I am using subprocess to print the output of ls.
output = subprocess.getoutput("ssh -i key.pem ubuntu#10.127.6.83 ls -l --time-style=long-iso /opt/databases | awk -F' ' '{print $6 $8}'")
lines = output.splitlines()
print(lines)
format = '%Y-%m-%d'
for line in lines:
if line != '':
date = datetime.strptime(line, format)
And when I print lines am getting a large list in the following format:
['', '2019-04-25friendship_graph_43458', '2019-07-18friendship_graph_45359', '2019-09-03friendship_graph_46553', '2019-10-02friendship_graph_46878']
I am trying to convert the above output to a dict with the dates in '%Y-%m-%d' format. So output would be something like:
{ '2019-04-25' : 'friendship_graph_43458',
'2019-07-18': 'friendship_graph_45359',
'2019-09-03': 'friendship_graph_46553' }
and so on, but not quite sure how to do so.
Technically if you don't want to use re if all dates are formatted the same then they will all be 10 characters long thus just slice the strings to make the dict in a comprehension:
data = ['', '2019-04-25friendship_graph_43458', '2019-07-18friendship_graph_45359', '2019-09-03friendship_graph_46553', '2019-10-02friendship_graph_46878']
output = {s[:10]: s[10:] for s in data if len(s) > 10}
{'2019-04-25': 'friendship_graph_43458', '2019-07-18': 'friendship_graph_45359', '2019-09-03': 'friendship_graph_46553', '2019-10-02': 'friendship_graph_46878'}
You could use a regular expression for each item in the list. For example:
(\d{4}-\d{2}-\d{2})(.*)
Then, you can just iterate through each item in the list and use the regular expression to the get the string in its two parts.
>>> import re
>>> regex = re.compile(r"(\d{4}-\d{2}-\d{2})(.*)")
>>> items = ['', '2019-04-25friendship_graph_43458', '2019-07-18friendship_graph_45359', '2019-09-03friendship_graph_46553', '2019-10-02friendship_graph_46878']
>>> items_dict = {}
>>> for i in items:
match = regex.search(i)
if match is None:
continue
items_dict[match.group(1)] = match.group(2)
>>> items_dict
{'2019-04-25': 'friendship_graph_43458', '2019-07-18': 'friendship_graph_45359', '2019-09-03': 'friendship_graph_46553', '2019-10-02': 'friendship_graph_46878'}
For lines that start with the date; use slices to separate the key from the value.
>>> s = '2019-04-25friendship_graph_43458'
>>> d = {}
>>> d[s[:10]] = s[10:]
>>> d
{'2019-04-25': 'friendship_graph_43458'}
>>>
Use re.findall and dictionary comprehension:
import re
lst = ['', '2019-04-25friendship_graph_43458', '2019-07-18friendship_graph_45359', '2019-09-03friendship_graph_46553', '2019-10-02friendship_graph_46878']
dct = {k: v for s in lst for k, v in re.findall(r'(\d\d\d\d-\d\d-\d\d)(.*)', s) }
print(dct)
# {'2019-04-25': 'friendship_graph_43458', '2019-07-18': 'friendship_graph_45359', '2019-09-03': 'friendship_graph_46553', '2019-10-02': 'friendship_graph_46878'}
Related
I got a list of strings. Those strings have all the two markers in. I would love to extract the string between those two markers for each string in that list.
example:
markers 'XXX' and 'YYY' --> therefore i want to extract 78665786 and 6866
['XXX78665786YYYjajk', 'XXX6866YYYz6767'....]
You can just loop over your list and grab the substring. You can do something like:
import re
my_list = ['XXX78665786YYYjajk', 'XXX6866YYYz6767']
output = []
for item in my_list:
output.append(re.search('XXX(.*)YYY', item).group(1))
print(output)
Output:
['78665786', '6866']
import re
l = ['XXX78665786YYYjajk', 'XXX6866YYYz6767'....]
l = [re.search(r'XXX(.*)YYY', i).group(1) for i in l]
This should work
Another solution would be:
import re
test_string=['XXX78665786YYYjajk','XXX78665783336YYYjajk']
int_val=[int(re.search(r'\d+', x).group()) for x in test_string]
the command split() splits a String into different parts.
list1 = ['XXX78665786YYYjajk', 'XXX6866YYYz6767']
list2 = []
for i in list1:
d = i.split("XXX")
for g in d:
d = g.split("YYY")
list2.append(d)
print(list2)
it's saved into a list
I am trying to make a list containing 2 strings:
List=["Hight = 7.2", "baselength = 8.32"]
But I am having a problem trying to extract the numbers from the strings:
For example:
If "Hight = 7.2" then the result should be: 7.2
or if the "Hight= 7.3232" then the result should be: 7.3232
Using re.findall :
>>> out = []
>>> for s in l:
out.append( float(re.findall('\d+(?:\.\d+)?', s)[0]) )
>>> out
=> [7.2, 8.0]
Or, without regex, using split,
>>> out = []
>>> for s in l:
num = s.replace(' ','').split('=')[1]
#note : removed whitespace so don't have to deal with cases like
# `n = 2` or `n=2`
out.append(float(num))
>>> out
=> [7.2, 8.0]
#driver values :
IN : l = ["Hight = 7.2","baselength = 8"]
How about this
[(item.split('=')[0],float(item.split('=')[1]) ) for item in List]
Output :
[('Hight ', 7.2), ('baselength ', 8.32)]
Having a label associated to a value is best managed with a dictionary, however if you must have each label=value pair as an entry in a list because perhaps you are reading it into Python from elsewhere, you could use the re module to extract the numeric value from each string in the list:
import re
list=["height = 7.2", "length = 8.32"]
for dim in list:
print(float(re.search('\d+.\d+', dim).group()))
You could convert your list to a dictionary using a comprehension:
import re
List=["Height = 7.2", "baselength = 8.32"]
rx = re.compile(r'(?P<key>\w+)\s*=\s*(?P<value>\d+(?:\.\d+)?)')
