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