I have a question to print values in list.
import time
strings = time.strftime("%Y,%m,%d")
t = strings.split(',')
date = [int(x) for x in t]
print date
then result is
[2016,5,15]
But I want to print values in date like this
20160515
How can I fix it?
What's wrong with doing it like this:
>>> strings = time.strftime("%Y%m%d")
>>> strings
'20160515'
you must just change your code :
import time
strings = time.strftime("%Y%m%d") # delete ','
print strings
Why don't just do:
time.strftime("%Y%m%d")
On the other hand, if you are just looking for a way to concatenate elements of a list, use join:
In [110]: s = time.strftime("%Y,%m,%d")
In [111]: sl = s.split(',')
In [112]: ''.join(sl)
Out[112]: '20160515'
Related
I have a list which looks like below
lis = '[-3.56568247e-02 -3.31957154e-02\n 7.04742894e-02\n 7.32413381e-02\n 1.74463019e-02]' (string type)
'\n' is also there in the list.
I need to convert this to actual list of integers
lis = [-3.56568247e-02,-3.31957154e-02 ,7.04742894e-02 ,7.32413381e-02, 1.74463019e-02] (list of integers)
I am doing the functionality, but it is failing
import as
res = ast.literal_eval(lis)
Can anyone tell me how to resolve this?
We can use re.findall along with a list comprehension:
lis = '[-3.56568247e-02 -3.31957154e-02 7.04742894e-02 7.32413381e-02\n 1.74463019e-02]'
output = [float(x) for x in re.findall(r'\d+(?:\.\d+)?(?:e[+-]\d+)?', lis)]
print(output)
# [0.0356568247, 0.0331957154, 0.0704742894, 0.0732413381, 0.0174463019]
You can try
[int(i) for i in lis.strip("[]").split(" ")]
You risk getting 1000 ways to do this.
This is a quick and easy way using only basic methods:
lis = '[1 2 3 4 5 77]'
elements = lis.replace('[','').replace(']','').split(' ')
my_ints = [int(e) for e in elements]
print(my_ints)
I have a data frame like
query
-----------
[]
[(apple,10),(orange,15)]
[(apple,2),(orange,5)]
python is reading this as a string instead of a list because when I do
df['query'].apply(lambda x: len(x))
I get 2 instead of 0 for the first row. Is there a way to convert this to a list.
You can use apply():
df['query'] = df['query'].apply(lambda x: x.strip('[]').split(','))
os, by list comprehension:
df['query'] = [x.strip('[]').split(',') for x in df['query']]
or, use ast.literal_eval():
import ast
df['query'] = df['query'].apply(lambda x: ast.literal_eval(x))
or,
df['query'] = df['query'].apply(ast.literal_eval)
I would like to separate my string every both commas but I can not, can you help me.
This is what I want: ['nb1,nb2','nb3,nb4','nb5,nb6']
Here is what I did :
a= 'nb1,nb2,nb3,nb4,nb5,nb6'
compteur=0
for i in a:
if i==',' :
compteur+=1
if compteur%2==0:
print compteur
test = a.split(',', compteur%2==0 )
print a
print test
The result:
2
4
nb1,nb2,nb3,nb4,nb5,nb6
['nb1', 'nb2,nb3,nb4,nb5,nb6']
Thanks you by advances for you answers
You can use regex
In [12]: re.findall(r'([\w]+,[\w]+)', 'nb1,nb2,nb3,nb4,nb5,nb6')
Out[12]: ['nb1,nb2', 'nb3,nb4', 'nb5,nb6']
A quick fix could be to simply first separate the elements by commas and then join the elements by two together again. Like:
sub_result = a.split(',')
result = [','.join(sub_result[i:i+2]) for i in range(0,len(sub_result),2)]
This gives:
>>> result
['nb1,nb2', 'nb3,nb4', 'nb5,nb6']
This will also work if the number of elements is odd. For example:
>>> a = 'nb1,nb2,nb3,nb4,nb5,nb6,nb7'
>>> sub_result = a.split(',')
>>> result = [','.join(sub_result[i:i+2]) for i in range(0,len(sub_result),2)]
>>> result
['nb1,nb2', 'nb3,nb4', 'nb5,nb6', 'nb7']
You use a zip operation of the list with itself to create pairs:
a = 'nb1,nb2,nb3,nb4,nb5,nb6'
parts = a.split(',')
# parts = ['nb1', 'nb2', 'nb3', 'nb4', 'nb5', 'nb6']
pairs = list(zip(parts, parts[1:]))
# pairs = [('nb1', 'nb2'), ('nb2', 'nb3'), ('nb3', 'nb4'), ('nb4', 'nb5'), ('nb5', 'nb6')]
Now you can simply join every other pair again for your output:
list(map(','.join, pairs[::2]))
# ['nb1,nb2', 'nb3,nb4', 'nb5,nb6']
Split the string by comma first, then apply the common idiom to partition an interable into sub-sequences of length n (where n is 2 in your case) with zip.
>>> s = 'nb1,nb2,nb3,nb4,nb5,nb6'
>>> [','.join(x) for x in zip(*[iter(s.split(','))]*2)]
['nb1,nb2', 'nb3,nb4', 'nb5,nb6']
I want to make list data to string.
