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calculate the median age for each region from frequency table with python

I have a dataframe that is similar to:
I would like to calculate the median age for each city but given that it is a frequency table I'm finding it somewhat tricky. Is there any function in pandas or other that would help me achieve this?

Maybe this works for you:
import numpy as np
import pandas as pd
# create dataframe
df = pd.DataFrame(
[
['Alabama', 34, 67, 89, 89, 67, 545, 4546, 3, 23],
['Georgia', 345, 65, 67, 32, 23, 567, 87, 647, 68]
],
columns=['City', 0, 1, 2, 3, 4, 5, 6, 7, 8]
).set_index('City')
print(df)
# calculate median for freq table
m = list() # median list
for index, row in df.iterrows():
v = list() # value list
z = zip(row.index, row.values)
for item in z:
for f in range(item[1]):
v.append(item[0])
m.append(np.median(v))
df_m = pd.DataFrame({'City': df.index, 'Median': m})
print(df_m)
Input:
0 1 2 3 4 5 6 7 8
City
Alabama 34 67 89 89 67 545 4546 3 23
Georgia 345 65 67 32 23 567 87 647 68
Output:
City Median
0 Alabama 6.0
1 Georgia 5.0

For each row, find the number of instances there are. Then take that number, divide by 2, and determine what age that would be by checking if the number of people have the age smaller than what we are looking for.
For example, for the row 'alabama', you would add 34 + 67 + ... + 23 = 5463. That, divided by 2, would be 2731.5 ==> 2731. Then, checking each age group, determine where the 2731th person would be.
At age 1, since 2731 > 34, check the next.
At age 2, since 2731 > 34 + 67, check the next.
At age 3, since 2731 > 34 + 67 + 89, check the next.
At age 4, since 2731 > 34 + 67 + 89 + 89, check the next.
At age 5, since 2731 > 34 + 67 + 89 + 89 + 67, check the next.
At age 6, since 2731 > 34 + 67 + 89 + 89 + 67 + 545, check the next.
At age 7, since 2731 < 34 + 67 + 89 + 89 + 67 + 545 + 4546, the median has to be in this age group.
Do this repeatedly for each city/state, and you should get the median for each one.

make a change in each group after groupby

Let we have a dataframe like
names = ['bob', 'alice', 'bob', 'bob', 'ali', 'alice', 'bob', 'ali', 'moji', 'ali', 'moji', 'ali', 'bob', 'bob', 'bob']
times = [21 , 34, 37, 40, 55, 65, 67, 84, 88, 90 , 91, 97, 104,105, 108]
df = pd.DataFrame({'name' : names , 'time_of_action' : times})
We would like to groupby it by name and in each group consider times from 0. That is in each group we want to subtract min-time_of_action in that group from all times of that group. how could we do this systematically with pandas?

If I am correct then you want this:
df['new time'] = df['time_of_action']-df.groupby('name')['time_of_action'].transform('min')
df:
name time_of_action new time
0 bob 21 0
1 alice 34 0
2 bob 37 16
3 bob 40 19
4 ali 55 0
5 alice 65 31
6 bob 67 46
7 ali 84 29
8 moji 88 0
9 ali 90 35
10 moji 91 3
11 ali 97 42
12 bob 104 83
13 bob 105 84
14 bob 108 87

Try this
df['new_time'] = df.groupby('name')['time_of_action'].apply(lambda x: x - x.min())
df
Output:
name time_of_action new_time
0 bob 21 0
1 alice 34 0
2 bob 37 16
3 bob 40 19
4 ali 55 0
5 alice 65 31

Others have already answered but here's mine
import pandas as pd
names = ['bob', 'alice', 'bob', 'bob', 'ali', 'alice', 'bob', 'ali', 'moji', 'ali', 'moji', 'ali', 'bob', 'bob', 'bob']
times = [21 , 34, 37, 40, 55, 65, 67, 84, 88, 90 , 91, 97, 104,105, 108]
df = pd.DataFrame({'name' : names , 'time_of_action' : times})
def subtract_min(df):
df['new_time'] = df['time_of_action'] - df['time_of_action'].min()
return df
df.groupby('name').apply(subtract_min).sort_values('name')
As others have said I am kind of guessing as well

