我试图找到某个值出现在一列中的次数.
我用data = pd.DataFrame.from_csv(‘data / DataSet2.csv’)制作了数据帧
现在我想找到某个列出现的次数.这是怎么做到的?
我以为是下面的,我在教育栏目中查看并计算时间?发生.
下面的代码显示我试图找到第9次出现的次数,错误是我运行代码时得到的
码
missing2 = df.education.value_counts()['9th']
print(missing2)
错误
KeyError: '9th'
最佳答案
您可以根据条件创建数据子集,然后使用
shape
或len:
print df
col1 education
0 a 9th
1 b 9th
2 c 8th
print df.education == '9th'
0 True
1 True
2 False
Name: education, dtype: bool
print df[df.education == '9th']
col1 education
0 a 9th
1 b 9th
print df[df.education == '9th'].shape[0]
2
print len(df[df['education'] == '9th'])
2
性能很有趣,最快的解决方案是比较numpy数组和sum:
码:
import perfplot, string
np.random.seed(123)
def shape(df):
return df[df.education == 'a'].shape[0]
def len_df(df):
return len(df[df['education'] == 'a'])
def query_count(df):
return df.query('education == "a"').education.count()
def sum_mask(df):
return (df.education == 'a').sum()
def sum_mask_numpy(df):
return (df.education.values == 'a').sum()
def make_df(n):
L = list(string.ascii_letters)
df = pd.DataFrame(np.random.choice(L, size=n), columns=['education'])
return df
perfplot.show(
setup=make_df,
kernels=[shape, len_df, query_count, sum_mask, sum_mask_numpy],
n_range=[2**k for k in range(2, 25)],
logx=True,
logy=True,
equality_check=False,
xlabel='len(df)')
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转载注明原文:Python Pandas计算特定值的出现次数 - 乐贴网