I am using Python 3.6 and I have a list of dictionaries like this:
list = [{'name': 'A', 'number':'1'}, {'name': 'B', 'number':'2'}, {'name': 'C', 'number':'3'}, {'name': 'D', 'number':'4'}]
I found out how to print the list in the desired format with:
for s in list:
name = s['name']
number = s['number']
print(name + " = "+ number)
Which gives:
A = 1
B = 2
C = 3
D = 4
I would like to be able to multiply the items 'number' by 2 for example and display:
A = 2
B = 4
C = 6
D = 8
Thank you!
Are you trying to temporarily multiply the values and print them out? Which in this case, you would change your last line to
print(name + " = "+ int(number) * 2)
However, if you want to multiply the values in your dictionary directly, you would go about it as so:
for s in list:
name = s['name']
s['number'] = str(int(s['number']) * 2) # multiply value by 2
number = s['number']
print(name + " = "+ number)
Note that your problem may arise from the fact that your dictionary values are stored as strings instead of integers, which means that to perform any kind of mathematical operation on them, you must convert them to an integer and back to a string.
You're able to multiply a number by using the * symbol 2 * 2 will output 4.
Because your values are stored as Strings you'll need to convert them to Integers first. int('2') * 2 == 4.
Then to print an Integer with a string you need to convert it back to a string.
for the last line change it to
print(name + " = "+ str(int(number)*2))
You can always iterate over your list to modify the value of its nested parts, i.e.:
your_list = [{'name': 'A', 'number': '1'},
{'name': 'B', 'number': '2'},
{'name': 'C', 'number': '3'},
{'name': 'D', 'number': '4'}]
for item in your_list: # iterate over each dictionary in your_list
# since your dict contains strings we have to convert the value into a number/integer
# before multiplying it by 2
item["number"] = int(item["number"]) * 2 # turn it back to a string with str() if needed
# now, let's demonstrate that the data changed in `your_list`:
for item in your_list: # iterate over each dictionary in your_list
print("{} = {}".format(item["name"], item["number"])) # print in your desired format
# A = 2
# B = 4
# C = 6
# D = 8
Related
There is a function in the python2 code that I am re-writing into python3
def abc(self, id):
if not isinstance(id, int):
id = int(id)
mask = self.programs['ID'] == id
assert sum(mask) > 0
name = self.programs[mask]['name'].values[0]
"id" here is a panda series where the index is strings and the column is int like the following
data = np.array(['1', '2', '3', '4', '5'])
# providing an index
ser = pd.Series(data, index =['a', 'b', 'c'])
print(ser)
self.programs['ID'] is a dataframe column where there is one row with integer data like '1'
import pandas as pd
# initialize list of lists
data = [[1, 'abc']]
# Create the pandas DataFrame
df = pd.DataFrame(data, columns = ['ID', 'name'])
I am really confused with the line "mask = self.programs['ID'] == id \ assert sum(mask) > 0". Could someone enlighten?
Basically, mask = self.programs['ID'] == id would return a series of boolean values, whether thoses 'ID' values are equal to id or not.
Then assert sum(mask) > 0 sums up the boolean series. Note that, bool True can be treated as 1 in python and 0 for False. So this asserts that, there is at least one case where programs['ID'] column has a value equal to id.
A string = 1 2 3 4
Program should return = [[1,2],[3,4]]
in python
I want the string to be converted into a list of every two element from string
You could go for something very simple such as:
s = "10 2 3 4 5 6 7 8"
l = []
i = 0
list_split_str = s.split() # splitting the string according to spaces
while i < len(s) - 1:
l.append([s[i], s[i + 1]])
i += 2
This should output:
[['10', '2'], ['3', '4'], ['5', '6'], ['7', '8']]
You could also do something a little more complex like this in a two-liner:
list_split = s.split() # stripping spaces from the string
l = [[a, b] for a, b in zip(list_split[0::2], list_split[1::2])]
The slice here means that the first list starts at index zero and has a step of two and so is equal to [10, 3, 5, ...]. The second means it starts at index 1 and has a step of two and so is equal to [2, 4, 6, ...]. So we iterate over the first list for the values of a and the second for those of b.
zip returns a list of tuples of the elements of each list. In this case, [('10', '2'), ('3', '4'), ('5', '6'), ...]. It allows us to group the elements of the lists two by two and iterate over them as such.
This also works on lists with odd lengths.
For example, with s = "10 2 3 4 5 6 7 ", the above code would output:
[['10', '2'], ['3', '4'], ['5', '6']]
disregarding the 7 since it doesn't have a buddy.
here is the solution if the numbers exact length is divisible by 2
def every_two_number(number_string):
num = number_string.split(' ')
templist = []
if len(num) % 2 == 0:
for i in range(0,len(num),2):
templist.append([int(num[i]),int(num[i+1])])
return templist
print(every_two_number('1 2 3 4'))
you can remove the if condition and enclosed the code in try and except if you want your string to still be convert even if the number of your list is not divisible by 2
def every_two_number(number_string):
num = number_string.split(' ')
templist = []
try:
for i in range(0,len(num),2):
templist.append([int(num[i]),int(num[i+1])])
except:
pass
return templist
print(every_two_number('1 2 3 4 5'))
I have a list,
A = ['A','B','C','D','E','F','G','H']
if user input x = 4, then I need an output that shows every value that is 4 distance away from each other.
