ValueError：无法在位置 0 (Pandas) 解析字符串“2.90 3.50 2.35”

Question

我正在尝试将下面“odds_df”的“1x2”列中的数据转换为浮点数据类型。

到目前为止，我已经走到了死胡同，似乎没有任何效果……就我的研究和尝试而言。

我需要帮助来克服这个问题，我很高兴我问了:)

ps！我在之前的帖子中忽略了代码 sn-p，希望这次我做对了。

在此处输入代码

import pandas as pd
import pickle
import numpy as np

odds_df = pd.read_csv('betodds.csv')

odds_df.head()

    Teams                               1x2
0   Leicester City\nManchester United   2.90\n3.50\n2.35
1   Aston Villa\nCrystal Palace         1.90\n3.70\n3.80
2   Fulham FC\nSouthampton FC           3.00\n3.40\n2.30
3   Arsenal FC\nChelsea FC              3.80\n3.70\n1.90
4   Manchester City\nNewcastle United   1.10\n9.50\n22.00

odds_df['1x2'] = odds_df['1x2'].replace('\n','  ', regex=True).values.tolist()

odds_df['1x2'] 
0      2.90  3.50  2.35
1      1.90  3.70  3.80
2      3.00  3.40  2.30
3      3.80  3.70  1.90
4     1.10  9.50  22.00
5      4.40  3.80  1.75
6      1.70  4.00  4.50
7      2.30  3.30  3.10
------------------------
Name: 1x2, dtype: object

odds_df['1x2'] = pd.to_numeric(odds_df['1x2'])
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
pandas\_libs\lib.pyx in pandas._libs.lib.maybe_convert_numeric()

ValueError: Unable to parse string "2.90  3.50  2.35"

During handling of the above exception, another exception occurred:

ValueError                                Traceback (most recent call last)
<ipython-input-74-20aa27752bb5> in <module>
----> 1 odds_df['1x2'] = pd.to_numeric(odds_df['1x2'])

D:\Anaconda\envs\web_scraping\lib\site-packages\pandas\core\tools\numeric.py in to_numeric(arg, errors, downcast)
    150         coerce_numeric = errors not in ("ignore", "raise")
    151         try:
--> 152             values = lib.maybe_convert_numeric(
    153                 values, set(), coerce_numeric=coerce_numeric
    154             )

pandas\_libs\lib.pyx in pandas._libs.lib.maybe_convert_numeric()

ValueError: Unable to parse string "2.90  3.50  2.35" at position 0

Answer 1

注意：您可能需要将 \\\\ 替换为 \\。我从pd.read_clipboard() 读取数据的方式让我不得不使用\\\\。对于OP，OP使用\\：

您的列仍然是由空格分隔的字符串。如您的错误所示，您不能将“2.90 3.50 2.35”设为数字。您可以将这三个数字分成 3 个新列，然后用.str.split('\\\\n', expand=True).astype(float) 转换为float。请注意，您需要转义 \n 正则表达式，但 \\n 不适用于拆分，但 \\\\n 确实有效。如果你做df.to_dict()，会有两个斜线，所以你需要 dto 再逃避这两个，这就是为什么你有\\\\：

odds_df[['1x2_a', '1x2_b', '1x2_c']] = odds_df['1x2'].str.split('\\\\n', expand=True).astype(float)


Out[1]: 
                               Teams                1x2  1x2_a  1x2_b  1x2_c
0  Leicester City\nManchester United   2.90\n3.50\n2.35    2.9    3.5   2.35
1        Aston Villa\nCrystal Palace   1.90\n3.70\n3.80    1.9    3.7    3.8
2          Fulham FC\nSouthampton FC   3.00\n3.40\n2.30    3.0    3.4    2.3
3             Arsenal FC\nChelsea FC   3.80\n3.70\n1.90    3.8    3.7    1.9
4  Manchester City\nNewcastle United  1.10\n9.50\n22.00    1.1    9.5   22.0

从那里，您可以将每一列彼此分开：

s = odds_df['1x2'].str.split('\\\\n', expand=True).astype(float)
odds_df['1x2'] = s[0] / s[1] / s[2]
odds_df
Out[1]: 
                               Teams                  1x2
0  Leicester City\nManchester United  0.35258358662613976
1        Aston Villa\nCrystal Palace  0.13513513513513514
2          Fulham FC\nSouthampton FC   0.3836317135549872
3             Arsenal FC\nChelsea FC   0.5405405405405406
4  Manchester City\nNewcastle United 0.005263157894736843