I have got a list containing nested lists like this :
[ [datetime.datetime(2000, 12, 10, 0, 0), 0.0011] , [datetime.datetime(2000, 12, 11, 0, 0), 0.0013 , [datetime.datetime(2000, 12, 12, 0, 0), 0.0014]]
etc..
How do I go about adding sub elements 2 by 2 like this :
sum(0.0011,0.0013) + 0.0014
then taking the result of this sum and adding it to the next sub element ?
I`m basically trying to compound the values .
thanks!
The easiest way to do this is with the sum() builtin and a generator expression:
>>>items = [[datetime.datetime(2000, 12, 10, 0, 0), 0.0011], [datetime.datetime(2000, 12, 11, 0, 0), 0.0013 ], [datetime.datetime(2000, 12, 12, 0, 0), 0.0014]]
>>>sum(item[1] for item in items)
0.0038000000000000004
Edit:
If you want to print out the result of each stage of the summation, you want to use functools.reduce() (which, in 2.x is the reduce builtin).
from functools import reduce
import datetime
items = [[datetime.datetime(2000, 12, 10, 0, 0), 0.0011], [datetime.datetime(2000, 12, 11, 0, 0), 0.0013 ], [datetime.datetime(2000, 12, 12, 0, 0), 0.0014]]
def add_printing_result(a, b):
total = a+b
print(total)
return total
reduce(add_printing_result, (item[1] for item in items))
Which gives us:
0.0024000000000000002
0.0038000000000000004
sum = 0, myarr = [ [datetime.datetime(2000, 12, 10, 0, 0), 0.0011] , [datetime.datetime(2000, 12, 11, 0, 0), 0.0013] , [datetime.datetime(2000, 12, 12, 0, 0), 0.0014]]
for(i in myarr):
sum+=i[1]
I'm sure there are better ways to do this (I'm no Python expert) but this should sum your values properly such that sum is the sub elements' sum.
Related
I need to do list comprehension of an array that I created:
array([['12/12/80', '0.513393'],
['12/15/80', '0.486607'],
['12/16/80', '0.450893'],
...,
['2/20/19', '172.029999'],
['2/21/19', '171.059998'],
['2/22/19', '172.970001']], dtype='<U10')
The output should look like this:
array([[datetime.datetime(1980, 12, 12, 0, 0), 0.513393],
[datetime.datetime(1980, 12, 15, 0, 0), 0.486607],
[datetime.datetime(1980, 12, 16, 0, 0), 0.450893],
[datetime.datetime(1980, 12, 17, 0, 0), 0.462054],
[datetime.datetime(1980, 12, 18, 0, 0), 0.475446]], dtype=object)
I am struggling for the right code to do the list comprehension. Any help would be highly appreciated.
Can you try the following:
[[datetime.datetime.strptime(dt, '%m/%d/%y'), float(val)] for dt, val in arr]
I am trying to plot data from temperature sensor with time steps. I have time steps in format "hh:mm:ss" after conversion from string to datetime format. First value in the list is "21:47:22" and the last one is "06:12:22" the next day.I have been trying to plot these values with order of indexes in the list however Python automaticaly sorting it from "00:00:00" to "24:00:00" on the x axis. Here is the image.
Could you please advice how to solve this issue? Below my code:
import matplotlib.pyplot as plt
import datetime
data = []
sensor1 = []
sensor2 = []
time = []
with open("output.txt","r") as f:
data = f.readlines()
first_sensor_len = len(data[0])
for var in data:
if var[2:7] == "First" and len(var) == first_sensor_len:
sensor1.append(var[28:33])
sensor2.append(var[75:80])
time.append(datetime.datetime.strptime(var[36:44], "%H:%M:%S"))
elif var[2:8] == "Second" and len(var) == first_sensor_len:
sensor2.append(var[29:34])
sensor1.append(var[75:80])
time.append(datetime.datetime.strptime(var[83:91], "%H:%M:%S"))
plt.plot(time, sensor1)
plt.show()
Supposed time looks like
timestr = ["21:47:22", "22:12:22", "23:12:22", "00:12:22", "01:12:22", "03:12:22", "06:12:22"]
time = [datetime.datetime.strptime(ts, "%H:%M:%S") for ts in timestr]
time
[datetime.datetime(1900, 1, 1, 21, 47, 22),
datetime.datetime(1900, 1, 1, 22, 12, 22),
datetime.datetime(1900, 1, 1, 23, 12, 22),
datetime.datetime(1900, 1, 1, 0, 12, 22),
datetime.datetime(1900, 1, 1, 1, 12, 22),
datetime.datetime(1900, 1, 1, 3, 12, 22),
datetime.datetime(1900, 1, 1, 6, 12, 22)]
You can use np.diff from numpy to mark every first time of a new day. If the difference of two consecutive time values is negative, there was midnight in between.
