Reputation: 97
Having a dataframe consisting of a person and the order...
person order elements
Alice [drink, snack, salad, fish, dessert] 5
Tom [drink, snack] 2
John [drink, snack, soup, chicken] 4
Mila [drink, snack, soup] 3
I want to known what customers had as a main meal. Thus, I want to add another column [main_meal] so that would be my df.
person order elements main_meal
Alice [drink, snack, salad, fish, dessert] 5 fish
Tom [drink, snack] 2 none
John [drink, snack, soup, chicken] 4 chicken
Mila [drink, snack, soup] 3 none
The rule is that if a customer ordered 4 or more meals, it means the 4th element is always the main dish, so I want to extract the 4th element from the list on the order column. If it contains less than 4 elements, then assign 'main_meal' to none. My code:
df['main_meal'] = ''
if df['elements'] >= 4:
df['main_meal'] = df.order[3]
else:
df['main_meal'] = 'none'
It doesn't work:
ValueError Traceback (most recent call last)
<ipython-input-100-39b7809cc669> in <module>()
1 df['main_meal'] = ''
2 df.head(5)
----> 3 if df['elements'] >= 4:
4 df['main_meal'] = df.order[3]
5 else:
~\Anaconda\lib\site-packages\pandas\core\generic.py in __nonzero__(self)
1571 raise ValueError("The truth value of a {0} is ambiguous. "
1572 "Use a.empty, a.bool(), a.item(), a.any() or
a.all()."
-> 1573 .format(self.__class__.__name__))
1574
1575 __bool__ = __nonzero__
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
What is wrong with my code?
Upvotes: 2
Views: 4975
Reputation: 354
You can also use the apply
and lambda functions.
df['main_meal'] = df['order'].apply(lambda r: r[3] if len(r) >= 4 else 'none')
It's slower than @jpp's answer for large datasets but faster (and more verbose) than both @jpp's and @Zero's answers for smaller ones (note that I added .fillna()
so they return the same result):
%timeit df['order'].apply(lambda r: r[3] if len(r) >= 4 else 'none') # 242 µs
%timeit pd.DataFrame(df['order'].values.tolist())[3].fillna('none') # 1.17 ms
%timeit df['order'].str[3].fillna('none') # 487 µs
# Large dataset
df = pd.concat([df]*100000)
%timeit df['order'].apply(lambda r: r[3] if len(r) >= 4 else 'none') # 118ms
%timeit pd.DataFrame(df['order'].values.tolist())[3].fillna('none') # 51.8ms
%timeit df['order'].str[3].fillna('none') # 309ms
And if you check their values they'll match.
x = df['order'].apply(lambda r: r[3] if len(r) >= 4 else 'none')
y = pd.DataFrame(df['order'].values.tolist())[3].fillna('none')
z = df['order'].str[3].fillna('none')
(x.values == y.values).all() # True
(x.values == z.values).all() # True
Python 3.6.6 | pandas 0.23.4
Upvotes: 0
Reputation: 164803
For small dataframes, you can use the str
accessor as per @Zero's solution. For larger dataframes, you may wish to use the NumPy representation to create a series:
# Benchmarking on Python 3.6, Pandas 0.19.2
df = pd.concat([df]*100000)
%timeit pd.DataFrame(df['order'].values.tolist())[3] # 125 ms per loop
%timeit df['order'].str[3] # 185 ms per loop
# check results align
x = pd.DataFrame(df['order'].values.tolist())[3].fillna('None').values
y = df['order'].str[3].fillna('None').values
assert (x == y).all()
Upvotes: 1
Reputation: 77027
Use str
method to slice
In [324]: df['order'].str[3]
Out[324]:
0 fish
1 NaN
2 chicken
3 NaN
Name: order, dtype: object
In [328]: df['main_meal'] = df['order'].str[3].fillna('none')
In [329]: df
Out[329]:
person order elements main_meal
0 Alice [drink, snack, salad, fish, dessert] 5 fish
1 Tom [drink, snack] 2 none
2 John [drink, snack, soup, chicken] 4 chicken
3 Mila [drink, snack, soup] 3 none
Upvotes: 4