Question

我们是新程序员，我们的问题似乎是 weight是一个对象，例如15'2，我们正试图将其变成一个数字，例如15.2，我们似乎无法弄清楚，我们尝试过的某些功能是

def value_to_int(df_value):
try:
    value = float(df_value[1:-1])
    suffix = df_value[-1:]
    if suffix == 'M':
        value = value * 1000000
    elif suffix == 'K':
        value = value * 1000
except ValueError:
    value = 0
return value
df['Value_float'] = df['Value'].apply(value_to_int)

TypeError                                 Traceback (most recent call last)
<ipython-input-143-9a5ed573b054> in <module>()
 14     return value
 15 
---> 16 df['Value_float'] = df['Height'].apply(value_to_int)

~\Anaconda3\lib\site-packages\pandas\core\series.py in apply(self, func, convert_dtype, args, **kwds)
   3192             else:
   3193                 values = self.astype(object).values
-> 3194                 mapped = lib.map_infer(values, f, convert=convert_dtype)
   3195 
   3196         if len(mapped) and isinstance(mapped[0], Series):

pandas/_libs/src\inference.pyx in pandas._libs.lib.map_infer()

<ipython-input-143-9a5ed573b054> in value_to_int(df_value)
  3 def value_to_int(df_value):
  4     try:
----> 5         value = float(df_value[1:-1])
  6         suffix = df_value[-1:]
  7 

TypeError: 'float' object is not subscriptable

Answer 1

如果要将“ 12'5”转换为浮点型，首先需要将“”替换为。因此您的代码将如下所示：

def value_to_int(df_value):
    try:
        value = float(df_value[1:-1].replace("'", "."))
        suffix = df_value[-1:]
        if suffix == 'M':
            value = value * 1000000
        elif suffix == 'K':
            value = value * 1000
    except ValueError:
        value = 0
    return value
df['Value_float'] = df['Value'].apply(value_to_int)

最好！

Answer 2

import pandas as pd
import re

df = pd.DataFrame(data={'val':["M15'2","K13'3", "17'4"]})
df['sex'] = df['val'].apply(lambda x: " ".join(re.findall("[a-zA-Z]+", x)))

# split string on sex
def split_str(row):
    if row['sex']:
        return row['val'].split(row['sex']).pop()
    else:
        return row['val']

df['weight'] = df.apply(lambda row: split_str(row), axis=1)
df['weight'] = df['weight'].apply(lambda x: x.replace("'", "."))
df['weight'] = df['weight'].astype(float)

# modify weight on sex
def multiply_weight(row):
    if row['sex'] == 'M':
        row['weight'] = row['weight'] * 1000000
    elif row['sex'] == 'K':
        row['weight'] = row['weight'] * 1000
    else:
        row['weight'] = 0
    return row

# return final df
df = df.apply(lambda row: multiply_weight(row), axis=1)

无法将另存为对象的重量测量值转换为int / double / numeric值

2 个答案: