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Most efficient way of applying a function based on condition

Suppose we have a master dictionary master_dict = {"a": df1, "b": df2, "c": df3}. Now suppose we have a list called condition_list. Suppose func is a function that returns a new dictionary that has the original keys of master_dict along with potentially new keys.
What is the best way to get the below code to work when the length of condition_list is greater than 2:
if(len(condition_list) == 1):
df = master_dict[condition_list[0]]
else:
df = func(master_dict(condition_list[0]))
df = df[condition_list[1]]

You need to ask clearly. Declare input and output. And try to make a demo code. Anyway, use a loop.
for i in range(len(condition_list)):
if i==0: df = master_dict[condition_list[i]]
else: df = func(df)[condition_list[i]];
If the "df" is a dataframe of pandas, the conditions can be applied at once. Search "select dataframe with multiple conditions"

pandas apply function to each group (output is not really an aggregation)

I have a list of time-series (=pandas dataframe) and want to calculate for each time-series (of a device) the matrixprofile.
One option is to iterate all the devices - which seems to be slow.
A second option would be to group by the devices - and apply a UDF. The problem is now, that the UDF will return 1:1 rows i.e. not a single scalar value per group but the same number of rows will be outputted as the input.
Is it still possible to somehow vectorize this calculation for reach group when 1:1 (or at least non scalar values) are returned?
import pandas as pd
df = pd.DataFrame({
'foo':[1,2,3], 'baz':[1.1, 0.5, 4], 'bar':[1,2,1]
})
display(df)
print('***************************')
# slow version retaining all the rows
for g in df.bar.unique():
print(g)
this_group = df[df.bar == g]
# perform a UDF which needs to have all the values per group
# i.e. for real I want to calculate the matrixprofile for each time-series of a device
this_group['result'] = this_group.baz.apply(lambda x: 1)
display(this_group)
print('***************************')
def my_non_scalar1_1_agg_function(x):
display(pd.DataFrame(x))
return x
# neatly vectorized application of a non_scalar function
# but this fails as: Must produce aggregated value
df = df.groupby(['bar']).baz.agg(my_non_scalar1_1_agg_function)
display(df)

For non-aggregated functions applied to each distinct group that does not return a non-scalar value, you need to iterate method across groups and then compile together.
Therefore, consider a list or dict comprehension using groupby(), followed by concat. Be sure method inputs and returns a full data frame, series, or ndarray.
# LIST COMPREHENSION
df_list = [ myfunction(sub) for index, sub in df.groupby(['group_column']) ]
final_df = pd.concat(df_list)
# DICT COMPREHENSION
df_dict = { index: myfunction(sub) for index, sub in df.groupby(['group_column']) }
final_df = pd.concat(df_dict, ignore_index=True)

Indeed this (see also the link above in the comment) is a way to get it to work in a faster/more desired way. Perhaps there is even a better alternative
import pandas as pd
df = pd.DataFrame({
'foo':[1,2,3], 'baz':[1.1, 0.5, 4], 'bar':[1,2,1]
})
display(df)
grouped_df = df.groupby(['bar'])
altered = []
for index, subframe in grouped_df:
display(subframe)
subframe = subframe# obviously we need to apply the UDF here - not the idempotent operation (=doing nothing)
altered.append(subframe)
print (index)
#print (subframe)
pd.concat(altered, ignore_index=True)
#pd.DataFrame(altered)

Create a dictionary from pandas empty dataframe with only column names

I have a pandas data frame with only two column names( single row, which can be also considered as headers).I want to make a dictionary out of this with the first column being the value and the second column being the key.I already tried the
to.dict() method, but it's not working as it's an empty dataframe.
Example
df=|Land |Norway| to {'Land': Norway}
I can change the pandas data frame to some other type and find my way around it, but this question is mostly to learn the best/different/efficient approach for this problem.
For now I have this as the solution :
dict(zip(a.iloc[0:0,0:1],a.iloc[0:0,1:2]))
Is there any other way to do this?