My list data like this :
[['data1'],['data2'],['data3']]
I want to convert to string like this :
"[data1] [data2] [data3]"
I try to use join like this :
data=[['data1'],['data2'],['data3']]
list=" ".join(data)
But get error like this :
string= " ".join(data)
TypeError: sequence item 0: expected string, list found
Can somebody help me?
Depending on how closely you want the output to conform to your sample, you have a few options, show here in ascending order of complexity:
>>> data=[['data1'],['data2'],['data3']]
>>> str(data)
"[['data1'], ['data2'], ['data3']]"
>>> ' '.join(map(str, data))
"['data1'] ['data2'] ['data3']"
>>> ' '.join(map(str, data)).replace("'", '')
'[data1] [data2] [data3]'
Keep in mind that, if your given sample of data doesn't match your actual data, these methods may or may not produce the desired results.
Have you tried?
data=[['data1'],['data2'],['data3']]
t = map(lambda x : str(x), data)
print(" ".join(t))
Live demo - https://repl.it/BOaS
In Python 3.x , the elements of the iterable for str.join() has to be a string .
The error you are getting - TypeError: sequence item 0: expected string, list found - is because the elements of the list you pass to str.join() is list (as data is a list of lists).
If you only have a single element per sublist, you can simply do -
" ".join(['[{}]'.format(x[0]) for x in data])
Demo -
>>> data=[['data1'],['data2'],['data3']]
>>> " ".join(['[{}]'.format(x[0]) for x in data])
'[data1] [data2] [data3]'
If the sublists can have multiple elements and in your output you want those multiple elements separated by a , . You can use a list comprehension inside str.join() to create a list of strings as you want. Example -
" ".join(['[{}]'.format(','.join(x)) for x in data])
For some other delimiter other than ',' , use that in - '<delimiter>'.join(x) .
Demo -
>>> data=[['data1'],['data2'],['data3']]
>>> " ".join(['[{}]'.format(','.join(x)) for x in data])
'[data1] [data2] [data3]'
For multiple elements in sublist -
>>> data=[['data1','data1.1'],['data2'],['data3','data3.1']]
>>> " ".join(['[{}]'.format(','.join(x)) for x in data])
'[data1,data1.1] [data2] [data3,data3.1]'
>>> import re
>>> l = [['data1'], ['data2'], ['data3']]
>>> s = ""
>>> for i in l:
s+= re.sub(r"\'", "", str(i))
>>> s
'[data1][data2][data3]'
How about this?
data = [['data1'], ['data2'], ['data3']]
result = " ".join('[' + a[0] + ']' for a in data)
print(result)
How about this:
In [13]: a = [['data1'],['data2'],['data3']]
In [14]: import json
In [15]: temp = " ".join([json.dumps(x) for x in a]).replace("\"", "")
In [16]: temp
Out[16]: '[data1] [data2] [data3]'
Try the following. This can also be achieved by "Reduce":
from functools import reduce
data = [['data1'], ['data2'], ['data3']]
print(list(reduce(lambda x,y : x+y, data)))
output: ['data1', 'data2', 'data3']
I have possible strings of prices like:
20.99, 20, 20.12
Sometimes the string could be sent to me wrongly by the user to something like this:
20.99.0, 20.0.0
These should be converted back to :
20.99, 20
So basically removing anything from the 2nd . if there is one.
Just to be clear, they would be alone, one at a time, so just one price in one string
Any nice one liner ideas?
For a one-liner, you can use .split() and .join():
>>> '.'.join('20.99.0'.split('.')[:2])
'20.99'
>>> '.'.join('20.99.1231.23'.split('.')[:2])
'20.99'
>>> '.'.join('20.99'.split('.')[:2])
'20.99'
>>> '.'.join('20'.split('.')[:2])
'20'
You could do something like this
>>> s = '20.99.0, 20.0.0'
>>> s.split(',')
['20.99.0', ' 20.0.0']
>>> map(lambda x: x[:x.find('.',x.find('.')+1)], s.split(','))
['20.99', ' 20.0']
Look at the inner expression of find. I am finding the first '.' and incrementing by 1 and then find the next '.' and leaving everything from that in the string slice operation.
Edit: Note that this solution will not discard everything from the second decimal point, but discard only the second point and keep additional digits. If you want to discard all digits, you could use e.g. #Blender's solution
It only qualifies as a one-liner if two instructions per line with a ; count, but here's what I came up with:
>>> x = "20.99.1234"
>>> s = x.split("."); x = s[0] + "." + "".join(s[1:])
>>> x
20.991234
It should be a little faster than scanning through the string multiple times, though. For a performance cost, you can do this:
>>> x = x.split(".")[0] + "." + "".join(x.split(".")[1:])
For a whole list:
>>> def numify(x):
>>> s = x.split(".")
>>> return float( s[0] + "." + "".join(s[1:]))
>>> x = ["123.4.56", "12.34", "12345.6.7.8.9"]
>>> [ numify(f) for f in x ]
[123.456, 12.34, 12345.6789]
>>> s = '20.99, 20, 20.99.23'
>>> ','.join(x if x.count('.') in [1,0] else x[:x.rfind('.')] for x in s.split(','))
'20.99, 20, 20.99'
If you are looking for a regex based solution and your intended behaviour is to discard everthing after the second .(decimal) than
>>> st = "20.99.123"
>>> string_decimal = re.findall(r'\d+\.\d+',st)
>>> float(''.join(string_decimal))
20.99