如何使用python在列表/数组中搜索字母？

Question

我有一个这样的列表/数组。

ch = [ eretail, retail, fax, xerox ]

我尝试使用单个/多个字母获取字符串。

例如，我发送了“e”字母作为输入。它将搜索整个字符串，如果它与字符串匹配，则字符串将显示为输出

例子：

1)Input:   e
  output:  eretail
           retail
           xerox
2)Input: x
  output:  fax
           xerox
3)Input: o
  output: xerox

4)Input: retail
  output: eretail
          retail

我需要上面的输出。

Answer 1

您可能需要这样做：

import re

ch = ['eretail', 'retail', 'fax', 'xerox']

input = ['e', 'x', 'o', 'retail']

for each in input:
    search = each + '+'
    output = []
    for i in ch:
        if(re.findall(each, i)):
            output.append(i)
    print('Input', each, '\n', 'output', output)

结果：

Input e 
 output ['eretail', 'retail', 'xerox']
Input x 
 output ['fax', 'xerox']
Input o 
 output ['xerox']
Input retail 
 output ['eretail', 'retail']

或者，以防万一输入不是数组：

import re

ch = ['eretail', 'retail', 'fax', 'xerox']

input = 'e'
search = input + '+'
output = []

for i in ch:
    if(re.findall(search, i)):
        output.append(i)

print('Input', input, '\n', 'output', output)

结果：

Input e 
 output ['eretail', 'retail', 'xerox']

请尝试上面的代码，我是 python 的初学者 - 如果需要增强，请编辑此答案。 re 也是 Python 的内置 Regex 模块，在这个例子中使用。

Answer 2

我不知道这是不是你想要的。

target = input()
for word in ch:
    if target in word:
        print(word)

如果没有，请为您的问题提供更多信息。 希望对您有所帮助。

Answer 3

您没有指定是否要在 python2 或 python3 上的代码 我是用python3写的，但这里没有太大的不同。

ch = ["eretail", "retail", "fax", "xerox"]

myLetter = str(input("Enter your letter: "))

for word in ch:
    if myLetter in word:
        print(word)

Answer 4

试试这个：

ch = ['eretail', 'retail', 'fax', 'xerox' ]
input_letter = "x"

new_list = []
for item in ch:
    if input_letter in item:
        new_list.append(item)
print(new_list)

Answer 5

def find(needle, haystack):
    """
    Descr:
    Checks for the presence of a string/character(needle) in the 
    items of an iterable object
    """
    if isinstance(haystack, list):
        for h in haystack:
            find(needle=needle, haystack=h)
    elif isinstance(haystack, dict):
        for k, v in haystack.items():
            find(needle=needle, haystack=v)
    else:
        if str(needle) in haystack:
            print(haystack)

# We can then call the function above as follows:
print("1. Search Results for e in '[eretail, retail, fax xerox]")
find(needle="e", haystack=[ "eretail", "retail", "fax" "xerox" ])

print("\n2. Search Results for x in '[eretail, retail, fax xerox]'")
find(needle="x", haystack=[ "eretail", "retail", "fax" "xerox" ])

print("\n3. Search Results for o in '[eretail, retail, fax xerox]'")
find(needle="o", haystack=[ "eretail", "retail", "fax" "xerox" ])

# Mixed Haystack illustration...
print("\n4. Search Results for cat in '[bobcat, [concatenate,boy, girl], dict(a=vacate, b=boy)]'")
find(needle="cat", haystack=["bobcat", ["concatenate", "boy", "girl"], dict(a="vacate", b="boy")])

# output:

1. Search Results for e in '[eretail, retail, fax xerox]
eretail
retail
faxxerox

2. Search Results for x in '[eretail, retail, fax xerox]'
faxxerox

3. Search Results for o in '[eretail, retail, fax xerox]'
faxxerox

4. Search Results for cat in '[bobcat, [concatenate,boy, girl], dict(a=vacate, b=boy)]'
bobcat
concatenate
vacate