我使用pandas让我的数据看起来像下面代码中的dict。
我想找到所有salsa类型,并将它们放在一个dict中,其中包含salsa类型为字典值的项目数。
这是在Python中。有没有办法在熊猫中做这样的事情?或者这是我应该迭代并使用plain-ole-Python的任务吗?
#!/usr/bin/env python3
import pandas as pd
items_df = pd.DataFrame({'choice_description': {0: '[Tomatillo Red Chili Salsa, [Fajita Vegetables, Black Beans, Pinto Beans, Cheese, Sour Cream, Guacamole, Lettuce]]', 1: '[Tomatillo-Red Chili Salsa (Hot), [Black Beans, Rice, Cheese, Sour Cream]]', 2: '[Fresh Tomato Salsa (Mild), [Rice, Cheese, Sour Cream, Guacamole, Lettuce]]', 3: '[Tomatillo Red Chili Salsa, [Fajita Vegetables, Black Beans, Pinto Beans, Cheese, Sour Cream, Guacamole, Lettuce]]'}, 'item_name': {0: 'Chips and Fresh Tomato Salsa', 1: 'Chips and Tomatillo-Green Chili Salsa', 2: 'Chicken Bowl', 3: 'Steak Burrito'}})
salsa_types_d = {}
for row in items_df.itertuples():
for food in row[1:]:
fixed_foods_l = food.replace("and",',').replace('[','').replace(']','').split(',')
fixed_foods_l = [f.strip() for f in fixed_foods_l if f.find("alsa") > -1]
for fixed_food in fixed_foods_l:
salsa_types_d[fixed_food] = salsa_types_d.get(fixed_food, 0) + 1
print('\n'.join("%-33s:%d" % (k,salsa_types_d[k]) for k in sorted(salsa_types_d,key=salsa_types_d.get,reverse=True)))
"""
Output:
Tomatillo Red Chili Salsa :2
Fresh Tomato Salsa :1
Fresh Tomato Salsa (Mild) :1
Tomatillo-Green Chili Salsa :1
Tomatillo-Red Chili Salsa (Hot) :1
---
Thank you for any insight.
Marilyn
"""
答案 0 :(得分:1)
这可以在不使用for循环的情况下完成,其中一种方法是通过stacking
列创建单独的df,然后replacing the values
之后dropping the values
创建不包含alsa
的{{1}} }。然后最后使用value_counts
来获取频率。
new_df = items_df.stack().reset_index(drop=True)
.replace(['and', '\[', '\]'],[',', '',''], regex=True).str.split(',')
.apply(lambda x: pd.Series([i.lstrip() for i in x if 'alsa' in i]))[0].value_counts()
输出:
Tomatillo Red Chili Salsa 2 Tomatillo-Green Chili Salsa 1 Tomatillo-Red Chili Salsa (Hot) 1 Fresh Tomato Salsa (Mild) 1 Fresh Tomato Salsa 1 Name: 0, dtype: int64