I'm trying to setting up pinout for IMX8 Hummingboard Mate by gpiod but I can't make this led script works propertly:
import gpiod
import sys
import time
if len(sys.argv) > 2:
LED_CHIP = sys.argv[1]
LED_LINE_OFFSET = int(sys.argv[2])
else:
print('''Usage:
python3 blink.py <chip> <line offset>''')
sys.exit()
chip = gpiod.chip(LED_CHIP)
led = chip.get_line(LED_LINE_OFFSET)
config = gpiod.line_request()
config.consumer = "Blink"
config.request_type = gpiod.line_request.DIRECTION_OUTPUT
led.request(config)
print(led.consumer)
while True:
led.set_value(0)
time.sleep(0.1)
led.set_value(1)
time.sleep(0.1)
I'm using this script this way:
test_led.py gpiochip3 23
Because the pin name is GPIO4_IO23 (https://solidrun.atlassian.net/wiki/spaces/developer/pages/396197889/GPIO+Pins+Control+-+HummingBoard+Pulse+Mate+i.MX8M+Plus+SOM), but as I understand this, GPIO4 is gpiochip3 because there is not gpiochip4 but there is gpiochip0. Anymay, this configuration it doesn't work.
My gpiochip and gpio line is this:
30240000.gpio gpiochip4
0
1
2
3
4
5
6
7
8
9
10
11
12
13 spi1 CS0
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
30230000.gpio gpiochip3
0
1
2
3
4
5
6
7
8
9
10
11
12 PHY reset
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
30220000.gpio gpiochip2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
30210000.gpio gpiochip1
0
1
2
3
4
5
6
7 device-wakeup
8 host-wakeup
9
10 shutdown
11 reset
12 cd
13
14
15
16
17
18
19 regulator-usdhc2
20
21
22
23
24
25
26
27
28
29
30
31
30200000.gpio gpiochip0
0
1 reset
2
3 interrupt
4
5 shutdown
6 reset
7 shutdown
8
9
10 regulator-m2-mpcie-pwr
11
12
13 shutdown
14 regulator-usb1-host-vbus
15 regulator-usb1-vbus
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
Related
I have a dataframe like this:
a\tb\tc d\te\tf g\th\ti
20\t21\t22 1\t2\t3 30\t31\t32
17\t18\t19 4\t5\t6 27\t28\t29
14\t15\t16 7\t8\t9 24\t25\t26
11\t12\t13 10\t11\t12 21\t22\t23
8\t9\t10 13\t14\t15 18\t19\t20
5\t6\t7 16\t17\t18 15\t16\t17
2\t3\t4 19\t20\t21 12\t13\t14
expected output:
a b c d e f g h i
0 20 21 22 1 2 3 30 31 32
1 17 18 19 4 5 6 27 28 29
2 14 15 16 7 8 9 24 25 26
3 11 12 13 10 11 12 21 22 23
4 8 9 10 13 14 15 18 19 20
5 5 6 7 16 17 18 15 16 17
6 2 3 4 19 20 21 12 13 14
My solution is:
l = list()
for column in df.columns:
columns = column.split()
d = df[column].str.split(expand=True)
l.append(d.rename(columns=dict(zip(range(len(columns)),columns))))
pd.concat(l,axis=1)
But this looks so complex.
Is there a simple way of doing this ?
Your approach looks good. You can simplify the rename part by just assigning the new names to .columns attribute:
def expand(col):
_df = df[col].str.split(expand=True)
_df.columns = col.split()
return _df
pd.concat(map(expand, df.columns), axis=1)
a b c d e f g h i
0 20 21 22 1 2 3 30 31 32
1 17 18 19 4 5 6 27 28 29
2 14 15 16 7 8 9 24 25 26
3 11 12 13 10 11 12 21 22 23
4 8 9 10 13 14 15 18 19 20
5 5 6 7 16 17 18 15 16 17
6 2 3 4 19 20 21 12 13 14
I have some code here:
for i in range(self.size):
print('{:6d}'.format(self.data[i], end=' '))
if (i + 1) % NUMBER_OF_COLUMNS == 0:
print()
Right now this prints as:
1
1
1
1
1
2
3
3
3
3
(whitespace)
3
3
3
etc.
It creates a new line when it hits 10 digits, but it doens't print the initial 10 in a row...
This is what I want-
1 1 1 1 1 1 1 2 2 3
3 3 3 3 3 4 4 4 4 5
However when it hits two digit numbers it gets messed up -
8 8 8 8 8 9 9 9 9 10
10 10 10 10 10 10 etc.
I want it to be right-aligned like this-
8 8 8 8 8 9
10 10 10 10 11 12 etc.
When I remove the format piece it will print the rows out, but there wont be the extra spacing in there of course!
