python-3.x 为什么第二个for循环不能正确执行？

zd287kbt 于 2023-01-14 发布在 Python

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我试着写两个for循环，为不同的输入返回一个分数，然后用新的分数创建一个新的字段，第一个循环运行良好，但是第二个循环从来没有返回正确的分数。

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)

当我运行脚本时，它返回以下内容：

print(score1)
     a  b  c
0  foo  1  0

print(score2)
     a  b
1  bar  3

为什么score2不创建新字段“c”或一个分数？

python-3.x

来源：https://stackoverflow.com/questions/75103671/why-wont-second-for-loop-execute-correctly

2条答案

按热度按时间

nimxete21#

避免使用for循环来有条件地更新DataFrame中不是Python列表的列。使用Pandas和Numpy的矢量化方法，比如numpy.select，它可以扩展到数百万行!记住这些数据科学工具的计算方式与一般使用的Python有很大的不同：

# LIST OF BOOLEAN CONDITIONS
conds = [
    score1['b'].lt(3),                            # EQUIVALENT TO < 3
    score1['b'].between(3, 4, inclusive="left"),  # EQUIVALENT TO >= 3 or < 4
    score1['b'].between(4, 5, inclusive="left"),  # EQUIVALENT TO >= 4 or < 5
    score1['b'].between(5, 8, inclusive="left"),  # EQUIVALENT TO >= 5 or < 8
    score1['b'].eq(8)                             # EQUIVALENT TO == 8
]   

# LIST OF VALUES
vals = [0, 1, 2, 3, 4]

# VECTORIZED ASSIGNMENT
score1['c'] = numpy.select(conds, vals, default=numpy.nan)

# LIST OF BOOLEAN CONDITIONS
conds = [
    score2['b'].lt(2),
    score2['b'].between(2, 4, inclusive="left"),
    score2['b'].between(4, 6, inclusive="left"),
    score2['b'].between(6, 8, inclusive="left"),
    score2['b'].eq(8)
]   

# LIST OF VALUES
vals = [0, 1, 2, 3, 4]

# VECTORIZED ASSIGNMENT
score2['c'] = numpy.select(conds, vals, default=numpy.nan)

赞(0）回复(0）举报 2023-01-14

vsnjm48y2#

在第二个for循环的第一次迭代中，j将在3中，因此您的条件不满足。

for j in score2['b']:
    if j < 3:
        score2['c'] = 0
    elif j <= 3 and i < 5:
        score2['c'] = 1
    elif j >= 5 and i < 7:
        score2['c'] = 2
    elif j >= 7 and i < 9:
        score2['c'] = 3
    elif j == 9:
        score2['c'] = 4

赞(0）回复(0）举报 2023-01-14

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python-3.x 为什么第二个for循环不能正确执行？

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