How to flatten pandas dataframe

Here is my pandas dataframe, and I would like to flatten. How can I do that ?
The input I have
key column
1 {'health_1': 45, 'health_2': 60, 'health_3': 34, 'health_4': 60, 'name': 'Tom'}
2 {'health_1': 28, 'health_2': 10, 'health_3': 42, 'health_4': 07, 'name': 'John'}
3 {'health_1': 86, 'health_2': 65, 'health_3': 14, 'health_4': 52, 'name': 'Adam'}
The expected output
All the health and name will become a column name of their own with their corresponding values. In no particular order.
health_1 health_2 health_3 health_4 name key
45 60 34 60 Tom 1
28 10 42 07 John 2
86 65 14 52 Adam 3

You can do it with one line solution,
df_expected = pd.concat([df, df['column'].apply(pd.Series)], axis = 1).drop('column', axis = 1)
Full version:
import pandas as pd
df = pd.DataFrame({"column":[
{'health_1': 45, 'health_2': 60, 'health_3': 34, 'health_4': 60, 'name': 'Tom'} ,
{'health_1': 28, 'health_2': 10, 'health_3': 42, 'health_4': 7, 'name': 'John'} ,
{'health_1': 86, 'health_2': 65, 'health_3': 14, 'health_4': 52, 'name': 'Adam'}
]})
df_expected = pd.concat([df, df['column'].apply(pd.Series)], axis = 1).drop('column', axis = 1)
print(df_expected)
DEMO: https://repl.it/repls/ButteryFrightenedFtpclient

This should work:
df['column'].apply(pd.Series)
Gives:
health_1 health_2 health_3 health_4 name
0 45 60 34 60 Tom
1 28 10 42 7 John
2 86 65 14 52 Adam

Try:
pd.concat([pd.DataFrame(i, index=[0]) for i in df.column], ignore_index=True)
Output:
health_1 health_2 health_3 health_4 name
0 45 60 34 60 Tom
1 28 10 42 7 John
2 86 65 14 52 Adam

The solutions using apply are going overboard. You can create your desired DataFrame using a list of dictionaries like you have in your column Series. You can easily get this list of dictionaries by using the tolist method:
res = pd.concat([df.key, pd.DataFrame(df.column.tolist())], axis=1)
print(res)
key health_1 health_2 health_3 health_4 name
0 1 45 60 34 60 Tom
1 2 28 10 42 7 John
2 3 86 65 14 52 Adam

Not sure I understand - This is the default format for a DataFrame?
import pandas as pd
df = pd.DataFrame([
{'health_1': 45, 'health_2': 60, 'health_3': 34, 'health_4': 60, 'name': 'Tom'} ,
{'health_1': 28, 'health_2': 10, 'health_3': 42, 'health_4': 7, 'name': 'John'} ,
{'health_1': 86, 'health_2': 65, 'health_3': 14, 'health_4': 52, 'name': 'Adam'}
])

Select every n values in pandas column

I would like to choose from a column every 3 values.
For example:
Input
12
73
56
33
16
output
12
73
56
------
73
56
33
-----
56
33
16
I have tried to add a key column and group by it, but my data frame is too large to perform the grouping. Here is my attempt:
df.groupby('key').agg(lambda x: x.tolist())

If use list type, you can do like this :
lst = [12,73,56,33,16]
slide_size=3
result = []
for i in range(0,len(lst)-slide_size+1):
result.append(lst[i:i+3])
result
# output : [[12, 73, 56], [73, 56, 33], [56, 33, 16]]
After this, you can transform the list to DataFrame

Pivot Table to Dictionary

I have this pivot table:
[in]:unit_d
[out]:
units
store_nbr item_nbr
1 9 27396
28 4893
40 254
47 2409
51 925
89 157
93 1103
99 492
2 5 55104
11 655
44 117125
85 106
93 653
I want to have a dictionary with 'store_nbr' as the key and 'item_nbr' as the values.
So, {'1': [9, 28, 40,...,99], '2': [5, 11 ,44, 85, 93], ...}

I'd use groupby here, after resetting the index to make it into columns:
>>> d = unit_d.reset_index()
>>> {k: v.tolist() for k, v in d.groupby("store_nbr")["item_nbr"]}
{1: [9, 28, 40, 47, 51, 89, 93, 99], 2: [5, 11, 44, 85, 93]}