使用 Pandas
替换字符串中缺少的最不频繁的空格
原文:https://www.geesforgeks.org/replace-missing-white-spaces-in-a-string-with-frequency-use-pandas/
让我们用 python 创建一个程序，使用 Pandas 库将字符串中的空格 替换为字符串中最少出现的字符 。
例 1:
String S = "akash loves gfg"
here:
'g' comes: 2 times
's' comes: 2 times
'a' comes: 2 times
'h' comes: 1 time
'o' comes: 1 time
'k' comes: 1 time
'v' comes: 1 time
'e' comes: 1 time
'f' comes: 1 time
'l' comes: 1 time
In this example, there are 7 characters with least frequency 1 so, there can be
7 valid outputs One of the possible output is given below:
So, the Output String will be: "akashlloveslgfg".
例 2:
string ="goodd noon"
here:
g comes: 1 time
o comes: 4 times
d comes: 2 times
n comes: 2 times
So the character with the least frequency 1 is g So here white spaces will be
replaced by the character g and the output will be:
"gooddgnoon"
现在，让我们看看实现:
蟒蛇 3
# importing pandas library
import pandas as pd
# taking string with white spaces
newstr1 = 'akash loves gfg'
# printing the original string
print("Original String given by user:",
newstr1)
# converting string into
# list of characters
ser = pd.Series(list(newstr1))
# counting the frequency
# of characters
element_freq = ser.value_counts()
# printing character and their
# respective frequency
print(element_freq)
current_freq = element_freq.dropna().index[-1]
# function element_freq.dropna()
# will Return a new Series with
# missing values removed
result = "".join(ser.replace(' ',
current_freq))
print(result)
输出:
