假设这样的数据:
>>> data
x
0 [wdq, sda, q]
1 [q, d, qasd]
2 [d, b, sdaaaa]
我想知道每个列表中有多少字符串包含a
,这意味着我需要这样的答案:
>>> data
x count_a
0 [wdq, sda, q] 1
1 [q, d, qasd] 1
2 [d, b, sdaaaa] 1
我怎么能在python中做到这一点?
答案 0 :(得分:2)
假设这是pandas.DataFrame
且x
是list
对象:
df['count_a'] = df['x'].apply(lambda x: sum('a' in e for e in x))
答案 1 :(得分:1)
你可以试试这个;
for i in my_series:
print (i.count('a'))
这给你的每个系列字母
答案 2 :(得分:0)
要了解每个String
中包含字母List
的{{1}}个数量,您可以使用以下内容:
a
这将打印以下内容:
l = ['wdq', 'sda', 'qaaa']
print(sum([1 for x in l if 'a' in x]))
答案 3 :(得分:0)
a = ['a', '12asf3', 'sdf']
b = ['gfdg5', ' ', 'vag gfd4']
c = [' fd4 ', 'sfsa fa', 'df4 a']
abc = [a, b, c]
for mem in abc:
counter = 0
for str in mem:
if 'a' in str:
counter += 1
print abc.index(mem), mem, counter
输出:
0 ['a', '12asf3', 'sdf'] 2
1 ['gfdg5', ' ', 'vag gfd4'] 1
2 [' fd4 ', 'sfsa fa', 'df4 a'] 2