如何在python中拆分和删除字符串中间的数字？

Question

DataFrame

# |Name                |Price    |24h   |Volume(24h)
50|Maker50MKR          |$1,096.96|4,52  |$351,617,227
36|Decentraland36MANA  |$0.9754  |4,11  |$265,949,302
47|Bitcoin SV47BSV     |$60.38   |4,08  |$50,895,114
86|1inch Network861INCH|$0.7637  |3,74  |$72,279,229
38|Hedera38HBAR        |$0.07594 |3,72  |$58,825,304

期望结果

# |Name         |Ticker|Price    |24h   |Volume(24h)
50|Maker        |MKR   |$1,096.96|4,52  |$351,617,227
36|Decentraland |MANA  |$0.9754  |4,11  |$265,949,302
47|Bitcoin SV   |BSV   |$60.38   |4,08  |$50,895,114
86|1inch Network|1INCH |$0.7637  |3,74  |$72,279,229
38|Hedera       |HBAR  |$0.07594 |3,72  |$58,825,304

问题是：

没有固定数字/数字的字符串(0-100)

• overlap (例如，1英寸)

• 没有固定的滴答键

Answer 1

pd.DataFrame(df.apply(lambda x: x.Name.split(str(x["#"])), axis=1).values.tolist())

Answer 2

因此，您可以根据#列将当前名称拆分为名称和代码。下面的代码可能不是最好的代码，也不是最优的，但它确实可以满足您的需要.

也许一位熊猫大师可以对此进行优化。我对此很感兴趣。

# insert Ticker column
df.insert(df.columns.get_loc("Name")+1, "Ticker", None)


for index, row in df.iterrows():
    # split the thing based on '#' column and update the columns
    df.at[index, "Name"], df.at[index, "Ticker"] = row["Name"].split(str(row["#"]))

print(df)

由此产生的df：

    #           Name Ticker      Price   24h   Volume(24h)
0  50          Maker    MKR  $1,096.96  4,52  $351,617,227
1  36   Decentraland   MANA    $0.9754  4,11  $265,949,302
2  47     Bitcoin SV    BSV     $60.38  4,08   $50,895,114
3  86  1inch Network  1INCH    $0.7637  3,74   $72,279,229
4  38         Hedera   HBAR   $0.07594  3,72   $58,825,304

Answer 3

df['Name'] = df['Name'].apply(lambda name: re.search(r"^[a-zA-Z\s]+", name).group())