You can align strings by "padding" values using a string's .rjust method. Using some dummy data:
NUMBER_OF_COLUMNS = 10
for i in range(100):
print("{}".format(i//2).rjust(3), end=' ')
#print("{:3}".format(i//2), end=' ') edit: this also works. Thanks AChampion
if (i + 1) % NUMBER_OF_COLUMNS == 0:
print()
#Output:
0 0 1 1 2 2 3 3 4 4
5 5 6 6 7 7 8 8 9 9
10 10 11 11 12 12 13 13 14 14
15 15 16 16 17 17 18 18 19 19
20 20 21 21 22 22 23 23 24 24
25 25 26 26 27 27 28 28 29 29
30 30 31 31 32 32 33 33 34 34
35 35 36 36 37 37 38 38 39 39
40 40 41 41 42 42 43 43 44 44
45 45 46 46 47 47 48 48 49 49
Another approach is to just chunk up the data into rows and print each row, e.g.:
def chunk(iterable, n):
return zip(*[iter(iterable)]*n)
for row in chunk(self.data, NUMBER_OF_COLUMNS):
print(' '.join(str(data).rjust(6) for data in row))
e.g:
In []:
for row in chunk(range(100), 10):
print(' '.join(str(data//2).rjust(3) for data in row))
Out[]:
0 0 1 1 2 2 3 3 4 4
5 5 6 6 7 7 8 8 9 9
10 10 11 11 12 12 13 13 14 14
15 15 16 16 17 17 18 18 19 19
20 20 21 21 22 22 23 23 24 24
25 25 26 26 27 27 28 28 29 29
30 30 31 31 32 32 33 33 34 34
35 35 36 36 37 37 38 38 39 39
40 40 41 41 42 42 43 43 44 44
45 45 46 46 47 47 48 48 49 49
I am using Calendar function in python for the first time this is my code:
import calendar
print ("The calender of year 2018 is : ")
print (calendar.calendar(2018,2,1,6))
print ("The starting day number in calendar is : ",end="")
print (calendar.firstweekday())
I am getting the followng output
The calender of year 2018 is :
Traceback (most recent call last):
File "C:/Users/AAKASH PATEL/Desktop/calendar.py", line 5, in <module>
import calendar
File "C:/Users/AAKASH PATEL/Desktop\calendar.py", line 10, in <module>
print (calendar.calendar(2018,2,1,6))
TypeError: 'module' object is not callable
How can i solve this
Rename your script name from calendar.py to calendarScript.py
"C:/Users/AAKASH PATEL/Desktop/calendar.py"
to
"C:/Users/AAKASH PATEL/Desktop/calendarScript.py",
Note: Do not name your script the same name as a module.
You named your script calendar.py, so it is that file that is imported. You can see this in the traceback:
File "C:/Users/AAKASH PATEL/Desktop/calendar.py", line 5, in <module>
import calendar
File "C:/Users/AAKASH PATEL/Desktop\calendar.py", line 10, in <module>
print (calendar.calendar(2018,2,1,6))
So import calendar imported calendar.py, which now has an attribute calendar, which is your script.
Rename your script to not mask the built-in library; for example, calendar_demo.py would be fine. Once you do, it works perfectly, on my system, it outputs:
The calender of year 2018 is :
2018
January February March
Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su
1 2 3 4 5 6 7 1 2 3 4 1 2 3 4
8 9 10 11 12 13 14 5 6 7 8 9 10 11 5 6 7 8 9 10 11
15 16 17 18 19 20 21 12 13 14 15 16 17 18 12 13 14 15 16 17 18
22 23 24 25 26 27 28 19 20 21 22 23 24 25 19 20 21 22 23 24 25
29 30 31 26 27 28 26 27 28 29 30 31
April May June
Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su
1 1 2 3 4 5 6 1 2 3
2 3 4 5 6 7 8 7 8 9 10 11 12 13 4 5 6 7 8 9 10
9 10 11 12 13 14 15 14 15 16 17 18 19 20 11 12 13 14 15 16 17
16 17 18 19 20 21 22 21 22 23 24 25 26 27 18 19 20 21 22 23 24
23 24 25 26 27 28 29 28 29 30 31 25 26 27 28 29 30
30
July August September
Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su
1 1 2 3 4 5 1 2
2 3 4 5 6 7 8 6 7 8 9 10 11 12 3 4 5 6 7 8 9
9 10 11 12 13 14 15 13 14 15 16 17 18 19 10 11 12 13 14 15 16
16 17 18 19 20 21 22 20 21 22 23 24 25 26 17 18 19 20 21 22 23
23 24 25 26 27 28 29 27 28 29 30 31 24 25 26 27 28 29 30
30 31
October November December
Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su
1 2 3 4 5 6 7 1 2 3 4 1 2
8 9 10 11 12 13 14 5 6 7 8 9 10 11 3 4 5 6 7 8 9
15 16 17 18 19 20 21 12 13 14 15 16 17 18 10 11 12 13 14 15 16
22 23 24 25 26 27 28 19 20 21 22 23 24 25 17 18 19 20 21 22 23
29 30 31 26 27 28 29 30 24 25 26 27 28 29 30
31
The starting day number in calendar is : 0
Note that calendar.calendar(2018) would suffice; the defaults for the next three arguments (w for the column width, l for lines per week, and c for the spacing between month columns) are 2, 1 and 6 respectively.
