Compare two columns using pandas

You could use np.where. If cond is a boolean array, and A and B are arrays, then

C = np.where(cond, A, B)

defines C to be equal to A where cond is True, and B where cond is False.

import numpy as np
import pandas as pd

a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])

df['que'] = np.where((df['one'] >= df['two']) & (df['one'] <= df['three'])
                     , df['one'], np.nan)

yields

  one  two three  que
0  10  1.2   4.2   10
1  15   70  0.03  NaN
2   8    5     0  NaN

If you have more than one condition, then you could use np.select instead. For example, if you wish df['que'] to equal df['two'] when df['one'] < df['two'], then

conditions = [
    (df['one'] >= df['two']) & (df['one'] <= df['three']), 
    df['one'] < df['two']]

choices = [df['one'], df['two']]

df['que'] = np.select(conditions, choices, default=np.nan)

yields

  one  two three  que
0  10  1.2   4.2   10
1  15   70  0.03   70
2   8    5     0  NaN

If we can assume that df['one'] >= df['two'] when df['one'] < df['two'] is False, then the conditions and choices could be simplified to

conditions = [
    df['one'] < df['two'],
    df['one'] <= df['three']]

choices = [df['two'], df['one']]

(The assumption may not be true if df['one'] or df['two'] contain NaNs.)

Note that

a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])

defines a DataFrame with string values. Since they look numeric, you might be better off converting those strings to floats:

df2 = df.astype(float)

This changes the results, however, since strings compare character-by-character, while floats are compared numerically.

In [61]: '10' <= '4.2'
Out[61]: True

In [62]: 10 <= 4.2
Out[62]: False

You could use apply() and do something like this

df['que'] = df.apply(lambda x : x['one'] if x['one'] >= x['two'] and x['one'] <= x['three'] else "", axis=1)

or if you prefer not to use a lambda

def que(x):
    if x['one'] >= x['two'] and x['one'] <= x['three']:
        return x['one']
    return ''
df['que'] = df.apply(que, axis=1)

One way is to use a Boolean series to index the column df['one']. This gives you a new column where the True entries have the same value as the same row as df['one'] and the False values are NaN.

The Boolean series is just given by your if statement (although it is necessary to use & instead of and):

>>> df['que'] = df['one'][(df['one'] >= df['two']) & (df['one'] <= df['three'])]
>>> df
    one two three   que
0   10  1.2 4.2      10
1   15  70  0.03    NaN
2   8   5   0       NaN

If you want the NaN values to be replaced by other values, you can use the fillna method on the new column que. I've used 0 instead of the empty string here:

>>> df['que'] = df['que'].fillna(0)
>>> df
    one two three   que
0   10  1.2   4.2    10
1   15   70  0.03     0
2    8    5     0     0

Wrap each individual condition in parentheses, and then use the & operator to combine the conditions:

df.loc[(df['one'] >= df['two']) & (df['one'] <= df['three']), 'que'] = df['one']

You can fill the non-matching rows by just using ~ (the "not" operator) to invert the match:

df.loc[~ ((df['one'] >= df['two']) & (df['one'] <= df['three'])), 'que'] = ''

You need to use & and ~ rather than and and not because the & and ~ operators work element-by-element.

The final result:

df
Out[8]: 
  one  two three que
0  10  1.2   4.2  10
1  15   70  0.03    
2   8    5     0