Here's a simple way convert the columns to a list and a list to a dictionary
def list_to_dict(a):
it = iter(a)
ret_dict = dict(zip(it, it))
return ret_dict
df = pd.DataFrame([], columns=['Land', 'Normway'])
dict_val = list_to_dict(df.columns.to_list())
dict_val # {'Land': 'Normway'}

Very manual solution
df = pd.DataFrame(columns=['Land', 'Norway'])
df = pd.DataFrame({df.columns[0]: df.columns[1]}, index=[0])
If you have any number of columns and you want each sequential pair to have this transformation, try:
df = pd.DataFrame(dict(zip(df.columns[::2], df.columns[1::2])), index=[0])
Note: You will get an error if your DataFrame does not have at least two columns.

Vectorized Flag Assignment in Dataframe

I have a dataframe with observations possessing a number of codes. I want to compare the codes present in a row with a list. If any codes are in that list, I wish to flag the row. I can accomplish this using the itertuples method as follows:
import pandas as pd
import numpy as np
df = pd.DataFrame({ 'id' : [1,2,3,4,5],
'cd1' : ['abc1', 'abc2', 'abc3','abc4','abc5'],
'cd2' : ['abc3','abc4','abc5','abc6',''],
'cd3' : ['abc10', '', '', '','']})
code_flags = ['abc1','abc6']
# initialize flag column
df['flag'] = 0
# itertuples method
for row in df.itertuples():
if any(df.iloc[row.Index, 1:4].isin(code_flags)):
df.at[row.Index, 'flag'] = 1
The output correctly adds a flag column with the appropriate flags, where 1 indicates a flagged entry.
However, on my actual use case, this takes hours to complete. I have attempted to vectorize this approach using numpy.where.
df['flag'] = 0 # reset
df['flag'] = np.where(any(df.iloc[:,1:4].isin(code_flags)),1,0)
Which appears to evaluate everything the same. I think I'm confused on how the vectorization treats the index. I can remove the semicolon and write df.iloc[1:4] and obtain the same result.
Am I misunderstanding the where function? Is my indexing incorrect and causing a True evaluation for all cases? Is there a better way to do this?

Using np.where with .any not any(..)
np.where((df.iloc[:,1:4].isin(code_flags)).any(1),1,0)

Reading values from Pandas dataframe rows into equations and entering result back into dataframe

I have a dataframe. For each row of the dataframe: I need to read values from two column indexes, pass these values to a set of equations, enter the result of each equation into its own column index in the same row, go to the next row and repeat.
After reading the responses to similar questions I tried:
import pandas as pd
DF = pd.read_csv("...")
Equation_1 = f(x, y)
Equation_2 = g(x, y)
for index, row in DF.iterrows():
a = DF[m]
b = DF[n]
DF[p] = Equation_1(a, b)
DF[q] = Equation_2(a, b)
Rather than iterating over DF, reading and entering new values for each row, this codes iterates over DF and enters the same values for each row. I am not sure what I am doing wrong here.
Also, from what I have read it is actually faster to treat the DF as a NumPy array and perform the calculation over the entire array at once rather than iterating. Not sure how I would go about this.
Thanks.

Turns out that this is extremely easy. All that must be done is to define two variables and assign the desired columns to them. Then set "the row to be replaced" equivalent to the equation containing the variables.
Pandas already knows that it must apply the equation to every row and return each value to its proper index. I didn't realize it would be this easy and was looking for more explicit code.
e.g.,
import pandas as pd
df = pd.read_csv("...") # df is a large 2D array
A = df[0]
B = df[1]
f(A,B) = ....
df[3] = f(A,B)

# If your equations are simple enough, do operations column-wise in Pandas:
import pandas as pd
test = pd.DataFrame([[1,2],[3,4],[5,6]])
test # Default column names are 0, 1
test[0] # This is column 0
test.icol(0) # This is COLUMN 0-indexed, returned as a Series
test.columns=(['S','Q']) # Column names are easier to use
test #Column names! Use them column-wise:
test['result'] = test.S**2 + test.Q
test # results stored in DataFrame
# For more complicated stuff, try apply, as in Python pandas apply on more columns :
def toyfun(df):
return df[0]-df[1]**2
test['out2']=test[['S','Q']].apply(toyfun, axis=1)
# You can also define the column names when you generate the DataFrame:
test2 = pd.DataFrame([[1,2],[3,4],[5,6]],columns = (list('AB')))