When copying your code and executing it, it works perfectly fine! But I see that it seems your script, where you have that code is called calendar.py! Python then thinks this is what you want to import! So rename your script to something else!
I am new to programming and have taken up learning python in an attempt to make some tasks I run in my research more efficient. I am running a PCA in the pandas module (I found a tutorial online) and have the script for this, but need to subselect part of a dataframe prior to the pca.
so far I have (just for example in reality I am reading a .csv file with a larger matrix)
x = np.random.randint(30, size=(8,8))
df = pd.DataFrame(x)
0 1 2 3 4 5 6 7
0 9 0 23 13 2 5 14 6
1 20 17 11 10 25 23 20 23
2 15 14 22 25 11 15 5 15
3 9 27 15 27 7 15 17 23
4 12 6 11 13 27 11 26 20
5 27 13 5 16 5 5 2 18
6 3 18 22 0 7 10 11 11
7 25 18 10 11 29 29 1 25
What I want to do is sub-select columns that satisfy a certain criteria in any of the rows, specifically I want every column that has at least one number =>27 (just for example) to produce a new dataframe
0 1 3 4 5
0 9 0 13 2 5
1 20 17 10 25 23
2 15 14 25 11 15
3 9 27 27 7 15
4 12 6 13 27 11
5 27 13 16 5 5
6 3 18 0 7 10
7 25 18 11 29 29
I have looked into the various slicing methods in pandas but none seem to do what I want (.loc and .iloc etc.).
The actual script I am using to read in thus far is
filename = 'Data.csv'
data = pd.read_csv(filename,sep = ',')
x = data.ix[:,1:] # variables - species
y = data.ix[:,0] # cases - age
so a sub dataframme of x is what I am after (as above).
Any advice is greatly appreciated.
Indexers like loc, iloc, and ix accept boolean arrays. For example if you have three columns, df.loc[:, [True, False, True]] will return all the rows and the columns 0 and 2 (when corresponding value is True). You can check whether any of the elements in a column is greater than or equal to 27 by (df>=27).any(). This will return True for the columns that has at least one value >=27. So you can slice the dataframe with:
df.loc[:, (df>=27).any()]
Out[34]:
0 1 3 4 5 7
0 8 2 28 9 14 21
1 24 26 23 17 0 0
2 3 24 7 15 4 28
3 29 17 12 7 7 6
4 5 3 10 24 29 14
5 23 21 0 16 23 13
6 22 10 27 1 7 24
7 9 27 2 27 17 12
And this is the initial dataframe:
df
Out[35]:
0 1 2 3 4 5 6 7
0 8 2 7 28 9 14 26 21
1 24 26 15 23 17 0 21 0
2 3 24 26 7 15 4 7 28
3 29 17 9 12 7 7 0 6
4 5 3 13 10 24 29 22 14
5 23 21 26 0 16 23 17 13
6 22 10 19 27 1 7 9 24
7 9 27 26 2 27 17 8 12
I have written a small program based on iterators to display a multicolumn calendar.
In that code I am using itertools.groupby to group the dates by month by the function group_by_months(). There I yield the month name and the grouped dates as a list for every month. However, when I let that function directly return the grouped dates as an iterator (instead of a list) the program leaves the days of all but the last column blank.
I can't figure out why that might be. Am I using groupby wrong? Can anyone help me to spot the place where the iterator is consumed or its output is ignored? Why is it especially the last column that "survives"?