If starting from 'A' after printing values that are 4 distance away from each other ie: {'A', 'E'}, the code should iterate back and start from 'B' to print all values from there ie: {'B', 'F'}
No number can be in more than one group
Any help is going to be appreciated since I am very new to python.
this is what I have done
x = input("enter the number to divide with: ")
A = ['A','B','C','D','E','F','G','H']
print("Team A is divided by " +x+ " groups")
print("---------------------")
out = [A[i] for i in range(0, len(A), int(x))]
print(out)
My code is printing only the following when user input x =4
{'A', 'E'}
But I need it to look like the following
{'A', 'E'}
{'B', 'F'}
{'C', 'G'}
{'D', 'H'}
what am I doing wrong?
Use zip:
out = list(zip(A, A[x:]))
For example:
x = 4 # int(input("enter the number to divide with: "))
A = ['A','B','C','D','E','F','G','H']
print(f"Team A is divided by {x} groups")
print("---------------------")
out = list(zip(A, A[x:]))
print(out)
Outputs:
[('A', 'E'), ('B', 'F'), ('C', 'G'), ('D', 'H')]
Here you have the live example
If you want to keep the comprehension:
out = [(A[i], A[i+x]) for i in range(0, len(A)-x)]
**You can find my answer below.
def goutham(alist):
for passchar in range(0,len(alist)-4):
i = alist[passchar]
j = alist[passchar+4]
print("{"+i+","+j+"}")
j = 0
alist = ['a','b','c','d','e','f','g','h']
goutham(alist)
Im trying to read input spread across multiline in form of dictionary and apply simple math operations on the value of dictionary . My code reads
d ={}
bal=0
text = input().split(",") #split the input text based on line'text'
print(text)
for i in range(5):
text1 = text[i].split(" ") #split the input text based on space & store in the list 'text1'
d[text1[0]] = int(text1[1]) #assign the 1st item to key and 2nd item to value of the dictionary
print(d)
for key in d:
if key=='D':
bal=bal+int(d[key])
#print(d[key])
elif key=='W':
bal=bal-int(d[key])
print(bal)
Input : W 300,W 200,D 100,D 400,D 600
output :{'D': 600, 'W': 200}
400
Expected Output: {'W':300,'W':200,'D':100,'D':400,'D':600}
600
ISSUE: The issue here is the code always reads 2 and last values only . For example in the above case output is
{'D': 600, 'W': 200}
400
Can someone let me know the issue with for loop .
Thanks in advance
You can try like this in a simpler way using your own approach. #Rakesh and #Sabesh suggested good. Dictionary is an unordered collection with unique and immutable keys. You can easily check this on your Python interactive console by executing help(dict).
You can check https://docs.python.org/2/library/collections.html#collections.defaultdict . Here you'll find number of examples on how to efficiently using dictionary.
>>> d = {}
>>> text = 'W 300,W 200,D 100,D 400,D 600'
>>>
>>> for item in text.split(","):
... arr = item.split()
... d.setdefault(arr[0], []).append(arr[1])
...
>>> d
{'W': ['300', '200'], 'D': ['100', '400', '600']}
>>>
>>> w = [int(n) for n in d['W']]
>>> d = [int(n) for n in d['D']]
>>>
>>> bal = sum(d) - sum(w)
>>> bal
600
>>>
Based on my little knowledge on pandas，pandas.Series.str.contains can search a specific str in pd.Series. But what if the dataframe is large and I just want to glance all kinds of str element in it before I do anything?
Example like this:
pd.DataFrame({'x1':[1,2,3,'+'],'x2':[2,'a','c','this is']})
x1 x2
0 1 2
1 2 a
2 3 c
3 + this is
I need a function to return ['+','a','c','this is']
If you are looking strictly at what are string values and performance is not a concern, then this is a very simple answer.
df.where(df.applymap(type).eq(str)).stack().tolist()
['a', 'c', '+', 'this is']
There are 2 possible ways - check numeric values saved as strings or not.
Check difference:
df = pd.DataFrame({'x1':[1,'2.78','3','+'],'x2':[2.8,'a','c','this is'], 'x3':[1,4,5,4]})
print (df)
x1 x2 x3
0 1 2.8 1
1 2.78 a 4 <-2.78 is float saved as string
2 3 c 5 <-3 is int saved as string
3 + this is 4
#flatten all values
ar = df.values.ravel()
#errors='coerce' parameter in pd.to_numeric return NaNs for non numeric
L = np.unique(ar[np.isnan(pd.to_numeric(ar, errors='coerce'))]).tolist()
print (L)
['+', 'a', 'c', 'this is']
Another solution is use custom function for check if possible convert to floats:
def is_not_float_try(str):
try:
float(str)
return False
except ValueError:
return True
s = df.stack()
L = s[s.apply(is_not_float_try)].unique().tolist()
print (L)
['a', 'c', '+', 'this is']
If need all values saved as strings use isinstance:
s = df.stack()
L = s[s.apply(lambda x: isinstance(x, str))].unique().tolist()
print (L)
['2.78', 'a', '3', 'c', '+', 'this is']
You can using str.isdigit with unstack
df[df.apply(lambda x : x.str.isdigit()).eq(0)].unstack().dropna().tolist()
Out[242]: ['+', 'a', 'c', 'this is']
Using regular expressions and set union, could try something like
>>> set.union(*[set(df[c][~df[c].str.findall('[^\d]+').isnull()].unique()) for c in df.columns])
{'+', 'a', 'c', 'this is'}
If you use a regular expression for a number in general, you could omit floating point numbers as well.