为何第二个for循环无法为score2生成字段c及正确分数？

尝试编写两个for循环，为不同输入计算分数并创建新字段存储该分数。第一个循环可正常运行，但第二个循环始终无法返回正确分数，甚至未生成字段c。

原代码 #

import pandas as pd

d = {'a':['foo','bar'], 'b':[1,3]}

df = pd.DataFrame(d)

score1 = df.loc[df['a'] == 'foo']
score2 = df.loc[df['a'] == 'bar']

for i in score1['b']:
    if i < 3:
        score1['c'] = 0
    elif i <= 3 and i < 4:
        score1['c'] = 1
    elif i >= 4 and i < 5:
        score1['c'] = 2
    elif i >= 5 and i < 8:
        score1['c'] = 3
    elif i == 8:
        score1['c'] = 4

for j in score2['b']:
    if j < 2:
        score2['c'] = 0
    elif j <= 2 and i < 4:
        score2['c'] = 1
    elif j >= 4 and i < 6:
        score2['c'] = 2
    elif j >= 6 and i < 8:
        score2['c'] = 3
    elif j == 8:
        score2['c'] = 4
        
print(score1)
print(score2)

运行结果 #

print(score1)
     a  b  c
0  foo  1  0

print(score2)
     a  b
1  bar  3

• 变量混淆错误：第二个循环的条件判断里，错误使用了第一个循环的变量i而非当前循环的j。比如elif j <= 2 and i < 4，此时i是第一个循环的最后值（1），但j的值是3，所有条件都不满足，导致score2['c']从未被赋值。
• 冗余循环逻辑：score1和score2各只有一行数据，完全没必要用for循环遍历，反而容易引发变量错误。
• 视图赋值隐患：score1 = df.loc[df['a'] == 'foo']返回的是原DataFrame的切片视图，直接赋值可能触发SettingWithCopyWarning，属于不规范写法。

方案1：修正变量并简化逻辑 #

把第二个循环里的所有i替换成j，同时去掉冗余的循环（直接取单值判断）：

import pandas as pd

d = {'a':['foo','bar'], 'b':[1,3]}
df = pd.DataFrame(d)

# 用copy()避免视图赋值问题
score1 = df.loc[df['a'] == 'foo'].copy()
score2 = df.loc[df['a'] == 'bar'].copy()

# 处理score1
i_val = score1['b'].iloc[0]
if i_val < 3:
    score1['c'] = 0
elif 3 <= i_val < 4:
    score1['c'] = 1
elif 4 <= i_val < 5:
    score1['c'] = 2
elif 5 <= i_val < 8:
    score1['c'] = 3
elif i_val == 8:
    score1['c'] = 4

# 处理score2，修正变量和条件逻辑
j_val = score2['b'].iloc[0]
if j_val < 2:
    score2['c'] = 0
elif 2 <= j_val < 4:
    score2['c'] = 1
elif 4 <= j_val < 6:
    score2['c'] = 2
elif 6 <= j_val < 8:
    score2['c'] = 3
elif j_val == 8:
    score2['c'] = 4

print(score1)
print(score2)

方案2：用pandas矢量化操作（更推荐） #

避免循环，用numpy.select实现批量判断，符合pandas的最佳实践：

import pandas as pd
import numpy as np

d = {'a':['foo','bar'], 'b':[1,3]}
df = pd.DataFrame(d)

# 定义foo的分数规则
foo_conditions = [
    (df['a'] == 'foo') & (df['b'] < 3),
    (df['a'] == 'foo') & (3 <= df['b'] < 4),
    (df['a'] == 'foo') & (4 <= df['b'] < 5),
    (df['a'] == 'foo') & (5 <= df['b'] < 8),
    (df['a'] == 'foo') & (df['b'] == 8)
]
foo_values = [0,1,2,3,4]

# 定义bar的分数规则
bar_conditions = [
    (df['a'] == 'bar') & (df['b'] < 2),
    (df['a'] == 'bar') & (2 <= df['b'] < 4),
    (df['a'] == 'bar') & (4 <= df['b'] < 6),
    (df['a'] == 'bar') & (6 <= df['b'] < 8),
    (df['a'] == 'bar') & (df['b'] == 8)
]
bar_values = [0,1,2,3,4]

# 合并规则并计算分数
all_conditions = foo_conditions + bar_conditions
all_values = foo_values + bar_values
df['c'] = np.select(all_conditions, all_values, default=np.nan)

print(df.loc[df['a'] == 'foo'])
print(df.loc[df['a'] == 'bar'])

修复后，score2会正确生成字段c，值为1。

内容的提问来源于stack exchange，提问作者SupaDupa