Compare Two Columns Using Pandas

You can use .equals for columns or entire dataframes.

df['col1'].equals(df['col2'])

If they're equal, that statement will return True, else 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  

I'd like to add this answer for those who are trying to compare the equality of values in two columns that have NaN values, and get False when both values are NaN. By definition, NaN != NaN (See: numpy.isnan(value) not the same as value == numpy.nan?).

If you want the two NaN comparison to return True, you can use:

df['compare'] = (df["col_1"] == df["col_2"]) | (df["col_1"].isna() & df["col_2"].isna())

Use np.select if you have multiple conditions to be checked from the dataframe and output a specific choice in a different column

conditions=[(condition1),(condition2)]
choices=["choice1","chocie2"]

df["new column"]=np.select=(condtion,choice,default=)

Note: No of conditions and no of choices should match, repeat text in choice if for two different conditions you have same choices

Use lambda expression:

df[df.apply(lambda x: x['col1'] != x['col2'], axis = 1)]

You can use the Series method where:

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

Result:

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

I think the closest to the OP's intuition is an inline if statement:

df['que'] = (df['one'] if ((df['one'] >= df['two']) and (df['one'] <= df['three']))