(This boolean array is appended to one initial False, which states that the first time value has always no day offset; the result of np.diff is generally one entry shorter than its input.)
import numpy as np
newday_marker = np.append(False, np.diff(time) < datetime.timedelta(0))
newday_marker
array([False, False, False, True, False, False, False], dtype=bool)
With np.cumsum this array can be transformed into the array of dayoffsets for each time value.
day_offset = np.cumsum(newday_marker)
day_offset
array([0, 0, 0, 1, 1, 1, 1], dtype=int32)
In the end this has to be converted to timedeltas and then can be added to the original list of time values:
date_offset = [datetime.timedelta(int(dt)) for dt in day_offset]
dtime = [t + dos for t, dos in zip(time, date_offset)]
dtime
[datetime.datetime(1900, 1, 1, 21, 47, 22),
datetime.datetime(1900, 1, 1, 22, 12, 22),
datetime.datetime(1900, 1, 1, 23, 12, 22),
datetime.datetime(1900, 1, 2, 0, 12, 22),
datetime.datetime(1900, 1, 2, 1, 12, 22),
datetime.datetime(1900, 1, 2, 3, 12, 22),
datetime.datetime(1900, 1, 2, 6, 12, 22)]
I have a situation where I need to get the third latest date, i.e
INPUT :
['14-04-2001', '29-12-2061', '21-10-2019',
'07-01-1973', '19-07-2014','11-03-1992','21-10-2019']
Also , INPUT
6
14-04-2001
29-12-2061
21-10-2019
07-01-1973
19-07-2014
11-03-1992
OUTPUT : 19-07-2014
import datetime
datelist = ['14-04-2001', '29-12-2061', '21-10-2019', '07-01-1973', '19-07-2014','11-03-1992','21-10-2019' ]
for d in datelist:
x = datetime.datetime.strptime(d,'%d-%m-%Y')
print x
How can i achieve this?
You can sort the list and take the 3rd element from it.
my_list = [datetime.datetime.strptime(d,'%d-%m-%Y') for d in list]
# [datetime.datetime(2001, 4, 14, 0, 0), datetime.datetime(2061, 12, 29, 0, 0), datetime.datetime(2019, 10, 21, 0, 0), datetime.datetime(1973, 1, 7, 0, 0), datetime.datetime(2014, 7, 19, 0, 0), datetime.datetime(1992, 3, 11, 0, 0), datetime.datetime(2019, 10, 21, 0, 0)]
my_list.sort(reverse=True)
my_list[2]
# datetime.datetime(2019, 10, 21, 0, 0)
Also, as per Kerorin's suggestion, if you don't need to sort in-place and just need the 3rd element always, you can simply do
sorted(my_list, reverse=True)[2]
Update
To remove the duplicates, taking inspiration from this answer, you can do the following -
import datetime
datelist = ['14-04-2001', '29-12-2061', '21-10-2019', '07-01-1973', '19-07-2014', '11-03-1992', '21-10-2019']
seen = set()
my_list = [datetime.datetime.strptime(d,'%d-%m-%Y')
for d in datelist
if d not in seen and not seen.add(d)]
my_list.sort(reverse=True)
You can use heapq.nlargest to do this.
