Why won't second for loop execute correctly?-CodePudding

I'm trying to write two for loops that will return a score for different inputs, and create a new field with the new score. The first loop works fine but the second loop never returns the correct score.

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)

When I run script it returns the following:

print(score1)
     a  b  c
0  foo  1  0

print(score2)
     a  b
1  bar  3

Why doesn't score2 create the new field "c" or a score?

CodePudding user response：

Avoid the use of for loops to conditionally update DataFrame columns which are not Python lists. Use vectorized methods of Pandas and Numpy such as numpy.select which scales to millions of rows! Remember these data science tools calculate much differently than general use Python:

# 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)

CodePudding user response：

On the first iteration of second for loop, j will be in 3. so that none your condition satisfies.

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