Dict = {m.group('key'): float(m.group('value'))
for item in List
for m in [rx.search(item)]}
print(Dict)
# {'Height': 7.2, 'baselength': 8.32}
Afterwards, you can access your values with e.g. Dict["Height"] (here: 7.2).
It's very simple. Use this method for any type of value
List=["Hight = 7.2", "baselength = 8.32"]
# showing example for one value , but you can loop the entire list
a = List[0].split("= ")[1] #accessing first element and split with "= "
print a
'7.2'
What is the best way to replace every string in the list?
For example if I have a list:
a = ['123.txt', '1234.txt', '654.txt']
and I would like to have:
a = ['123', '1234', '654']
Assuming that sample input is similar to what you actually have, use os.path.splitext() to remove file extensions:
>>> import os
>>> a = ['123.txt', '1234.txt', '654.txt']
>>> [os.path.splitext(item)[0] for item in a]
['123', '1234', '654']
Use a list comprehension as follows:
a = ['123.txt', '1234.txt', '654.txt']
answer = [item.replace('.txt', '') for item in a]
print(answer)
Output
['123', '1234', '654']
Assuming that all your strings end with '.txt', just slice the last four characters off.
>>> a = ['123.txt', '1234.txt', '654.txt']
>>> a = [x[:-4] for x in a]
>>> a
['123', '1234', '654']
This will also work if you have some weird names like 'some.txtfile.txt'
You could split you with . separator and get first item:
In [486]: [x.split('.')[0] for x in a]
Out[486]: ['123', '1234', '654']
Another way to do this:
a = [x[: -len("txt")-1] for x in a]
What is the best way to replace every string in the list?
That completely depends on how you define 'best'. I, for example, like regular expressions:
import re
a = ['123.txt', '1234.txt', '654.txt']
answer = [re.sub('^(\w+)\..*', '\g<1>', item) for item in a]
#print(answer)
#['123', '1234', '654']
Depending on the content of the strings, you could adjust it:
\w+ vs [0-9]+ for only digits
\..* vs \.txt if all strings end with .txt
data.colname = [item.replace('anythingtoreplace', 'desiredoutput') for item in data.colname]
Please note here 'data' is the dataframe, 'colname' is the column name you might have in that dataframe. Even the spaces are accounted, if you want to remove them from a string or number. This was quite useful for me. Also this does not change the datatype of the column so you might have to do that separately if required.
I have a string :
datalist = "popupWin:'http://www.adachikan.net/index.php/catalog/product/gallery/id/3049/image/1861/', useZoom: 'cloudZoom', smallImage: 'http://www.adachikan.net/media/catalog/product/cache/1/image/315x400/9df78eab33525d08d6e5fb8d27136e95/1/8/187267-875-40.jpg'"
And i want to convert this string into Python Dict. I can use split to break the string and form list:
datalist.split(',') = ["popupWin:'http://www.adachikan.net/index.php/catalog/product/gallery/id/3049/image/1861/'",
" useZoom: 'cloudZoom'",
" smallImage: 'http://www.adachikan.net/media/catalog/product/cache/1/image/315x400/9df78eab33525d08d6e5fb8d27136e95/1/8/187267-875-40.jpg'"]
And so on to get the desired result...
I there any easy way to use this datalist as a dict like : datalist['smallImage'] etc..
Split each item after splitting by ',', by ':'. like so:
datadict = {}
for item in datalist.split(','):
key, value = item.split(':', 1)
datadict[key.strip()] = value.strip()
Use a dict comprehension with str.split and str.strip:
>>> dic = {k.strip(): v.strip().strip("'")
for k,v in (x.split(':',1) for x in datalist.split(','))}
>>> dic
{'useZoom': 'cloudZoom', 'popupWin': 'http://www.adachikan.net/index.php/catalog/product/gallery/id/3049/image/1861/', 'smallImage': 'http://www.adachikan.net/media/catalog/product/cache/1/image/315x400/9df78eab33525d08d6e5fb8d27136e95/1/8/187267-875-40.jpg'}
>>> dic['smallImage']
'http://www.adachikan.net/media/catalog/product/cache/1/image/315x400/9df78eab33525d08d6e5fb8d27136e95/1/8/187267-875-40.jpg'
I am new to Python so I have lots of doubts. For instance I have a string:
string = "xtpo, example1=x, example2, example3=thisValue"
For example, is it possible to get the values next to the equals in example1 and example3? knowing only the keywords, not what comes after the = ?
You can use regex:
>>> import re
>>> strs = "xtpo, example1=x, example2, example3=thisValue"
>>> key = 'example1'
>>> re.search(r'{}=(\w+)'.format(key), strs).group(1)
'x'
>>> key = 'example3'
>>> re.search(r'{}=(\w+)'.format(key), strs).group(1)
'thisValue'
Spacing things out for clarity
>>> Sstring = "xtpo, example1=x, example2, example3=thisValue"
>>> items = Sstring.split(',') # Get the comma separated items
>>> for i in items:
... Pair = i.split('=') # Try splitting on =
... if len(Pair) > 1: # Did split
... print Pair # or whatever you would like to do
...
[' example1', 'x']
[' example3', 'thisValue']
>>>