Create all pairs combinations of columns names, loop and divide to new columns:

from  itertools import combinations

for a, b in combinations(df.columns, 2):
    df[f'{a}/{b}'] = df[a].div(df[b])

Or use list comprehension, join together by concat and add original columns by join:

df = df.join(pd.concat([df[a].div(df[b]).rename(f'{a}/{b}') 
                         for a, b in combinations(df.columns, 2)], 1))

print (df)
   x1  x2  x3  x4     x1/x2     x1/x3     x1/x4     x2/x3     x2/x4     x3/x4
0   4   7   1   5  0.571429  4.000000  0.800000  7.000000  1.400000  0.200000
1   5   8   3   3  0.625000  1.666667  1.666667  2.666667  2.666667  1.000000
2   4   9   5   6  0.444444  0.800000  0.666667  1.800000  1.500000  0.833333
3   5   4   7   9  1.250000  0.714286  0.555556  0.571429  0.444444  0.777778
4   5   2   1   2  2.500000  5.000000  2.500000  2.000000  1.000000  0.500000
5   4   3   0   4  1.333333       inf  1.000000       inf  0.750000  0.000000

We can take advantage of the way that Boolean values are handled mathematically (True being 1 and False being 0) and use 3 aggregation functions sum, count and mean per group (groupby aggregate). We can also take advantage of Named Aggregation to both create and rename the columns in one step:

df = (
    df.groupby('playerId', as_index=False)
        .agg(wins=('winner', 'sum'),
             totalCount=('winner', 'count'),
             winPct=('winner', 'mean'))
)
# Scale up winPct
df['winPct'] *= 100

df:

   playerId  wins  totalCount  winPct
0      1848     1           2    50.0
1      1988     0           2     0.0
2      3543     1           1   100.0

DataFrame and imports:

import pandas as pd

df = pd.DataFrame({
    'playerId': [1848, 1988, 3543, 1848, 1988],
    'winner': [True, False, True, False, False]
})

Try (with the original df):

df.groupby("Department")["Status"].value_counts(normalize=True).mul(100)

Outputs:

Department  Status
Finance     Contractor    100.000000
Marketing   Contractor     50.000000
            Employee       50.000000
Sales       Employee       66.666667
            Contractor     33.333333
Name: Status, dtype: float64

This should do:

df['price_per_sqft'] = df['SalePrice']/df['LotArea']

or you can use pd.assign

df.assign(price_per_sqft = df.SalePrice/df.LotArea)

If your columns are all named consistently using the exact pattern in your question, you can do something like this:

for i in range(1, 101):
    df['ratio_b{}'.format(i)] = df['b{}_amt'.format(i)] / df['total_balance']

I think you need convert string columns to float or int, because their type is string (but looks like numbers):

Apple_farm['Good_apples'] = Apple_farm['Good_apples'].astype(float)
Apple_farm['Total_apples'] = Apple_farm['Total_apples'].astype(float)

Apple_farm['Good_apples'] = Apple_farm['Good_apples'].astype(int)
Apple_farm['Total_apples'] = Apple_farm['Total_apples'].astype(int)

Sample:

import pandas as pd

Good_apples = ["10", "20", "3", "7", "9"]
Total_apples = ["20", "80", "30", "70", "90"]
d = {"Good_apples": Good_apples, "Total_apples": Total_apples}
Apple_farm = pd.DataFrame(d)
print Apple_farm 
  Good_apples Total_apples
0          10           20
1          20           80
2           3           30
3           7           70
4           9           90

print Apple_farm.dtypes
Good_apples     object
Total_apples    object
dtype: object

print Apple_farm.at[0,'Good_apples']
10

print type(Apple_farm.at[0,'Good_apples'])
<type 'str'>

Apple_farm['Good_apples'] = Apple_farm['Good_apples'].astype(int)
Apple_farm['Total_apples'] = Apple_farm['Total_apples'].astype(int)

print Apple_farm.dtypes
Good_apples     int32
Total_apples    int32
dtype: object

print Apple_farm.at[0,'Good_apples']
10

print type(Apple_farm.at[0,'Good_apples'])
<type 'numpy.int32'>

Apple_farm['Perc_Good'] = (Apple_farm['Good_apples'] / Apple_farm['Total_apples']) *100

print Apple_farm
   Good_apples  Total_apples  Perc_Good
0           10            20       50.0
1           20            80       25.0
2            3            30       10.0
3            7            70       10.0
4            9            90       10.0

If I understand you correctly, you are describing the following:

Setup

d = {'id1': ['x', 'x'], 'id2': ['z','w'], 'metric': [100,10] }
df = pd.DataFrame(data=d)
df

Solution

# Manually choose the value by which to scale the column 'metric'
scaler = df.loc[(df['id1'] == 'x') & (df['id2'] == 'z'), 'metric'].values

# Divide all 'metric' values by the above scaler value
df['result'] = df['metric'] / scaler

df
  id1 id2  metric  result
0   x   z     100     1.0
1   x   w      10     0.1

Looks like you want the ratio of y1 to the total instead. Use groupby + value_counts:

v = df.groupby('AAA').BBB.value_counts().unstack()
df['RATIO'] = df.AAA.map(v.y2 / (v.y2 + v.y1))

  AAA BBB CCC  DDD     RATIO
0  x1  y1  t1   10  0.333333
1  x1  y1  t2   11  0.333333
2  x1  y2  t3   18  0.333333
3  x2  y2  t1   17  0.666667
4  x2  y2  t1   21  0.666667
5  x2  y1  t1   30  0.666667

To generalise for many groups, you may use

df['RATIO'] = df.AAA.map(v.y2 / v.sum(axis=1))

Idea is create MultiIndex and then add one year for datetimes in rename, divide columns and add suffix by DataFrame.add_suffix, last add to original by DataFrame.join:

df1 = df.set_index(['Date','Route'])

df2 = df1.rename(lambda x: x + pd.offsets.DateOffset(years=1), level=0)

df3 = df1[['Col1','Col2']].div(df2[['Col1','Col2']]).add_suffix('_ratio')

df = df.join(df3, on=['Date','Route'])
print (df)
         Date Route  Col1  Col2  Col1_ratio  Col2_ratio
0  2020-01-01     A    99    26         NaN         NaN
1  2020-01-02     A    37    96         NaN         NaN
2  2020-01-03     A    20    83         NaN         NaN
3  2021-01-01     A    50    79    0.505051    3.038462
4  2021-01-02     A    16    50    0.432432    0.520833
5  2021-01-03     A    52    33    2.600000    0.397590
6  2020-01-01     B    48    44         NaN         NaN
7  2020-01-02     B    96    30         NaN         NaN
8  2020-01-03     B    17    42         NaN         NaN
9  2021-01-01     B    34    74    0.708333    1.681818
10 2021-01-02     B    49    37    0.510417    1.233333
11 2021-01-03     B    70    12    4.117647    0.285714

Just subset your data frame using that ratio:

df[df$Male / df$Female > 0.5, ]

     Cars Male Female
2 vw golf   76     12
3 ford ka   34     55

Demo