我在一列中有一些字串，最初使用逗號作為千位和小數的分隔符，我需要將此字串轉換為浮點數，我該怎么做？
我首先嘗試將所有逗號替換為點：
df['min'] = df['min'].str.replace(',', '.')
并嘗試轉換為浮點數：
df['min']= df['min'].astype(float)
但它回傳了以下錯誤：
ValueError Traceback (most recent call last)
<ipython-input-29-5716d326493c> in <module>
----> 1 df['min']= df['min'].astype(float)
2 #df['mcom']= df['mcom'].astype(float)
3 #df['max']= df['max'].astype(float)
~\anaconda3\lib\site-packages\pandas\core\generic.py in astype(self, dtype, copy, errors)
5544 else:
5545 # else, only a single dtype is given
-> 5546 new_data = self._mgr.astype(dtype=dtype, copy=copy, errors=errors,)
5547 return self._constructor(new_data).__finalize__(self, method="astype")
5548
~\anaconda3\lib\site-packages\pandas\core\internals\managers.py in astype(self, dtype, copy, errors)
593 self, dtype, copy: bool = False, errors: str = "raise"
594 ) -> "BlockManager":
--> 595 return self.apply("astype", dtype=dtype, copy=copy, errors=errors)
596
597 def convert(
~\anaconda3\lib\site-packages\pandas\core\internals\managers.py in apply(self, f, align_keys, **kwargs)
404 applied = b.apply(f, **kwargs)
405 else:
--> 406 applied = getattr(b, f)(**kwargs)
407 result_blocks = _extend_blocks(applied, result_blocks)
408
~\anaconda3\lib\site-packages\pandas\core\internals\blocks.py in astype(self, dtype, copy, errors)
593 vals1d = values.ravel()
594 try:
--> 595 values = astype_nansafe(vals1d, dtype, copy=True)
596 except (ValueError, TypeError):
597 # e.g. astype_nansafe can fail on object-dtype of strings
~\anaconda3\lib\site-packages\pandas\core\dtypes\cast.py in astype_nansafe(arr, dtype, copy, skipna)
993 if copy or is_object_dtype(arr) or is_object_dtype(dtype):
994 # Explicit copy, or required since NumPy can't view from / to object.
--> 995 return arr.astype(dtype, copy=True)
996
997 return arr.view(dtype)
ValueError: could not convert string to float: '1.199.75'
如果可能的話，我想洗掉所有點和逗號，然后在轉換為浮點數之前在變數的最后兩個字符之前添加點。
輸入：
df['min'].head()
9.50
10.00
3.45
1.095.50
13.25
預期輸出：
9.50
10.00
3.45
1095.50
13.25
uj5u.com熱心網友回復：
如果你總是有 2 位小數：
df['min'] = pd.to_numeric(df['min'].str.replace('.', '', regex=False)).div(100)
輸出（為清楚起見作為新列 min2）：
min min2
0 9.50 9.50
1 10.00 10.00
2 3.45 3.45
3 1.095.50 1095.50
4 13.25 13.25
uj5u.com熱心網友回復：
嘗試這個。
df['min'] = df['min'].str.replace(',', '')
df['min'] = df['min'].str[:-2] '.' df['min'].str[-2:]
df['min']= df['min'].astype(float)
希望能幫助到你
轉載請註明出處，本文鏈接：https://www.uj5u.com/yidong/491385.html