如何提高计算速度str到float

Question

df['distance'].iloc[0]

output: '0.02790952'

type(df['distance'].iloc[0])

output: str

df.shape

(118884, 40)

I try to parse a string to a float

for i in tqdm(range(len(df['distance']))):
    df['distance'].iloc[i] = float(df['distance'].iloc[i])

Answer 1

Here is a quick comparison between several methods. I used a single column DataFrame, with 1 million rows.

Using .apply(np.float)

%%timeit 
df[0].apply(np.float)
476 ms ± 6.65 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Using .astype(float)

%%timeit
df[0].astype(float)
336 ms ± 2.66 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Using pd.to_numeric

%%timeit
pd.to_numeric(df[0])
244 ms ± 2.28 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Using pd.Series and dtype argument

%%timeit
pd.Series(df[0], dtype=np.float64)
333 ms ± 2.88 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

The winner is pd.to_numeric , To anyone who reads this, if you think of a faster way, please comment!