Here's the code:
import datetime
from itertools import zip_longest, groupby
def grouper(iterable, n, fillvalue=None):
"""\
copied from the docs:
https://docs.python.org/3.4/library/itertools.html#itertools-recipes
"""
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
def generate_dates(start_date, end_date, step=datetime.timedelta(days=1)):
while start_date < end_date:
yield start_date
start_date += step
def group_by_months(seq):
for k,v in groupby(seq, key=lambda x:x.strftime("%B")):
yield k, v # Why does it only work when list(v) is yielded here?
def group_by_weeks(seq):
yield from groupby(seq, key=lambda x:x.strftime("%2U"))
def format_month(month, dates_of_month):
def format_week(weeknum, dates_of_week):
def format_day(d):
return d.strftime("%3e")
weekdays = {d.weekday(): format_day(d) for d in dates_of_week}
return "{0} {7} {1} {2} {3} {4} {5} {6}".format(
weeknum, *[weekdays.get(i, " ") for i in range(7)])
yield "{:^30}".format(month)
weeks = group_by_weeks(dates_of_month)
yield from map(lambda x:format_week(*x), weeks)
start, end = datetime.date(2016,1,1), datetime.date(2017,1,1)
dates = generate_dates(start, end)
months = group_by_months(dates)
formatted_months = map(lambda x: (format_month(*x)), months)
ncolumns = 3
quarters = grouper(formatted_months, ncolumns)
interleaved = map(lambda x: zip_longest(*x, fillvalue=" "*30), quarters)
formatted = map(lambda x: "\n".join(map(" ".join, x)), interleaved)
list(map(print, formatted))
This is the failing output:
January February March
09 1 2 3 4 5
10 6 7 8 9 10 11 12
11 13 14 15 16 17 18 19
12 20 21 22 23 24 25 26
13 27 28 29 30 31
April May June
22 1 2 3 4
23 5 6 7 8 9 10 11
24 12 13 14 15 16 17 18
25 19 20 21 22 23 24 25
26 26 27 28 29 30
July August September
35 1 2 3
36 4 5 6 7 8 9 10
37 11 12 13 14 15 16 17
38 18 19 20 21 22 23 24
39 25 26 27 28 29 30
October November December
48 1 2 3
49 4 5 6 7 8 9 10
50 11 12 13 14 15 16 17
51 18 19 20 21 22 23 24
52 25 26 27 28 29 30 31
This is the expected output:
January February March
00 1 2 05 1 2 3 4 5 6 09 1 2 3 4 5
01 3 4 5 6 7 8 9 06 7 8 9 10 11 12 13 10 6 7 8 9 10 11 12
02 10 11 12 13 14 15 16 07 14 15 16 17 18 19 20 11 13 14 15 16 17 18 19
03 17 18 19 20 21 22 23 08 21 22 23 24 25 26 27 12 20 21 22 23 24 25 26
04 24 25 26 27 28 29 30 09 28 29 13 27 28 29 30 31
05 31
April May June
13 1 2 18 1 2 3 4 5 6 7 22 1 2 3 4
14 3 4 5 6 7 8 9 19 8 9 10 11 12 13 14 23 5 6 7 8 9 10 11
15 10 11 12 13 14 15 16 20 15 16 17 18 19 20 21 24 12 13 14 15 16 17 18
16 17 18 19 20 21 22 23 21 22 23 24 25 26 27 28 25 19 20 21 22 23 24 25
17 24 25 26 27 28 29 30 22 29 30 31 26 26 27 28 29 30
July August September
26 1 2 31 1 2 3 4 5 6 35 1 2 3
27 3 4 5 6 7 8 9 32 7 8 9 10 11 12 13 36 4 5 6 7 8 9 10
28 10 11 12 13 14 15 16 33 14 15 16 17 18 19 20 37 11 12 13 14 15 16 17
29 17 18 19 20 21 22 23 34 21 22 23 24 25 26 27 38 18 19 20 21 22 23 24
30 24 25 26 27 28 29 30 35 28 29 30 31 39 25 26 27 28 29 30
31 31
October November December
39 1 44 1 2 3 4 5 48 1 2 3
40 2 3 4 5 6 7 8 45 6 7 8 9 10 11 12 49 4 5 6 7 8 9 10
41 9 10 11 12 13 14 15 46 13 14 15 16 17 18 19 50 11 12 13 14 15 16 17
42 16 17 18 19 20 21 22 47 20 21 22 23 24 25 26 51 18 19 20 21 22 23 24
43 23 24 25 26 27 28 29 48 27 28 29 30 52 25 26 27 28 29 30 31
As the docs state (c.f.):
when the groupby() object is advanced, the previous group is no longer visible. So, if that data is needed later, it should be stored as a list
That means the iterators are consumed, when the code later accesses the returned iterators out of order, i.e., when the groupby is actually iterated. The iteration happens out of order because of the chunking and interleaving that is done here.
We observe this specific pattern (i.e., only the last column is fully displayed) because of the way we iterate. That is:
The month names for the first line are printed. Thereby the iterators for up to the last column's month are consumed (and their content discarded). The groupby() object produces the last column's month name only after the first columns' data.
We print the first week line. Thereby the already exhausted iterators for the first columns are filled up automatically using the default value passed to zip_longest(). Only the last column still provides actual data.
The same happens for the next lines of month names.