pandas 如何将数据帧列乘以浮点常量？
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How do I multiply a dataframe column by a float constant?
提问by Krishna Nevase
I'm trying to multiply a column by a float. I have the code for it here:
我正在尝试将一列乘以一个浮点数。我在这里有它的代码：
if str(cMachineName)==str("K42"):
df_temp.loc[:, "P"] *= float((105.0* 59.0*math.pi*0.95/1000)/3540)
But it gives me this error:
但它给了我这个错误：
TypeError: can't multiply sequence by non-int of type 'float'.
How do I solve it?
我该如何解决？
回答by jezrael
I think problem is some non numeric values like 45as string:
我认为问题是一些非数字值，如45字符串：
Solution is converting to float, intby astype:
解决方案正在转换为float，int通过astype：
df_temp = pd.DataFrame({'P':[1,2.5,'45']})
print (df_temp['P'].dtype)
object
df_temp["P"] = df_temp["P"].astype(float)
df_temp["P"] *= float((105.0* 59.0*math.pi*0.95/1000)/3540)
print (df_temp)
P
0 0.005223
1 0.013057
2 0.235030
Another problem is non numeric data like gh, for converting is necessary to_numericwith errors='coerce'for converting them to NaNs:
另一个问题是非数字数据，如gh，转换是必要的to_numeric，errors='coerce'以便将它们转换为NaNs：
df_temp = pd.DataFrame({'P':[1,2.5,'gh']})
print (df_temp['P'].dtype)
object
df_temp["P"] = pd.to_numeric(df_temp["P"], errors='coerce')
print (df_temp)
P
0 1.0
1 2.5
2 NaN
df_temp["P"] *= float((105.0* 59.0*math.pi*0.95/1000)/3540)
print (df_temp)
P
0 0.005223
1 0.013057
2 NaN
