 3 7 1 1.0
4 x5 4 'lion' 1 1 3 1 1 0.0
5 x6 8 'cat' 10 10 9 7 1 0.0
I wanted to change the target_class values based on some conditions and assign values of the above list.
I believe need replace NaNs values to 1 only for indexes specified in list idx:
mask = df['target_class'].isnull()
idx = [1,2,3]
df.loc[mask, 'target_class'] = df[mask].index.isin(idx).astype(int)
print (df)
points sundar cate king varun vicky john charlie target_class
1 x2 5 'cat' 4 10 3 2 1 1.0
2 x3 3 'cat' 1 2 3 1 1 1.0
3 x4 6 'lion' 8 4 3 7 1 1.0
4 x5 4 'lion' 1 1 3 1 1 0.0
5 x6 8 'cat' 10 10 9 7 1 0.0
Or:
idx = [1,2,3]
s = pd.Series(df.index.isin(idx).astype(int), index=df.index)
df['target_class'] = df['target_class'].fillna(s)
EDIT:
From comments solution is assign values by index and columns values with DataFrame.loc:
df2.loc['x2', 'target_class'] = list1[0]
I suppose your conditions for imputing the nan values does not depend on the number of them in a column. In the code below I stored all the imputation rules in one function that receives as parameters the entire row (containing the nan) and the column you are investigating for. If you also need all the dataframe for the imputation rules, just pass it through the replace_nan function. In the example I imputate the col element with the mean values of the other columns.
import pandas as pd
import numpy as np
def replace_nan(row, col):
row[col] = row.drop(col).mean()
return row
df = pd.DataFrame(np.random.rand(5,3), columns = ['col1', 'col2', 'col3'])
col_to_impute = 'col1'
df.loc[[1, 3], col_to_impute] = np.nan
df = df.apply(lambda x: replace_nan(x, col_to_impute) if np.isnan(x[col_to_impute]) else x, axis=1)
The only thing that you should do is making the right assignation. That is, make an assignation in the rows that contain nulls.
Example dataset:
,event_id,type,timestamp,label
0,asd12e,click,12322232,0.0
1,asj123,click,212312312,0.0
2,asd321,touch,12312323,0.0
3,asdas3,click,33332233,
4,sdsaa3,touch,33211333,
Note: The last two rows contains nulls in column: 'label'. Then, we load the dataset:
df = pd.read_csv('dataset.csv')
Now, we make the appropiate condition:
cond = df['label'].isnull()
Now, we make the assignation over these rows (I don't know the logical of assignation. Therefore I assign 1 value to NaN's):
df1.loc[cond,'label'] = 1
There are another more accurate approaches. fillna() method could be used. You should provide the logical in order to help you.
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)