import heapq
from datetime import datetime
datelist = [
'14-04-2001',
'29-12-2061',
'21-10-2019',
'07-01-1973',
'19-07-2014',
'11-03-1992',
'21-10-2019'
]
heapq.nlargest(3, {datetime.strptime(d, "%d-%m-%Y") for d in datelist})[-1]
This return datetime.datetime(2014, 7, 19, 0, 0)
What's the best R equivalent of the Python 2-variable list comprehension
[datetime(y,m,15) for y in xrange(2000,2020) for m in [3,6,9,12]]
The result
[datetime.datetime(2000, 3, 15, 0, 0),
datetime.datetime(2000, 6, 15, 0, 0),
datetime.datetime(2000, 9, 15, 0, 0),
datetime.datetime(2000, 12, 15, 0, 0),
datetime.datetime(2001, 3, 15, 0, 0) ... ]
This will produce equivalent results in R
with(expand.grid(m=c(3,6,9,12), y=2000:2020), ISOdate(y,m,15))
We use expand.grid to get all combinations of year and month, and then we just use the vectorized ISOdate function to get the values.
Another way to express the desired result using the listcompr package:
library(listcompr)
gen.list(ISOdate(y, m, 15), y = 2000:2019, m = c(3, 6, 9, 12))
Note that range(a,b) (or xrange(a, b) in Python2) is equivalent to a:(b-1) (stop element b excluded) in R (and not a:b, which includes the stop element).
I have two lists. One list name 'date' has dates in it which are related to persons birth date.
data = [ datetime.datetime(1958, 3, 15, 0, 0), datetime.datetime(1958, 9, 15, 0, 0), datetime.datetime(1930, 10, 23, 0, 0), datetime.datetime(1928, 9, 15, 0, 0), datetime.datetime(1928, 1, 23, 0, 0), datetime.datetime(1925, 11, 15, 0, 0), datetime.datetime(1962, 7, 20, 0, 0),datetime.datetime(1960, 12, 14, 0, 0), datetime.datetime(1960, 5, 10, 0, 0),datetime.datetime(1963, 9, 7, 0, 0), datetime.datetime(1956, 3, 10, 0, 0), datetime.datetime(1955, 2, 15, 0, 0),datetime.datetime(1958, 11, 14, 0, 0),datetime.datetime(1956, 8, 24, 0, 0),datetime.datetime(1990, 4, 30, 0, 0)]
Now next list contains marriage dates.
marriage = [ datetime.datetime(1985, 5, 14, 0, 0),datetime.datetime(1945, 6, 15, 0, 0), datetime.datetime(1938, 6, 11, 0, 0), datetime.datetime(1995, 4, 5, 0, 0), datetime.datetime(1987, 2, 26, 0, 0), datetime.datetime(1983, 12, 13, 0, 0), datetime.datetime(1980, 9, 16, 0, 0), datetime.datetime(2011, 6, 19, 0, 0)]
each date from the 'marriage' list is related to 2 dates from 'date' list. Now, I want to compare one date from marriage list to two dates from date list so that i can print"birth date is less than marriage.
How can accomplish this task using loop? confused with this one.
Please note that I used import datetime, import re to accomplish date comparison.
for i in range(len(data)):
if data[i] < marriage[i]:
print "birthdate is lt marriage date"
else:
print "birthdate is gt or eq to marriage date"
I'm not sure what you are trying to accomplish here... Also you don't need re for date comparison, you can use normal < > == <= >= operators.
This also sounds like a job for a hash(dictionary)...
marriage = {
'marriage1' : {
'1' : <birthday>,
'2' : <birthday>,
'marriage-date' : <marriage-date>
},
'marriage2' : {
'1' : <birthday>,
'2' : <birthday>,
'marriage-date' : <marriage-date>
}
}
A hash(dictionary) will make comparisons much easier with lists that don't contain the same number of values.
This assumes that the marriage and birth dates are in the same order (i.e., the first two birth dates correspond to the first marriage date and the next 2 birth dates correspond to the second marriage date)
for i in range(len(marriage)):
if marriage[i] > data[i*2] and marriage[i] > data[(i*2)+1]:
print "Both birthdates less than marriage data"
I believe my assumption is correct because there are twice as many entries in the data list as there are in the marriage list.