Python & Pandas:构造lambda参数

Question

我正在用Python和Pandas编写一个脚本，它使用lambda语句在csv列中根据分配给每一行的数字等级编写预先格式化的注释。虽然我可以做一些系列，但我有困难的一个案例。

这里是csv的is结构：

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下面是编写新列composition_comment的工作代码。(我确信有一种更简洁的表达方式，但我仍然在学习Python和Pandas。)

import pandas as pd

df = pd.read_csv('stack.csv')
composition_score_value = 40  #calculated by another process

composition_comment_a_level = "Good work." # For scores falling between 100 and 90 percent of composition_score_value.
composition_comment_b_level = "Satisfactory work." # For scores between 89 and 80.
composition_comment_c_level = "Improvement needed." # For scores between 79 and 70.
composition_comment_d_level = "Unsatisfactory work." # For scores below 69.

df['composition_comment'] = df['composition_score'].apply(lambda element: composition_comment_a_level if element <= (composition_score_value * 1) else element >= (composition_score_value *.90))
df['composition_comment'] = df['composition_score'].apply(lambda element: composition_comment_b_level if element <= (composition_score_value *.899) else element >= (composition_score_value *.80))
df['composition_comment'] = df['composition_score'].apply(lambda element: composition_comment_c_level if element <= (composition_score_value *.799) else element >= (composition_score_value *.70))
df['composition_comment'] = df['composition_score'].apply(lambda element: composition_comment_d_level if element <= (composition_score_value *.699) else element >= (composition_score_value *.001))

df 

df.to_csv('stack.csv', index=False)

预期产出如下：

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但实际产出是：

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对于为什么要编写True值，以及为什么最后一行处理正确，有什么想法吗？任何帮助都很感激。

Answer 1

虽然许多其他选项显示了如何改进apply操作，但我建议使用pd.cut

df['composition_comment'] = pd.cut(
    df['composition_score'] / composition_score_value,  # Divide to get percent
    bins=[0, 0.7, 0.8, 0.9, np.inf],                    # Set Bounds
    labels=[composition_comment_d_level,                # Set Labels
            composition_comment_c_level,
            composition_comment_b_level,
            composition_comment_a_level],
    right=False                                         # Set Lower bound inclusive
)

df

   composition_score   composition_comment
0                 40            Good work.
1                 35    Satisfactory work.
2                 31   Improvement needed.
3                 27  Unsatisfactory work.

*设置right=False使低界包含，这意味着回收箱是：

[0.0, 0.7)  # 0.0 (inclusive) up to 0.7 (not inclusive)
[0.7, 0.8)  # 0.7 (inclusive) up to 0.8 (not inclusive)
[0.8, 0.9)  # 0.8 (inclusive) up to 0.9 (not inclusive)
[0.9, inf)  # 0.9 (inclusive) up to infinity

备注：

如果有一个集合上界，可以修改

1. inf。如果小于0的值是预期的1.
2. np.NINF，则1将不作为right=False的上限工作，因为1严格地不小于

可以使用的值而不是下限。

主要的好处是，有一个明确的，这是返回。这意味着像sort_values这样的操作将排序，而不是按字母顺序排序，而是按类别排序。

['Unsatisfactory work.' < 'Improvement needed.' < 'Satisfactory work.' < 'Good work.']

df = df.sort_values('composition_comment')

df

   composition_score   composition_comment
3                 27  Unsatisfactory work.
2                 31   Improvement needed.
1                 35    Satisfactory work.
0                 40            Good work.

程序设置：

import numpy as np
import pandas as pd

df = pd.DataFrame({'composition_score': [40, 35, 31, 27]})
composition_score_value = 40  # calculated by another process

# For scores falling between 100 and 90 percent of composition_score_value.
composition_comment_a_level = "Good work."
# For scores between 89 and 80.
composition_comment_b_level = "Satisfactory work."
# For scores between 79 and 70.
composition_comment_c_level = "Improvement needed."
# For Scores below 70
composition_comment_d_level = "Unsatisfactory work."

Answer 2

它覆盖了您在以上代码行中所做的工作。当它看到满足element <= (composition_score_value *.699)的值时，它返回composition_comment_d_level。如果值不满足这一要求，则返回element >= (composition_score_value *.001)，它本质上是一个布尔值，为True或False。

这一项应能发挥作用：

def composition_comment(element):
    composition_score_value = 40
    if element <= (composition_score_value *.699) :
        return "Unsatisfactory work."  
    elif element <= (composition_score_value *.799):
        return "Improvement needed."
    elif  element <= (composition_score_value *.899):
        return "Satisfactory work."
    elif element <= (composition_score_value * 1):
        return "Good work."
    else:
        return None

df['composition_comment'] = df['composition_score'].apply(composition_comment)
df

Answer 3

您可以使用pd.cut()来执行映射，这比在if-else语句中显式写出每个条件要好得多：

import pandas as pd

df = pd.DataFrame({'composition_score': [40, 35, 31, 27]})
composition_score_value = 40

bins = pd.IntervalIndex.from_tuples([(0, 70), (70,80), (80,90), (90,100)])
labels = ['Unsatisfactory work.','Improvement needed.','Satisfactory work.','Good work.']
d = dict(zip(bins,labels))
x = pd.cut(df['composition_score']/composition_score_value*100, bins, right=False).map(d)

产量：

0              Good work.
1      Satisfactory work.
2     Improvement needed.
3    Unsatisfactory work.
Name: composition_score, dtype: category
Categories (4, object): ['Unsatisfactory work.' < 'Improvement needed.' < 'Satisfactory work.' < 'Good work.']