您可能想尝试这样的事情:
# first extract precipitation data for later use
precipitation = [data[i][1] for i in xrange(0, rows)]
# then test the range (i, i+m)
all_dry = all([(data==0) for data in precipitation[i:i+m]])
all_wet = not any([(data==0) for data in precipitation[i:i+m]])
# of course you can also use
all_wet = all([(data>0) for data in precipitation[i:i+m]])
但是请注意,这种方法在测试相邻天数时会引入冗余计算,因此可能不适合处理大量数据。
编辑:
好的,这次让我们寻找更有效的方法。
# still extract precipitation data for later use first
precipitation = [data[i][1] for i in xrange(0, rows)]
# let's start our calculations by counting the longest consecutive dry days
consecutive_dry = [1 if data == 0 else 0 for data in precipitation]
for i in xrange(1, len(consecutive_dry))
if consecutive_dry[i] == 1:
consecutive_dry[i] += consecutive_dry[i - 1]
# then you will see, if till day i there're m consecutive dry days, then:
consecutive_dry[i] >= m # here is the test
# ...and it would be same for wet day testings.
这显然比上面的方法更有效:为了测试总共 N 天和 M 个连续范围,前一个需要 O(N * M) 操作来计算,而这个需要 O(N)。
再次编辑:
这是原始代码的编辑版本。由于您的代码可以运行,这也应在您的 PC 或其他设备上运行。
import numpy as np
data = np.loadtxt('Data Series.txt', usecols=(1,3))
dry = np.zeros(12)
wet = np.zeros(12)
rows,cols = data.shape #reading number of rows and columns into variables
# prepare
precipitation = [data[i][1] for i in xrange(0, rows)]
# collecting data for consecutive dry days
consecutive_dry = [1 if data == 0 else 1 for data in precipitation]
for i in xrange(1, len(consecutive_dry))
if consecutive_dry[i] == 1:
consecutive_dry[i] += consecutive_dry[i - 1]
# ...and for wet days
consecutive_wet = [1 if data > 0 else 0 for data in precipitation]
for i in xrange(1, len(consecutive_wet))
if consecutive_wet[i] == 1:
consecutive_wet[i] += consecutive_wet[i - 1]
# set your day range here.
day_range = 3
for i in xrange (0,rows):
if consecutive_dry[i] >= day_range:
month_id = data[i,0]
dry[month_id - 1] += 1
if consecutive_wet[i] >= day_range:
month_id = data[i,0]
wet[month_id - 1] += 1
print '3 Days Dry Spell\n', dry
print '3 Days Wet Spell\n', wet
请试试这个,如果有任何问题,请告诉我。