如何在python中计算数据帧中的行数？

Question

import numpy as np
import pandas as pd 

year=np.array([1851.203,1851.632,1851.969,1851.975,1852.314,1852.347,1852.358,1852.385,1852.977,1853.196,1853.229,1853.319,1853.5,1854.135,1856.396,1856.506,1856.539,1856.618,1857.138,1857.404])
coal=pd.DataFrame(year)

def disaster(year1,year2):
    disaster = (coal >= year1) & (coal <= year2)
    print year.sum()

disaster(1852,1856)
>>>360806.155

我想写一个函数，在两年之间给我多少年。就像我把年份18521856加上所有的年份，但是我想知道在这个间隔之间有多少行（整数）。在

例如，1851是4次，1852是5次，当我把间隔（18511852）放在一起时，它会加起来，并给出9。我想要这个两年。在

有人能帮忙吗？在

Answer 1

def disaster(year1,year2):
    return  np.sum((coal >= year1) & (coal <= year2+1))

print disaster(1851,1852)

>>> 
0    9
dtype: int64

Answer 2

我根据一些假设修改了你的代码。我想你要做的是用你的面具蒙版成原始的测向图，打印或求和，另外得到长度。要获得原始df使用掩码所需的长度，请调用dropna()删除NaN行，然后对结果df调用len：

In [16]:

year=np.array([1851.203,1851.632,1851.969,1851.975,1852.314,1852.347,1852.358,1852.385,1852.977,1853.196,1853.229,1853.319,1853.5,1854.135,1856.396,1856.506,1856.539,1856.618,1857.138,1857.404])
coal=pd.DataFrame(year)

def disaster(year1,year2):
    total = (coal >= year1) & (coal <= year2)
    print (coal[total].sum())
    print(len(coal[total].dropna()))

disaster(1852,1856)
0    18529.76
dtype: float64
10